This is very much overkill but I love it! A simple way to see why it is 11 is as follows: b has to be even, since 2a and 100 are. b has to be less than/equal to 20. There are 11 even numbers between 0 and 20. a is then uniquely determined by (100-5b)/2 = 5(20-b)/2 which is a positive number.
That's exactly how i did it! Because if b was odd 5b would end with 5 and then 2a would have to end with 5 which is impossible for an integer value of a. I was so confused when i skipped to the end of the video to see the solution😅
This is what I love about this channel. The problem can be solved in your head, but when you go through the video, you learn so much about different ways to apply to more general cases.
Alternate formulation: > From 2a + 5b = 100, we have 2a = 100 - 5b = 5(20 - b), and we must conclude that a has to be a multiple of 5; therefore, a = 5c > Plugging that back in our equation, we have: 10c + 5b = 100. Dividing both sides by 5 yields 2c + b = 20; now, this gives us b = 20 - 2c = 2(10 - c), and therefore b is an even number, so we can write b = 2d > From 2c + b = 20, we get 2c + 2d = 20, or c + d = 10. We can show that this equation has 11 solutions for non-negative integer values of c and d, and since this is a transformed version of the original equation, we can conclude that the original equation must have 11 solutions as well.
I didn't even have to plot down anything on paper to solve this. I just say that the solutions are enumerated by a = 50 down to 0, with step size 5, i.e. a=50, 45, 40, ..., 0, or a/5 = 10, ..., 0, which are 11 values.
Easier answer: reframe this as a discrete linear equation with a being y and b being x. So this way, it would be y = -5/2x + 50. The slope is negative and the y-intercept is positive, so in the first quadrant of the plane (positive x and y) you have a finite number of integer values for x and y. Start at the y intercept (0,50). The slope is -5/2, so every 5 units vertically there will be an integer solution. Divide 50/5=10 for the rest of the solutions, and add the extra point where you started, the y-intercept. 10+1=11. Done. Here are all the valid coordinates for the solution: (0,50) (2,45) (4,40) (6,35) (8,30) (10,25) (12,20) (14,15) (16,10) (18,5) (20,0)
This works really well for this simple example, but the nice thing about the method that Michael shows is that his method works for any number of coefficients added together in linear combination to reach a final sum. For example, this method would work for 2a+5b+7c+11d+13e+... = 10000 or something like that.
I'm sure we all got the answer by just doing it in our heads. The goal here was to use generating functions because the general question will not be as simple as 2a+5b=100.
Q. 2a+5b=100 a,bE non- negative integers Ans. as 5b will be multiple of 5, and 2a will always be even. for sum to be even (100), 5b must be even for sum to be multiple of 10 (100), 2a must be divisible by 10. Hence, a=5m. and b=2n. m,n E non-negative integers. Now, 10{ n+m}=100 n+m=10 So, starting from 0 till 10 for n, and 10 till 0 for m. So, there will be 11 ordered pairs for (m,n) or (a,b) This is how a kid would do it. gratitude for watching pro's do it.
Yes it is.but i think that the idea is to learn this genarating function tool.this question was easy but he has one video when the question is hard and this tool helps
you prolly dont care but if you're bored like me during the covid times you can watch all the latest movies and series on instaflixxer. Been binge watching with my gf recently =)
Well, a simpler solution would be as follows: note that if (a,b) is a solution, then (a+5t, b-2t) is also a solution, and all the solutions are of this form. Take the maximal solution (0,20), then we have 11 solutions coresponding to (0,20), (5,18),...,(50,0).
dude i'm confused, the video was uploaded 8 min ago, the video is 18 min long, you posted the comment 45 sec ago, my question is : did you watch all the video? Or you just come to post the comment and go?
@@elie.makdissi actually he uploads his videos and keeps them unlisted , so if you check the playlists regularly then you might find the unlisted videos , I think that this dude might have watched the video and then commented or as you said, he/she might have just watched the beginning of the video and realised that he used a different method
@@elie.makdissi I saw the problem and solved it, then i speed run the video and saw his solution and checked that his was different from mine and that mine was simpler.
Watching this series has made me appreciate: 1. How limited my mathematics exposure was in high school and college 2. How many different techniques professional mathematicians can be proficient in.
when you add text annotations to the video, i recommend adding a black background to the text with some transparency, or outlining the text in black so it's easier to read. it doesnt contrast with the stuff on the board very well. dunno what editing software you use but I have to imagine theres an easy way to do it.
Taking (mod 5) we have that a is multiple of 5 so a=5k for a non-negative integer k and then taking (mod 2) we have that b is a multiple of 2 so b=5m for non-negative integer m. So the equation becomes 10k+10m=100 or k+m=10 which gives us 11 pairs for k, m and because k and m are integers it follows that a, b are also gonna be integers for each pair as they are multiples of them so we get 11 pairs for a, b.
Note that (a, b) = (0, 20) is a solution to 2a + 5b = 100. Using a well-known result, solutions can be written parametrically as a = 5t and b = -2t + 20; because a and b are required to be non-negative integers, the least (integer) value for t is 0 and the greatest t-value is 10. Thus, there are 11 solutions.
There can be no doubt that the material presented here is an example of powerful methods being applied to a specific problem. My admiration for all things intellectual admires that unequivocally. The practical side of my nature which also admires people who work with their hands, is a background noise which questions what possible use the high power math techniques can claim in the real world, It sees the power of intellect being applied to trivial ends.
That's what I was thinking, but he says out loud the thing he's writing, and then there's an edit before "+ g", and he keeps it. So he must have meant it!
A great little problem. I solved it on my way to work today. Here's how I did it: First, I noticed that 5b can end only with the digits 0 or 5, depending on whether b is even or odd. That means that 100-5b will also end in either 0 or 5. However, this number is equal to 2a, and for a to be an integer it must be divisible by 2. So any b values that lead to 100-5b ending with 5 can be discarded. In other words, b can only be an even number. But which even numbers? The condition that a and b must be positive integers means that b can at most be 20, because then 2a=0 -> a=0, which meets the condition. So b must be any even integer greater than or equal to 0 and less than or equal to 10. There are 11 such integers, and each one gives only one possible value for a. Therefore, the size of the solution set is 9. The solution set itself is {(0,20), (5,18), (10,16), (15,14), (20,12), (25,10), (30,8), (35,6), (40,4), (45,2), (50,0)}
I read the thumbnail problem wrong as Z_>0 not Z_>=0 I like that I ended up doing a combinatoric solution, with counting partitions. Initial Solution: 2a + 5b = 100 LHS is divisible by 2 and 5 so a must be divisible by 5 and b must be divisible by 2 consider new equation with 5n=a and 2m=b then 10n+10m = 100 then n+m =10 there are nine partitions of ten into two positive integers seen from there being nine spots to choose the split point n=1, m=9 n=2, m=8 n=3, m=7 n=4, m=6 n=5, m=5 and the same with n and m flipped generating the solutions to the initial problem a=5, b= 18 a=10, b= 16 a=15, b=14 etc... Mistakes corner: noticed while watching was problem was for non negative integers for nonnegative solutions you have to add a partition at 0 and 10, getting 11 solutions instead. ie these two solutions where missing from my original soultion n=0, m =10 n=10, m= 0 a= 0, b=20 a=50, b=0
There is also 0+10 and 10+0, since it says "Non-negative" so 0 included. So in the end we have a=0/b=20 a=5/b=18 a=10/b=16 a=15/b=14 a=20/b=12 . . . a=45/b=2 a=50/b=0
Everybody else has commented how much more simply this problem could be solved to show there are 11 solutions. In fact using those more elementary methods you can easily generalize this result: the number of non-negative solutions to 2a+5b=10m is m+1, or more generally the number of non-negative solutions to 2a+5b=n is floor(3n/5) - ceiling(n/2) + 1. I wouldn't want to try getting that using Michael's method!
That does seem a bit overkill - I just said 2a+5b=100 implies a = 50 - 5/2 b, so we have b = 0 (mod 2). But we also know a >= 0 and solving that inequality gives us b = 0, and for each b there is a distinct a (b injective in a), we can rephrase our problem from "how many (a,b) pairs s.t. 2a+..." to "how many even b in the interval [0,20]", which gives us eleven, that's 0, 2, 4... 19, 20.
I knew something was up when Prof Penn posted a posted a problem that can be solved in your head, but took 18+ minutes to solve in the video! That is some serious overkill! (as Samuel Marger pointed out already)
Write 5b=100-2a. As a runs from 0 to 50, the right hand side is an arbitrary even number between 0 and 100. Thus, the right hand side is an even number which is a multiple of 5, i.e., 5b must be a multiple of 10. Thus, a can be only one of the 11 values: a=0,5,10,15,20,25,30,35,40,45,50. For each one of these values of a, there is exactly one value of b. In total, 11 solutions (a,b).
That is an interesting method, albeit a complex one. I saw it shown to us once somewhere (I still remember these 13/20ths) and was very impressed. Generating functions kind of make combinatorics into bearable nearly smooth-function math. :)
interesting, i solved this as a diophantine equation ( solved using linear congruence method) and hence used an inequality to narrow down the results to non negative integers. i got a = 50-5k and b=2k . Although i got that there is only 10 solutions and so im not sure why im missing an extra solution. Edit: stupid mistake, k goes up to 10 but when we include k =0 solution there are 11 solutions
probably k = 0 or k = 10 check your inequality allows 50 - 5k and k be zero your inequality probably is too tight as every k = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] works and giving exactly 11 soultions
Simplifying 2a + 5b = 100 Solving 2a + 5b = 100 Solving for variable 'a'. Move all terms containing a to the left, all other terms to the right. Add '-5b' to each side of the equation. 2a + 5b + -5b = 100 + -5b Combine like terms: 5b + -5b = 0 2a + 0 = 100 + -5b 2a = 100 + -5b Divide each side by '2'. a = 50 + -2.5b Simplifying a = 50 + -2.5b
If 2a + 5b = 100 then a = 50 - (5b/2) which is a non-negative integer. For a to satisfy this condition 1) b must less than or equal to 20 and 2) 2 must divide b. Therefore, the total number of solutions is the number of multiples of 2 within 0 and 20 (including both). By counting by hand, you get 11 different solutions.
You can also solve using Diophantine equations: first you get one (arbitrary) solution to the equation: a0=0 and b0=20. Then you can parametrize all solutions by letting a= a0+5*k and b = b0 - 2*k. Since a and b must be positive, k can be only the numbers from 0 to 10. This gives a total of 11 solutions.
It has a remarkable solution the following way.Observation of the problem reveals that "a" has a maximum value of "50" when "b" is zero.And "b" has a maximum value of "20" when "a" is zero.Let us add "5t" to "a" and subtract "2t" from "b" without altering the given problem. We will get 2(a+5t) + 5(b-2t)=100 Now put a=0 & corresponding b=20, getting 2(5t)+5(20-2t)=100 Without loss of generality we may treat "5t" as "a" & "(20-2t) " as "b". 5t can have only 11 values when t = 0, 1,2,3,4,5,6,7,8,9,10. Hence, 2a+5b =100 has only 11 solutions.
After noticing that 2 and 5 are both factors of 100, while 2 and 5 are co-primes and cannot divide each other in N. Thus 5 divides a, while 2 divides b. When simplifying 2a + 5b = 100, w/ a = 5x and b = 2y, w/ (x,y) belong to N, 2*5x + 5*2y = 100 x + y = 10. So, x = 0, 1, 2, ..., 10; hence a total of 11 solutions.
I would rather use the method he mentioned in the beginning of the video. The general solution of a and b are as follows: a=5k-200 b=-2k+100 Substitute different values of k and we can get different values of a and b. Obviously, k at least equals 40 to make a positive. The eleven solutions are as follows: (0,20), (5,18), (10,16), (15,14), (20,12), (25,10), (30,8), (35,6), (40,4), (45,2), (50,0).
Great job, your videos are really great. But I must say that this is a lot of work if you ask me. You could solve this with very little work, here is the solution that I am talking about: 2a+5b=100,for some (a,b)ϵN_0×N_0 (where N_0={0,1,2,3,…} well at least in Croatia where I'm from) 2a=100-5b=5(20-b) Now since the righthand side is divisible by 5 and two is obviously not a multiple of 5 we conclude: a=5x,for some xϵN_0 so: 10x=5(20-b) by cancelling and rearranging: b=2(10-x) since the righthand side is a multiple of 2, b must also be, so: b=2y,for some yϵN_0 Now we have: x+y=10,for some (x,y)ϵN_0×N_0 We know that there are 11 solutions to that equation in that collection of numbers and they are: x 0 1 2 3 4 5 6 7 8 9 10 y 10 9 8 7 6 5 4 3 2 1 0 We can trivially show that there is a one to one map from (x,y) to (a,b) (Pf: M(x,y)=(a,b)=(5x,2y) which is an injective multivariable function). From that we can conclude that the number of valid (x,y) is equal to the number of valid (a,b) and therefor there are 11 solutions. See easy. Better than doing partial fractions decomposition (I would literally do anything to avoid doing that). Anyway thanks for your effort.
@@Last_Resort991 Look I could write this solution, and so could you, in about 4 to 5 steps, the thing is to make it rigorous you need to do more, the solution represented in this video uses partial fractions decomposition of very complicated fractions, and most of the work isn't shown (because then the video would be 20 minutes longer). Anyway, I like to be rigorous when I solve problems, that is why it looks longer.
Why going so complicated I just did it as follows 2a+5b=100 2a=100-5b a=(100-5b)÷2 since a,b>=0 We have a>=0 when b is 0,2,4,6,8,10,12,14,16,18and 20 because 100-5b must be even therefore b must be even and b=
I must admit I got lost with this solution. I looked at it and bounded a at 50 and b at 20. Then noted a=50-5/2*b, so a is an integer iff b is even. Then, trivially the (a,b) pairs are (50,0), (45,5), (40,10), ... , (0,20). So 11 in total. Appreciate Michael was showing us a different technique but this one seemed like early high school maths to solve!
Since 2a is even and 100 is even, 5b must also be even then b is even... the lower bound for b is when b = 0, and the upper bound for b is when a = 0 which is b = 20, we get b = { x : x is even and 0
This is a great way to prove that 1/4 + 101/10 + 13/20 = 11 without doing the calculation (11 being obtained with the usual way to solve this problem). :-)
mdc(2,5)=1 então 2.(-1)+5.1=1 ... 2(-100)+5.100=100. Logo, (a,b)=(-100,100) é uma solução particular da equação diofantina 2a+5b=100, cuja solução geral é : a=-100+5t e b=100-2t. Por fim, impondo que a>0 e b>0, obtemos t>20 e t
This can be solved in one’s head in a few seconds. The lcm of 2 and 5 is 10. There are only two solutions to add to 10 using 2 and 5: 22222 or 55. There are 10 such chunks in 100. There are only 11 unique solutions (order doesn’t matter) to fill 10 slots with 2 choices each. (one solution is zero of the 2nd choice, one solution is one of the 2nd choice, up to one solution is ten of the 2nd choice). So the answer is 11.
In summery: there are (K+ij)/ij ways to solve the equation ia+jb=K when a and b are restricted to be like a, b ≥ 0. And K is a multiple of ij. Let's solve 2a+5b = 100 with it. (100+2×5)/2×5 (100+10)/10 110/10 11 Let's try 4a+5b = 100 (100+4×5)/4×5 (100+20)/20 120/20 6 Solution 1: (0, 20) 100 = 2×5((0+1)+(0+1)...+(0+1)) Solution 2: (1, 19) 100 = 2×5((0+1)+(0+1)...+(1+0)) (1-x_n = y_n) 100 = 2×5((x_1+y_1)+(x_2+y_2)...+(x_10+y_10)) The first way is to set all x_n = 0. The second is to set x_1 to x_9 = 0 and x_10 = 1 The nth way is to set x_1 to x_11-n = 0 and x_12-n to x_10 = 1 There are 11 ways because the x_n after the last x_n is x_11 Now let's try 4a+5b = 100 Solution 1: (0, 25) 100 = 4×5((0+1)+(0+1)...+(0+1)) Solution 2: (4, 20) 100 = 4×5((0+1)+(0+1)...+(1+0)) (1-x_n = y_n) 100 = 4×5((x_1+y_1)+(x_2+y_2)...+(x_5+y_5)) The first way is to set all x_n = 0. The second is to set x_1 to x_4 = 0 and x_5 = 1 The nth way is to set x_1 to x_6-n = 0 and x_7-n to x_5 = 1 There are 6 ways because the x_n after the last x_n is x_6 Now let's try ia+jb = 100 100 = ij((0+1)+(0+1)...+(0+1)) 100 = ij((0+1)+(0+1)...+(1+0)) (1-x_n = y_n) 100 = ij((x_1+y_1)+(x_2+y_2)...+(x_(100/ij)+y_(100/ij))) The first way is to set all x_n = 0. The second is to set x_1 to x_((100-ij)/ij) = 0 and x_(100/ij) = 1 The nth way is to set x_1 to x_((100+ij)/ij)-n = 0 and x_((100+2ij)/ij)-n to x_(100/ij) = 1 There are (100+ij)/ij ways because the x_n after the last x_n is x_((100+ij)/ij) Now let's try ia+jb=K K = ij((0+1)+(0+1)...+(0+1)) K = ij((0+1)+(0+1)...+(1+0)) (1-x_n = y_n) K = ij((x_1+y_1)+(x_2+y_2)...+(x_(K/ij)+y_(K/ij))) The first way is to set all x_n = 0. The second is to set x_1 to x_((K-ij)/ij) = 0 and x_(K/ij) = 1 The nth way is to set x_1 to x_((K+ij)/ij)-n = 0 and x_((K+2ij)/ij)-n to x_(K/ij) = 1 There are (K+ij)/ij ways because the x_n after the last x_n is x_((K+ij)/ij)
2a + 5b = 103 Hence, a = (103 - 5b)/2 Hence, (103 - 5b)/2 > b Hence, 7b < 103 Hence, b < 103/7 = 14.71…. For a to be a positive integer, b is also an odd, positive integer So, we could have b = 1→a = 49, b = 3→a = 44, b = 5→a = 39,…..,b= 13→a= 19 Hence, 7 pairs, which are (49,1), (44,3),(39,5),……(19,13)
I don't understand these generating functions and the complexity of the solution made my jaw drop. I approached this very differently: Since 2a is always even, 5b must also be even so that the sum 2a+5b = 100 (even). 5b is only even, if b is even. The smallest value for b = 0 making a = 50. The largest value for b = 20 making a = 0. So, b = 0, 2, 4,... ...,20 and the corresponding values of a are 50, 45, ... ...,0. That's 11 pairs of (a, b).
im not a math expert but..... 2a+5b=100 2a=100-5b a=50-5b/2 note that b=even 0≤50-5b/2 5b/2≤50 5b≤100 0≤b≤20 counting all even integers between 20 and zero would give us 20-0/2 +1=11 #### did he want it to do it the hard why or maybe my method is a bit risky
Every second comment is stating it’s overkill. You’re not wrong but you must understand two things. Firstly, the method is refined for demonstration purposes, and secondly this allows solutions to arbitrary n values which is very powerful
@@magnusPurblind Explanation: Dividing n by 5 shows the range of possible b (i.e. with n = 99, there are 20 possibilities for b: 0, 1, ..., 19). Because 2a is always even, we have to further divide n/5 by 2. Therefore you get n/(5*2). This is not always integer, so we take the floor function. If n is divisible by 5, there is always the additional solution a = 0, b = n/5 which we have to add, therefore we take the ceiling function in that case.
I think it is even possible to generalize the number of solutions for arbitrary coefficients before a and b: v*a + w*b = n => number of solutions is floor(n/(w*v)) (+1 if n is divisible by w).
i am no match for the incredible skill at play with this, but i'll post my simplistic method regardless: first we ask: what are the factors of b? if b is odd, we get odd=even, therefore b must be even. The same divisibility rule applies for a with 5, so we know we can express: a = 5m, b = 2n. therefore we get the equation m+n = 10, for m, n nonnegative integers. for this, there are exactly 11 solutions.
An easier solution would be that since a and b are integers we get that 2 divides b and 5 divides a (take the equation 2a+5b=100 modulo 2 and modulo 5) so there are n,m with a=5n and b=2m. The number of pairs (a,b) is the same as the number of pairs (n,m) and we have 10n+10m=100 or n+m=10. Now it is obvious that there are 11 pairs (n,m) with n+m=10
Overkill. Notice b must be even for the equation to have a solution. There are only 11 values from 0 to 20 that satisfy this condition. Then solve for a=(100-5b)/2 That being said, I still love the generalized solution which is instructive.
HOMEWORK : Find all bases of logarithms in which a real positive number can be equal to its logarithm or prove that none exist. SOURCE : IMO 1979 - Shortlist problem
Let a be the positive number, and b the base. We have that log_b(a) = a, or alternatively, b^a = a. This means that b = a^(1/a), and we can show using calculus that if a is a positive number, 0 < b
Log(base b)x = x => b^x=x If we compare both of their graphs, until they intersect at a tangent, the equation will have two solutions. At the tangent it will have one solution and after that no solutions. So slopes are equal => d/dx(b^x) = d/dx(x) => b^x*ln(b)=1 And we have the intersection equation b^x=x => ln(x^1/x)=1/x => ln(x) = 1 or 1/x= 0 (not possible) So x= e And b = e^1/e The thing given in question cannot be possible for b>e^1/e
SOLUTION *(0,1) ∪ (1, e^(1/e)]* Let assume that a ∈ R \ {1} is such that there exist a and x such that x = log_a(x), or equivalently f(x) := (ln x)/x = ln(a). Then a is a value of the function f(x) for x ∈ R+ \ {1}, and the converse also holds. First we observe that f(x) tends to -∞ as x→0 and f(x) tends to 0 as x→1. Since f(x) > 0 for x > 1, the function f(x) takes its maximum at a point x for which f'(x) = (1 - ln(x))/x² = 0. Hence max f(x) = f(e) = e^(1/e). It follows that the set of values of f(x) for x ∈ R+ is the interval (-∞, e^(1/e)), and consequently the desired set of bases a of logarithms is (0,1) ∪ (1, e^(1/e)].
I jump to the end after solving and suddenly see serieses😮 Idk if it valid but it works Since it's only integers and 0 obviously works we can just check how many multiples of 5 are there that are an even number away from 100(/how many even numbers are there that are a multiple of 5 away from 100) and the only multiples of 5 that satisfy this condition are multiples of 10 And then you just realized it's 100/10+1 Where the +1 accounts for the 0
And here I was hoping for a closed form for the coefficients of the power series for C(x)... I came up with 11 as soon as I realized that any solution would require that 10 divides both 2 a and 5 b. Alexandre Ribeiro's solution gives the details as to why.
First one I've done from the thumbnail. The numbers are coprime and are both factors of the final answer. So a must be a multiple of 5 and b, a multiple of 2. They must also be less than or equal to 100 divided by the other variable, and there are fewer multiples of 5. And for every a there is only one b. So the answer is just the number of multiples of 5 that are less than 100/2 (and greater than or equal to zero). So 50/5+1= 11. It actually takes more time to solve, IMO. To avoid just listing every answer, I'd have to think about how to write it as an expression using nonnegative integers.
It seems as if the solution was made hard intentionally although it was beautiful. My solution is if px+qy+r=0 has a solution (x,y) then we can rewrite the eqn as p(x+q)+q(y-p)+r=0 or p(x-q)+q(y+p)+r=0 and say if (x,y) is a solution then (x+q, y-p) and (x--q ,y+p) are also solutions. The solutions form an AP. we can find non negative terms in the ap to get the answer and we can derive general formula for obtaining all the infinte solutions.
2a and 5b must be multiples of 10. a is the multiple of 5 from zero to 50, and b is the reverse multiple of 2 from 20 to 0, and they are 11. (0,20), (5, 18), (10, 16), …, (40, 4), (45, 2), (50, 0).
Isn't it easier to write a=(100-5b)/2. Considering b is a positive integer, a is a positive integer iff 100-5b is even and positive => b is even and b=
Of course there is just 11 solutions since 100-5b is even and nonnegative integer just for b=0,...,10 (only nonnegative integer b are allowed). Michael just showed us a general approach. And also that a difficult calculation can bring many possible typos....
2a is always even, 100 is even, so 5b must be even. So let's say b=2d. We then have 2a+10d = 100 or a = 50-5d. For any value of d we choose we can find exactly one value of a = 50-5d such that the equation is true, and this value is always integer as d is integer. Considering both a and d must be positive, 0
So much headache! Why not this: a mod 5 and b mod 2 are zeros. So a and b can be written as 5m and 2n respectively. Therefore n+m=10. Solving for non-zero solutions, gives you (0,10), (1,9), ...,(9,1) & (10,1).
well.... since 100-5b can be divide by 5 thats mean 2a must be the multiplier of 5 (and 2a have to be even) 5b cannot be bigger than 100 , so max value of b is 20 when b is odd, 2a will be odd and contradict the setting so the possible solution of b will be{20, 18, 16, ..., 4, 2, 0} thats 11 solutions. Or since 2a must be the multiplier of 5 (and 2a have to be even) so 2a will be the multiplier of 10 since 2a+5b>=0 and there is only 11 integers which can express in 10N (where N c- Z+) hence 11 solutions. the vid is great. but i would not spend 18 mins on this particular question.
100 consists of 10 decades. each decade is either five 2s or two 5s. So there can be 0..10 decades from 2s, the remainder from 5s, so there are 11 ordered pairs. I didn’t understand the video
You always do these in the most complicated way possible. If 2a + 5b = 2c + 5d, then 2(a-c) = 5(d-b), so a-c = 5k and d-b = 2k for some k. In other words, c = a - 5k and d = b + 2k. So starting with (a,b) = (50,0) the solutions are (50,0), (45,2), (40,4),...,(0,20), for a total of 11 solutions.
Easy sol. If b is odd, 5b is odd then 2a = 100 - 5b will be odd. But it isn't possible, so b is even. No. of even values of 5b from 0 to 100 is 11. For any even value of 5b, 2a will be even which can be satisfied as 2a | even for any a. No. of sol is 11. Here's a good place to stop
why so complicated? How about: 2a +5b = 100 a = 50 + 5/2 b implies: b is even (as 5/2 b needs to be an integer), further a,b nonnegative implies tha b is in the set {0,2,4,6....20} (as for b > 20 we get a
B can't be an odd number because it makes 2a an odd number which is impossible. That means b has to be an even number from 0 to 20. And that is the number of solutions.
You did need 18min to solve that?! It is pretty clear, that a must be a multiple of 5 and b must be a multiple of 2, because 100's prime factors are only 2 and 5. Also clear: if you choose either a or b, the other one is determined as well. Looking at a, it can only have the values 0, 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50 (resulting in the corresponding values for b being 20, 18, 16, 14, 12, 10, 8, 6, 4, 2 and 0).
Crazy that the generating formula is equivalent to this (if I didn't make any mistake): floor(n/10) + 1 (-1 if ending in 1 or 3, because these are odd and
The other cool thing I noticed is that the difference in the fractional part of the generating function by incrementing n is as follows: (1/2) from (1/4)(-1)^n 1/10 from (n+1)/10 8/20 from the 5 weird numbers for a total of (10+2+8)/20 = 20/20, which keeps the solution an integer.
11:53 Hang on, doesn't the b term on the right turn out as b(1 + 2x + 2x^2 + 2x^3 + 2x^4 + x^5) ? b/(1 - x)^2 is multiplied by (1 - x^2)(1 - x^5) which equals b/(1 - x)^2 times (1 - x)(1 + x)(1 - x)(1 + x + x^2 + x^3 + x^4) . The 1/(1 - x)^2 is cancelled by the two (1 - x) factors leaving us with b(1 + x)(1 + x + x^2 + x^3 + x^4) or b(1 + 2x + 2x^2 + 2x^3 + 2x^4 + x^5) . Edit: so your correction isn't quite right either, as it gives (1 + x^2 + x^3 + x^4) instead of (1 + x + x^2 + x^3 + x^4) .
So I solved this problem my own, but was afraid I was overlooking something. Heres what I done. So first, given that both 2a and 100 are even numbers, then 5b must also be positive. So we make b=2c. That gives us: 2a + 10c = 100 => a = 5 * (10 - c) Given that both "a" and "c" are non-negative integers, therefore c∈{0, 1, 2, ..., 10}. Therefore there are 11 possible answers. (a, b) ∈ { (50,0) ; (45,2) ; (40,4) ; (35,6) ; (30,8) ; (25,10) ; (20,12) ; (15,14) ; (10,16) ; (5,18) ; (0,20) } So I was afraid this was too simple, but Professor Penn did an absolute overkill and got the same result, so my line of thinking is probably correct (I think).
This is very much overkill but I love it!
A simple way to see why it is 11 is as follows:
b has to be even, since 2a and 100 are.
b has to be less than/equal to 20.
There are 11 even numbers between 0 and 20.
a is then uniquely determined by (100-5b)/2 = 5(20-b)/2 which is a positive number.
That's exactly how i did it! Because if b was odd 5b would end with 5 and then 2a would have to end with 5 which is impossible for an integer value of a. I was so confused when i skipped to the end of the video to see the solution😅
This is what I love about this channel. The problem can be solved in your head, but when you go through the video, you learn so much about different ways to apply to more general cases.
Everyone: 11!
Micheal Penn: 18 minutes of Generating Functions!
Me: So you said everyone think there are 39916800 solutions huh?
What? It's just 11, not 11!
@@dufadufdil4778 I was waiting for this
Alternate formulation:
> From 2a + 5b = 100, we have 2a = 100 - 5b = 5(20 - b), and we must conclude that a has to be a multiple of 5; therefore, a = 5c
> Plugging that back in our equation, we have: 10c + 5b = 100. Dividing both sides by 5 yields 2c + b = 20; now, this gives us b = 20 - 2c = 2(10 - c), and therefore b is an even number, so we can write b = 2d
> From 2c + b = 20, we get 2c + 2d = 20, or c + d = 10. We can show that this equation has 11 solutions for non-negative integer values of c and d, and since this is a transformed version of the original equation, we can conclude that the original equation must have 11 solutions as well.
You're right, and this doesn't even take long to check all of the solutions.
Hey I know you from Quora! You write excellent answers :)
@@pbj4184 wow, I'm internet famous now haha so glad someone recognized me!
I didn't even have to plot down anything on paper to solve this. I just say that the solutions are enumerated by a = 50 down to 0, with step size 5, i.e. a=50, 45, 40, ..., 0, or a/5 = 10, ..., 0, which are 11 values.
Why do it simply if you can over-complcate it? :-D
Easier answer: reframe this as a discrete linear equation with a being y and b being x. So this way, it would be y = -5/2x + 50.
The slope is negative and the y-intercept is positive, so in the first quadrant of the plane (positive x and y) you have a finite number of integer values for x and y.
Start at the y intercept (0,50). The slope is -5/2, so every 5 units vertically there will be an integer solution.
Divide 50/5=10 for the rest of the solutions, and add the extra point where you started, the y-intercept. 10+1=11. Done.
Here are all the valid coordinates for the solution: (0,50) (2,45) (4,40) (6,35) (8,30) (10,25) (12,20) (14,15) (16,10) (18,5) (20,0)
This works really well for this simple example, but the nice thing about the method that Michael shows is that his method works for any number of coefficients added together in linear combination to reach a final sum. For example, this method would work for 2a+5b+7c+11d+13e+... = 10000 or something like that.
I'm sure we all got the answer by just doing it in our heads. The goal here was to use generating functions because the general question will not be as simple as 2a+5b=100.
That's how I wld have solved... Coz I hate combinatorics
@@keithmasumoto9698 yes, we all solved it in our heads. Yes.
i love that three random fractions: 1/4, 101/10 and 13/20 all add up to a nice integer. That's just so good...
Answer was already 11...
Q. 2a+5b=100 a,bE non- negative integers
Ans. as 5b will be multiple of 5, and 2a will always be even.
for sum to be even (100), 5b must be even
for sum to be multiple of 10 (100), 2a must be divisible by 10.
Hence, a=5m. and b=2n. m,n E non-negative integers.
Now, 10{ n+m}=100
n+m=10
So, starting from 0 till 10 for n, and 10 till 0 for m.
So, there will be 11 ordered pairs for (m,n) or (a,b)
This is how a kid would do it.
gratitude for watching pro's do it.
That's how I solved it in my head when I saw the thumbnail. And I visualized it as the points on a line in a-b-coordinates.
Michael: We'll use *Generating Functions* which is pretty standard...
Me: Yeah! I know what are they
The solution could be found easier, as everybody says, but this is amazing way to find it in general case. Great video!
That's a bit of a big overkill...
Yes it is.but i think that the idea is to learn this genarating function tool.this question was easy but he has one video when the question is hard and this tool helps
@@yoav613 That doesn't stop me from thinking this should belong in his 'overkill' series, though.
@@samuelmarger9031 i think he gave up of this over kill series but i agree this video should be there
you prolly dont care but if you're bored like me during the covid times you can watch all the latest movies and series on instaflixxer. Been binge watching with my gf recently =)
@Dax Victor Yea, have been watching on instaflixxer for months myself =)
you take on the most challenging problems bro, you worked this easily, great job
Dude this isn't that hard.
He just solves it with an overkill tool.
@@subpopulations exactly
Well, a simpler solution would be as follows: note that if (a,b) is a solution, then (a+5t, b-2t) is also a solution, and all the solutions are of this form. Take the maximal solution (0,20), then we have 11 solutions coresponding to (0,20), (5,18),...,(50,0).
dude i'm confused, the video was uploaded 8 min ago, the video is 18 min long, you posted the comment 45 sec ago, my question is : did you watch all the video? Or you just come to post the comment and go?
@@elie.makdissi actually he uploads his videos and keeps them unlisted , so if you check the playlists regularly then you might find the unlisted videos , I think that this dude might have watched the video and then commented or as you said, he/she might have just watched the beginning of the video and realised that he used a different method
@@elie.makdissi I saw the problem and solved it, then i speed run the video and saw his solution and checked that his was different from mine and that mine was simpler.
@@aa-lr1jk ah okay
@@notananimenerd1333 ah okay
Watching this series has made me appreciate:
1. How limited my mathematics exposure was in high school and college
2. How many different techniques professional mathematicians can be proficient in.
when you add text annotations to the video, i recommend adding a black background to the text with some transparency, or outlining the text in black so it's easier to read. it doesnt contrast with the stuff on the board very well. dunno what editing software you use but I have to imagine theres an easy way to do it.
Taking (mod 5) we have that a is multiple of 5 so a=5k for a non-negative integer k and then taking (mod 2) we have that b is a multiple of 2 so b=5m for non-negative integer m. So the equation becomes 10k+10m=100 or k+m=10 which gives us 11 pairs for k, m and because k and m are integers it follows that a, b are also gonna be integers for each pair as they are multiples of them so we get 11 pairs for a, b.
Note that (a, b) = (0, 20) is a solution to 2a + 5b = 100. Using a well-known result, solutions can be written parametrically as a = 5t and b = -2t + 20; because a and b are required to be non-negative integers, the least (integer) value for t is 0 and the greatest t-value is 10. Thus, there are 11 solutions.
Yep, approaches like that are very familiar from DEs
This solution exhibits the power of generating functions in solving these partition problems. This is very informative.
There can be no doubt that the material presented here is an example of powerful methods being applied to a specific problem. My admiration for all things intellectual admires that unequivocally. The practical side of my nature which also admires people who work with their hands, is a background noise which questions what possible use the high power math techniques can claim in the real world, It sees the power of intellect being applied to trivial ends.
9:39 The fraction with the quartic in the denominator should read dx^3+ex^2+fx+g in the numerator, no?
That's what I was thinking, but he says out loud the thing he's writing, and then there's an edit before "+ g", and he keeps it. So he must have meant it!
A great little problem. I solved it on my way to work today. Here's how I did it:
First, I noticed that 5b can end only with the digits 0 or 5, depending on whether b is even or odd. That means that 100-5b will also end in either 0 or 5.
However, this number is equal to 2a, and for a to be an integer it must be divisible by 2. So any b values that lead to 100-5b ending with 5 can be discarded. In other words, b can only be an even number.
But which even numbers? The condition that a and b must be positive integers means that b can at most be 20, because then 2a=0 -> a=0, which meets the condition. So b must be any even integer greater than or equal to 0 and less than or equal to 10. There are 11 such integers, and each one gives only one possible value for a.
Therefore, the size of the solution set is 9. The solution set itself is {(0,20), (5,18), (10,16), (15,14), (20,12), (25,10), (30,8), (35,6), (40,4), (45,2), (50,0)}
18:37
I read the thumbnail problem wrong as Z_>0 not Z_>=0
I like that I ended up doing a combinatoric solution, with counting partitions.
Initial Solution:
2a + 5b = 100
LHS is divisible by 2 and 5 so a must be divisible by 5 and b must be divisible by 2
consider new equation with 5n=a and 2m=b
then 10n+10m = 100
then n+m =10
there are nine partitions of ten into two positive integers seen from there being nine spots to choose the split point
n=1, m=9
n=2, m=8
n=3, m=7
n=4, m=6
n=5, m=5
and the same with n and m flipped
generating the solutions to the initial problem
a=5, b= 18
a=10, b= 16
a=15, b=14
etc...
Mistakes corner:
noticed while watching was problem was for non negative integers
for nonnegative solutions you have to add a partition at 0 and 10, getting 11 solutions instead.
ie these two solutions where missing from my original soultion
n=0, m =10
n=10, m= 0
a= 0, b=20
a=50, b=0
There is also 0+10 and 10+0, since it says "Non-negative" so 0 included.
So in the end we have
a=0/b=20
a=5/b=18
a=10/b=16
a=15/b=14
a=20/b=12
.
.
.
a=45/b=2
a=50/b=0
I misread the thumbnail and had to edit my solution, too.
Everybody else has commented how much more simply this problem could be solved to show there are 11 solutions. In fact using those more elementary methods you can easily generalize this result: the number of non-negative solutions to 2a+5b=10m is m+1, or more generally the number of non-negative solutions to 2a+5b=n is floor(3n/5) - ceiling(n/2) + 1. I wouldn't want to try getting that using Michael's method!
That does seem a bit overkill - I just said 2a+5b=100 implies a = 50 - 5/2 b, so we have b = 0 (mod 2). But we also know a >= 0 and solving that inequality gives us b = 0, and for each b there is a distinct a (b injective in a), we can rephrase our problem from "how many (a,b) pairs s.t. 2a+..." to "how many even b in the interval [0,20]", which gives us eleven, that's 0, 2, 4... 19, 20.
I knew something was up when Prof Penn posted a posted a problem that can be solved in your head, but took 18+ minutes to solve in the video! That is some serious overkill! (as Samuel Marger pointed out already)
Write 5b=100-2a. As a runs from 0 to 50, the right hand side is an arbitrary even number between 0 and 100. Thus, the right hand side is an even number which is a multiple of 5, i.e., 5b must be a multiple of 10. Thus, a can be only one of the 11 values: a=0,5,10,15,20,25,30,35,40,45,50. For each one of these values of a, there is exactly one value of b. In total, 11 solutions (a,b).
There is a very easy and intuitive way to solve this. Just look at b. b has to be between 0 and 20 (so 5b
That is an interesting method, albeit a complex one. I saw it shown to us once somewhere (I still remember these 13/20ths) and was very impressed. Generating functions kind of make combinatorics into bearable nearly smooth-function math. :)
Another great video, lots to learn
interesting, i solved this as a diophantine equation ( solved using linear congruence method) and hence used an inequality to narrow down the results to non negative integers. i got a = 50-5k and b=2k . Although i got that there is only 10 solutions and so im not sure why im missing an extra solution.
Edit: stupid mistake, k goes up to 10 but when we include k =0 solution there are 11 solutions
Umm, you're not. I did it the same way. Since a and b are non negative, the parameter k has to vary from 0 to 10, giving us 11 unique solution sets
@@utkarshsharma9563 yep dont worry, i forgot to count k=0 solution *facepalm*
@@domc3743 I thought you'd do that. I did it too
You actually do have 11 there, k can be anything from 0 to 10 (inclusive). That's 11
probably k = 0 or k = 10 check your inequality allows 50 - 5k and k be zero
your inequality probably is too tight
as every k = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] works and giving exactly 11 soultions
Awesome generalization to the problem! Love to learn new maths from unexpected places.
Simplifying
2a + 5b = 100
Solving
2a + 5b = 100
Solving for variable 'a'.
Move all terms containing a to the left, all other terms to the right.
Add '-5b' to each side of the equation.
2a + 5b + -5b = 100 + -5b
Combine like terms: 5b + -5b = 0
2a + 0 = 100 + -5b
2a = 100 + -5b
Divide each side by '2'.
a = 50 + -2.5b
Simplifying
a = 50 + -2.5b
If 2a + 5b = 100 then a = 50 - (5b/2) which is a non-negative integer. For a to satisfy this condition 1) b must less than or equal to 20 and 2) 2 must divide b.
Therefore, the total number of solutions is the number of multiples of 2 within 0 and 20 (including both). By counting by hand, you get 11 different solutions.
You can also solve using Diophantine equations: first you get one (arbitrary) solution to the equation: a0=0 and b0=20. Then you can parametrize all solutions by letting a= a0+5*k and b = b0 - 2*k. Since a and b must be positive, k can be only the numbers from 0 to 10. This gives a total of 11 solutions.
Hi Prof Penn...Thanks for another GREAT VIDEO!!
It has a remarkable solution the following way.Observation of the problem reveals that "a" has a maximum value of "50" when "b" is zero.And "b" has a maximum value of "20" when "a" is zero.Let us add "5t" to "a" and subtract "2t" from "b" without altering the given problem. We will get
2(a+5t) + 5(b-2t)=100
Now put a=0 & corresponding b=20, getting
2(5t)+5(20-2t)=100
Without loss of generality we may treat "5t" as "a" & "(20-2t) " as "b".
5t can have only 11 values when t = 0,
1,2,3,4,5,6,7,8,9,10.
Hence, 2a+5b =100 has only 11 solutions.
After noticing that 2 and 5 are both factors of 100, while 2 and 5 are co-primes and cannot divide each other in N. Thus 5 divides a, while 2 divides b. When simplifying 2a + 5b = 100, w/ a = 5x and b = 2y, w/ (x,y) belong to N, 2*5x + 5*2y = 100 x + y = 10.
So, x = 0, 1, 2, ..., 10; hence a total of 11 solutions.
I would rather use the method he mentioned in the beginning of the video.
The general solution of a and b are as follows:
a=5k-200
b=-2k+100
Substitute different values of k and we can get different values of a and b. Obviously, k at least equals 40 to make a positive. The eleven solutions are as follows:
(0,20), (5,18), (10,16), (15,14), (20,12), (25,10), (30,8), (35,6), (40,4), (45,2), (50,0).
Great job, your videos are really great. But I must say that this is a lot of work if you ask me. You could solve this with very little work, here is the solution that I am talking about:
2a+5b=100,for some (a,b)ϵN_0×N_0 (where N_0={0,1,2,3,…} well at least in Croatia where I'm from)
2a=100-5b=5(20-b)
Now since the righthand side is divisible by 5 and two is obviously not a multiple of 5 we conclude:
a=5x,for some xϵN_0
so:
10x=5(20-b)
by cancelling and rearranging:
b=2(10-x)
since the righthand side is a multiple of 2, b must also be, so:
b=2y,for some yϵN_0
Now we have:
x+y=10,for some (x,y)ϵN_0×N_0
We know that there are 11 solutions to that equation in that collection of numbers and they are:
x 0 1 2 3 4 5 6 7 8 9 10
y 10 9 8 7 6 5 4 3 2 1 0
We can trivially show that there is a one to one map from (x,y) to (a,b) (Pf: M(x,y)=(a,b)=(5x,2y) which is an injective multivariable function). From that we can conclude that the number of valid (x,y) is equal to the number of valid (a,b) and therefor there are 11 solutions.
See easy. Better than doing partial fractions decomposition (I would literally do anything to avoid doing that). Anyway thanks for your effort.
yeah nice
Talking about an easy solution and writes a book.
@@Last_Resort991 Look I could write this solution, and so could you, in about 4 to 5 steps, the thing is to make it rigorous you need to do more, the solution represented in this video uses partial fractions decomposition of very complicated fractions, and most of the work isn't shown (because then the video would be 20 minutes longer). Anyway, I like to be rigorous when I solve problems, that is why it looks longer.
Why going so complicated I just did it as follows
2a+5b=100
2a=100-5b
a=(100-5b)÷2
since a,b>=0
We have a>=0 when b is 0,2,4,6,8,10,12,14,16,18and 20
because 100-5b must be even therefore b must be even and b=
I must admit I got lost with this solution. I looked at it and bounded a at 50 and b at 20. Then noted a=50-5/2*b, so a is an integer iff b is even. Then, trivially the (a,b) pairs are (50,0), (45,5), (40,10), ... , (0,20). So 11 in total. Appreciate Michael was showing us a different technique but this one seemed like early high school maths to solve!
Since 2a is even and 100 is even, 5b must also be even then b is even... the lower bound for b is when b = 0, and the upper bound for b is when a = 0 which is b = 20, we get b = { x : x is even and 0
This is a great way to prove that 1/4 + 101/10 + 13/20 = 11 without doing the calculation (11 being obtained with the usual way to solve this problem). :-)
mdc(2,5)=1 então 2.(-1)+5.1=1 ... 2(-100)+5.100=100. Logo, (a,b)=(-100,100) é uma solução particular da equação diofantina 2a+5b=100, cuja solução geral é : a=-100+5t e b=100-2t.
Por fim, impondo que a>0 e b>0, obtemos t>20 e t
2a+5b=100, for 2a must be even so 5b must be even as well to get an even sum. Therefore, b must be even numbers which meets the requirement 5b
This can be solved in one’s head in a few seconds. The lcm of 2 and 5 is 10. There are only two solutions to add to 10 using 2 and 5: 22222 or 55. There are 10 such chunks in 100.
There are only 11 unique solutions (order doesn’t matter) to fill 10 slots with 2 choices each. (one solution is zero of the 2nd choice, one solution is one of the 2nd choice, up to one solution is ten of the 2nd choice).
So the answer is 11.
In summery: there are (K+ij)/ij ways to solve the equation ia+jb=K when a and b are restricted to be like a, b ≥ 0. And K is a multiple of ij.
Let's solve 2a+5b = 100 with it.
(100+2×5)/2×5
(100+10)/10
110/10
11
Let's try 4a+5b = 100
(100+4×5)/4×5
(100+20)/20
120/20
6
Solution 1: (0, 20)
100 = 2×5((0+1)+(0+1)...+(0+1))
Solution 2: (1, 19)
100 = 2×5((0+1)+(0+1)...+(1+0))
(1-x_n = y_n)
100 = 2×5((x_1+y_1)+(x_2+y_2)...+(x_10+y_10))
The first way is to set all x_n = 0.
The second is to set x_1 to x_9 = 0 and x_10 = 1
The nth way is to set x_1 to x_11-n = 0 and x_12-n to x_10 = 1
There are 11 ways because the x_n after the last x_n is x_11
Now let's try 4a+5b = 100
Solution 1: (0, 25)
100 = 4×5((0+1)+(0+1)...+(0+1))
Solution 2: (4, 20)
100 = 4×5((0+1)+(0+1)...+(1+0))
(1-x_n = y_n)
100 = 4×5((x_1+y_1)+(x_2+y_2)...+(x_5+y_5))
The first way is to set all x_n = 0.
The second is to set x_1 to x_4 = 0 and x_5 = 1
The nth way is to set x_1 to x_6-n = 0 and x_7-n to x_5 = 1
There are 6 ways because the x_n after the last x_n is x_6
Now let's try ia+jb = 100
100 = ij((0+1)+(0+1)...+(0+1))
100 = ij((0+1)+(0+1)...+(1+0))
(1-x_n = y_n)
100 = ij((x_1+y_1)+(x_2+y_2)...+(x_(100/ij)+y_(100/ij)))
The first way is to set all x_n = 0.
The second is to set x_1 to x_((100-ij)/ij) = 0 and x_(100/ij) = 1
The nth way is to set x_1 to x_((100+ij)/ij)-n = 0 and x_((100+2ij)/ij)-n to x_(100/ij) = 1
There are (100+ij)/ij ways because the x_n after the last x_n is x_((100+ij)/ij)
Now let's try ia+jb=K
K = ij((0+1)+(0+1)...+(0+1))
K = ij((0+1)+(0+1)...+(1+0))
(1-x_n = y_n)
K = ij((x_1+y_1)+(x_2+y_2)...+(x_(K/ij)+y_(K/ij)))
The first way is to set all x_n = 0.
The second is to set x_1 to x_((K-ij)/ij) = 0 and x_(K/ij) = 1
The nth way is to set x_1 to x_((K+ij)/ij)-n = 0 and x_((K+2ij)/ij)-n to x_(K/ij) = 1
There are (K+ij)/ij ways because the x_n after the last x_n is x_((K+ij)/ij)
Hey, Michael please make videos about olympiad inequalities too. I am pretty sure you are expert at those as well.
Very interesting way of solving this! I did not have a pen when trying it so here is my way :
2a + 5b = 100 thus we see that a
2a + 5b = 103
Hence, a = (103 - 5b)/2
Hence, (103 - 5b)/2 > b
Hence, 7b < 103
Hence, b < 103/7 = 14.71….
For a to be a positive integer, b is also an odd, positive integer
So, we could have b = 1→a = 49, b = 3→a = 44, b = 5→a = 39,…..,b= 13→a= 19
Hence, 7 pairs, which are (49,1), (44,3),(39,5),……(19,13)
I don't understand these generating functions and the complexity of the solution made my jaw drop. I approached this very differently: Since 2a is always even, 5b must also be even so that the sum 2a+5b = 100 (even). 5b is only even, if b is even. The smallest value for b = 0 making a = 50. The largest value for b = 20 making a = 0. So, b = 0, 2, 4,... ...,20 and the corresponding values of a are 50, 45, ... ...,0. That's 11 pairs of (a, b).
im not a math expert but.....
2a+5b=100
2a=100-5b
a=50-5b/2
note that b=even
0≤50-5b/2
5b/2≤50
5b≤100
0≤b≤20
counting all even integers between 20 and zero would give us
20-0/2 +1=11 ####
did he want it to do it the hard why or maybe my method is a bit risky
Every second comment is stating it’s overkill. You’re not wrong but you must understand two things. Firstly, the method is refined for demonstration purposes, and secondly this allows solutions to arbitrary n values which is very powerful
Most "easy" solutions in here also work with arbitrary n.
@@magnusPurblind if n is divisible by 5: ceiling(n/(5*2)) else floor(n/(5*2))
@@magnusPurblind Explanation: Dividing n by 5 shows the range of possible b (i.e. with n = 99, there are 20 possibilities for b: 0, 1, ..., 19). Because 2a is always even, we have to further divide n/5 by 2. Therefore you get n/(5*2). This is not always integer, so we take the floor function. If n is divisible by 5, there is always the additional solution a = 0, b = n/5 which we have to add, therefore we take the ceiling function in that case.
@@magnusPurblind What are you talking about? The formula is right for 11 and 21, which are both 1 mod 10.
I think it is even possible to generalize the number of solutions for arbitrary coefficients before a and b:
v*a + w*b = n
=> number of solutions is floor(n/(w*v)) (+1 if n is divisible by w).
i am no match for the incredible skill at play with this, but i'll post my simplistic method regardless:
first we ask: what are the factors of b? if b is odd, we get odd=even, therefore b must be even. The same divisibility rule applies for a with 5, so we know we can express: a = 5m, b = 2n. therefore we get the equation m+n = 10, for m, n nonnegative integers. for this, there are exactly 11 solutions.
An easier solution would be that since a and b are integers we get that 2 divides b and 5 divides a (take the equation 2a+5b=100 modulo 2 and modulo 5) so there are n,m with a=5n and b=2m. The number of pairs (a,b) is the same as the number of pairs (n,m) and we have 10n+10m=100 or n+m=10. Now it is obvious that there are 11 pairs (n,m) with n+m=10
funny , i did the same solution with (m,n). hahahaha
0
Overkill. Notice b must be even for the equation to have a solution. There are only 11 values from 0 to 20 that satisfy this condition. Then solve for a=(100-5b)/2
That being said, I still love the generalized solution which is instructive.
2a+5b=100 a/5 +b/2 = 10 a' + b' = 10 ( easy to see that a' , b' are natural ) So number of solutions is 11
HOMEWORK : Find all bases of logarithms in which a real positive number can be equal to its logarithm or prove that none exist.
SOURCE : IMO 1979 - Shortlist problem
Let a be the positive number, and b the base. We have that log_b(a) = a, or alternatively, b^a = a. This means that b = a^(1/a), and we can show using calculus that if a is a positive number, 0 < b
Log(base b)x = x
=> b^x=x
If we compare both of their graphs, until they intersect at a tangent, the equation will have two solutions. At the tangent it will have one solution and after that no solutions.
So slopes are equal
=> d/dx(b^x) = d/dx(x)
=> b^x*ln(b)=1
And we have the intersection equation b^x=x
=> ln(x^1/x)=1/x
=> ln(x) = 1 or 1/x= 0 (not possible)
So x= e
And b = e^1/e
The thing given in question cannot be possible for b>e^1/e
SOLUTION
*(0,1) ∪ (1, e^(1/e)]*
Let assume that a ∈ R \ {1} is such that there exist a and x such that x = log_a(x), or equivalently f(x) := (ln x)/x = ln(a). Then a is a value of the function f(x) for x ∈ R+ \ {1}, and the converse also holds.
First we observe that f(x) tends to -∞ as x→0 and f(x) tends to 0 as x→1. Since f(x) > 0 for x > 1, the function f(x) takes its maximum at a point x for which f'(x) = (1 - ln(x))/x² = 0. Hence max f(x) = f(e) = e^(1/e).
It follows that the set of values of f(x) for x ∈ R+ is the interval (-∞, e^(1/e)), and consequently the desired set of bases a of logarithms is (0,1) ∪ (1, e^(1/e)].
Nice method
@@goodplacetostop2973 damn, forgot about that pesky 1.
I nominate this video for induction into the overkill video series hall of fame
I jump to the end after solving and suddenly see serieses😮
Idk if it valid but it works
Since it's only integers and 0 obviously works we can just check how many multiples of 5 are there that are an even number away from 100(/how many even numbers are there that are a multiple of 5 away from 100) and the only multiples of 5 that satisfy this condition are multiples of 10
And then you just realized it's 100/10+1
Where the +1 accounts for the 0
And here I was hoping for a closed form for the coefficients of the power series for C(x)... I came up with 11 as soon as I realized that any solution would require that 10 divides both 2 a and 5 b. Alexandre Ribeiro's solution gives the details as to why.
How many distinct ordered pairs of natural numbers (a,b) satisfying 2a+5b=100?
Such a sweet solution
First one I've done from the thumbnail. The numbers are coprime and are both factors of the final answer. So a must be a multiple of 5 and b, a multiple of 2. They must also be less than or equal to 100 divided by the other variable, and there are fewer multiples of 5. And for every a there is only one b. So the answer is just the number of multiples of 5 that are less than 100/2 (and greater than or equal to zero). So 50/5+1= 11.
It actually takes more time to solve, IMO. To avoid just listing every answer, I'd have to think about how to write it as an expression using nonnegative integers.
It seems as if the solution was made hard intentionally although it was beautiful. My solution is
if px+qy+r=0 has a solution (x,y) then we can rewrite the eqn as p(x+q)+q(y-p)+r=0 or p(x-q)+q(y+p)+r=0 and say if (x,y) is a solution then (x+q, y-p) and (x--q ,y+p) are also solutions. The solutions form an AP. we can find non negative terms in the ap to get the answer and we can derive general formula for obtaining all the infinte solutions.
2a and 5b must be multiples of 10.
a is the multiple of 5 from zero to 50, and b is the reverse multiple of 2 from 20 to 0, and they are 11.
(0,20), (5, 18), (10, 16), …, (40, 4), (45, 2), (50, 0).
numerator of irreducible quartic should be dx^3 + ex^2 + fx + g
This problem can be easily done using the theory of continued fractions
Isn't it easier to write a=(100-5b)/2. Considering b is a positive integer, a is a positive integer iff 100-5b is even and positive => b is even and b=
x/lcm(a,b)+1 if x is divisible by a and b
Of course there is just 11 solutions since 100-5b is even and nonnegative integer just for b=0,...,10 (only nonnegative integer b are allowed). Michael just showed us a general approach. And also that a difficult calculation can bring many possible typos....
b cannot be odd and can only take on values between 0 and 20, hence there are 20/2+1 = 11 solutions
I have done this: we find an obvious solution say a=50 and b=0. Then the general solution is a-5x, b+2x. Both a-5x and b+2x must be non negative.
2a is always even, 100 is even, so 5b must be even. So let's say b=2d. We then have 2a+10d = 100 or a = 50-5d. For any value of d we choose we can find exactly one value of a = 50-5d such that the equation is true, and this value is always integer as d is integer. Considering both a and d must be positive, 0
So much headache! Why not this: a mod 5 and b mod 2 are zeros. So a and b can be written as 5m and 2n respectively. Therefore n+m=10. Solving for non-zero solutions, gives you (0,10), (1,9), ...,(9,1) & (10,1).
well.... since 100-5b can be divide by 5
thats mean 2a must be the multiplier of 5 (and 2a have to be even)
5b cannot be bigger than 100 , so max value of b is 20
when b is odd, 2a will be odd and contradict the setting
so the possible solution of b will be{20, 18, 16, ..., 4, 2, 0}
thats 11 solutions.
Or since 2a must be the multiplier of 5 (and 2a have to be even)
so 2a will be the multiplier of 10
since 2a+5b>=0
and there is only 11 integers which can express in 10N (where N c- Z+)
hence 11 solutions.
the vid is great. but i would not spend 18 mins on this particular question.
100 consists of 10 decades. each decade is either five 2s or two 5s. So there can be 0..10 decades from 2s, the remainder from 5s, so there are 11 ordered pairs. I didn’t understand the video
Very good.
You always do these in the most complicated way possible. If 2a + 5b = 2c + 5d, then 2(a-c) = 5(d-b), so a-c = 5k and d-b = 2k for some k. In other words, c = a - 5k and d = b + 2k. So starting with (a,b) = (50,0) the solutions are (50,0), (45,2), (40,4),...,(0,20), for a total of 11 solutions.
If you are working out a general purpose approach, like Generating Functions, it helps to have an example you can cross check manually.
Easy sol.
If b is odd, 5b is odd then 2a = 100 - 5b will be odd.
But it isn't possible, so b is even.
No. of even values of 5b from 0 to 100 is 11.
For any even value of 5b, 2a will be even which can be satisfied as 2a | even for any a.
No. of sol is 11.
Here's a good place to stop
2a is even, so 5b must also be even so 5b any of {0,10,20,30,40,50,60,70,80,90,100}. So answer is 11.
why so complicated? How about: 2a +5b = 100 a = 50 + 5/2 b
implies: b is even (as 5/2 b needs to be an integer), further a,b nonnegative implies tha b is in the set {0,2,4,6....20} (as for b > 20 we get a
Consider 2 a + 5 b = m, m>=0
If m=10k+1 or m=10k+3 #solutions=k else #solutions=k+1
I love overkill series.
I'd look at this like seeing how many ways you can get one pound with 5p and 2p coins.
5 divides a, so a = 5n, n >= 0
max(10n) = 100
so n = 0, 1, 2, ... , 10
11 solutions.
Quite the overkill
9 solutions
(a;b) = (5;18), (10;16), (15;14), (20;12), (25;10), (30;8), (35;6), (40;4) or (45;2)
Another nice overkill video
Isn't it a diophantine equation?
B can't be an odd number because it makes 2a an odd number which is impossible. That means b has to be an even number from 0 to 20. And that is the number of solutions.
is there an easier way to do this without generating functions?
a = (100 - 5b) / 2
b = 2k
a = 50 - 5k
k = 0, ..., 10
11 solutions :)
Looks like I am late to the table for saying there's a quicker solution.
Just by looking at it we can see a = 0,5,...,50.
You did need 18min to solve that?!
It is pretty clear, that a must be a multiple of 5 and b must be a multiple of 2, because 100's prime factors are only 2 and 5. Also clear: if you choose either a or b, the other one is determined as well. Looking at a, it can only have the values 0, 5, 10, 15, 20, 25, 30, 35, 40, 45 and 50 (resulting in the corresponding values for b being 20, 18, 16, 14, 12, 10, 8, 6, 4, 2 and 0).
Holy Shit!
Crazy that the generating formula is equivalent to this (if I didn't make any mistake):
floor(n/10) + 1 (-1 if ending in 1 or 3, because these are odd and
The other cool thing I noticed is that the difference in the fractional part of the generating function by incrementing n is as follows:
(1/2) from (1/4)(-1)^n
1/10 from (n+1)/10
8/20 from the 5 weird numbers
for a total of (10+2+8)/20 = 20/20, which keeps the solution an integer.
Solve this! Pa+P1b=P2, a,b e Z, and P123 are any primes!
Well I did it the dumb way. Start with a = 50 and b = 0. Then take away 5 copies of a and add 2 copies of b. Repeat until you have all b and no a.
11:53
Hang on, doesn't the b term on the right turn out as b(1 + 2x + 2x^2 + 2x^3 + 2x^4 + x^5) ?
b/(1 - x)^2 is multiplied by (1 - x^2)(1 - x^5) which equals b/(1 - x)^2 times (1 - x)(1 + x)(1 - x)(1 + x + x^2 + x^3 + x^4) .
The 1/(1 - x)^2 is cancelled by the two (1 - x) factors leaving us with b(1 + x)(1 + x + x^2 + x^3 + x^4) or b(1 + 2x + 2x^2 + 2x^3 + 2x^4 + x^5) .
Edit: so your correction isn't quite right either, as it gives (1 + x^2 + x^3 + x^4) instead of (1 + x + x^2 + x^3 + x^4) .
So I solved this problem my own, but was afraid I was overlooking something. Heres what I done. So first, given that both 2a and 100 are even numbers, then 5b must also be positive. So we make b=2c. That gives us:
2a + 10c = 100
=> a = 5 * (10 - c)
Given that both "a" and "c" are non-negative integers, therefore c∈{0, 1, 2, ..., 10}. Therefore there are 11 possible answers.
(a, b) ∈ { (50,0) ; (45,2) ; (40,4) ; (35,6) ; (30,8) ; (25,10) ; (20,12) ; (15,14) ; (10,16) ; (5,18) ; (0,20) }
So I was afraid this was too simple, but Professor Penn did an absolute overkill and got the same result, so my line of thinking is probably correct (I think).