4^8+26^3 is not divisible by __?___ Modular arithmetic, Reddit r/askmath
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- čas přidán 7. 03. 2024
- We have to figure out the number 4^8+26^3 is NOT divisible by which of the following?
(A) 3 (B) 4 (C) 5 (D) 6 We cannot use a calculator for this so how can we solve this number theory problem without doing much calculation? I will demonstrate a solution by using modular arithmetic, i.e. "work with the remainder".
This question is from Reddit r/askmath.
Original post / sh9elafz9x
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I love the your standard math professor approach. However, I was able to reason the answer quickly.
1) Any number ending in 6, then raised to any power will always still end in a 6. Remember, this is the units digit. We will need this later.
2) 4 raised to any power will always end in either 4 or 6. If it's raised to an odd power, it will end in 4. If it's raised to an even power, it ends in 6. In this case, 4^8 ends in 6.
3) Add the two units we just figured out, 6+6=12, and 12 is NOT divisible by 5.
Answer choice C.
this is way better than my solution. 😃
Bro 😳, I thought it in the same way 😳. What a coincidence
True I thought that too
@@bprpmathbasicsMaybe "better" in this particular case, but I just learned a hell of a lot about modular math that I never learned in any undergrad math course I took. Dr Penn uses modular math a bit in his videos, and I admit I never took the time to learn that "weird" notation, so just have taken his explanations at face value. *Now* I understand, and can go back to his channel and actually get it all now. I can't thank you enough! :D
I went for the units also, but hadn't realized this 4, 6 pattern.
I made it 2^16. As I already know 2^10, I started from there: 1024, 2048, 16, 32, 64, 128, 256. Then did the 6+6.
But the modular arithmetic lesson was great. And learning this pattern is great also.
the last digit of 4^(2n) is always 6 and 6^(n) is always 6
so the last digit will be 6+6 will give 1(2) which isn't divisible by 5
That's how I did it, too.
Compact, but that's it.
That's just doing the computation with an extra trip through mod 10, though. The modular arithmetic approach is much more general.
Exactly how I did it
@@arbitrary54 modular arithmetic is the generalization of it, but unfortunately isn't taught in a lot of the american education system in my experience, and can trip people up when first learning about it. so staying in base 10, but talking about how we really only need to evaluate the sum of the last digits. plus as a bonus you can use this problem as an example on how modular arithmetic does generalize the process! just my perspective as a past tutor
I did not learn modular arithmetic in school, this was very interesting! Great video, thank you!
You can do 26 faster when you're working modulo 3 if you think of it as −1. (Similarly when you work with 4 modulo 5.)
I wouldn’t say it’s faster because of how easy it is. For 4 modulo 5 however yeah it’s lot easier by using -1. Otherwise you would have to calculate 4 to the power of 8 which is kinda hard or take the mod successively which is easy but takes a bit of time.
In any prime mod p you can reduce exponents mod p-1 due to fermat's little theorem
@@ryznak4814 4⁸ is just (2²)⁸ = 2¹⁶, (so 16 bit for computer nerds) which is 65536.
@@sophiastern2719 : Yes, or more generally in any modulus n, you can reduce exponents by the totient of n. (Except that you can't always reduce them all the way to 0.)
what? where does -1 come from?
As both summands are even, the number is divisible by 6 if and only if it's divisible by 3. So it must be divisible by both, assuming the question only has one correct answer. That leaves only 4 and 5 as possible answers. 4 can be eliminated since the first summand is a power of 4, and the second has a factor of 2^3 which is 8. So the only choice left is 5.
That is exactly what I did, except I started with 4.
Yep. Any even number raised to the power of n or higher must by definition be divisible by 2ⁿ. As must any sum of such numbers.
Three is the only number you really need to do mod calculations for (though I'm sure there's probably some way to get around even that). Six is confirmed once three is, since the final sum must be even, and four and five can be more easily confirmed or rejected respectively by other means.
The multiple choice gives things away a little bit, 3 and 6 can be eliminated since one would imply the other and they cant both be true on a question like this
Edit to clarify for my fellow nerds: 6 divides x implies 3 divides x, but 3 divides x implies 6 divides x only if 2 divides x. Since 2 divides 4^8+26^3, which we can see since its of the form (2(integer a))^(integer b)+(2(integer c))^(integer d), were free to use the biconditional and eliminate both 3 and 6 based on the assumption that only one answer can be correct. As an aside, we can also eliminate 4 since any integer power higher than 1 of an even number is divisible by 4, and adding two numbers divisible by 4 results in a number divisible by 4.
Not entirely correct. If a number is divisible by 6 that implies that is also divisible by 3, but if a number is divisible by 3 it may not be divisible by 6 (for example 9).
@@luisaleman9512yeah, but in this case, 4^8+26^3 is an even number, so if 6 is wrong, it means 3 is also wrong (and vice versa but pretty sure you got that already)
@@CarolCanGame yes, of course in the context of this problem your comment is correct. I was thinking of a general case.
@@luisaleman9512Yes, but if show 4 first, then 3 & 6 are in the same boat and either are both in or both out.
@@luisaleman9512 The number is a sum of two even numbers, and thus even as well. So, yes, if it is divisible by 3, it is divisible by 6, too.
since this is a multiple choice question, it doesn’t really need modular arithmetic for it. the thought process is like this:
1. 3 and 6: since the addition results is an even number, if it is divisible by 3, then it MUST be divisible by 6. since the question is an MC, both 3 and 6 are ruled out;
2. 4: 4 to whatever power is obviously divisible by 4, so only need to consider 26^3; as 26 is an even number, any power larger than 2 must at least have “2 copies of 2s” which is also divisible by 4. thus the addition result is also divisible by 4;
3. only one answer left, so it must be 5
I like this solution but I feel like 99% of math teachers would mark this as wrong if "showing your work" was part of it. Math teachers tend to hate logical arguments.
@@Phyrre56 They don't necessarily hate logical arguments, they only dislike it if it ends up being one that is only useful in certain scenarios. If the numbers were changed around to not include 3, 6, and 4 then the logical argument would be irrelevant and you'd still have to figure it out without a calculator.
@@Phyrre56it would be wrong depending on the level of education
@@MikehMike01 Less about the level of education, and more about the format of the test. If it's a multiple choice question where you only need to key in the answer (like, say, the SAT) then you can use all the tricks in your arsenal, and it's worth doing so because the test is timed. If it's a formal math class where you are expected to show work, then the answer is likely only partial or no credit.
@@Phyrre56 99%? What's your sample size? I'd hate to think this is true. Logical arguments are what math IS.
I didn't know anything about modular arithmetic but I followed this easily and found it fascinating!
r/whoosh for me!
Still love this channel. I never would've known modular arithmetic existed.
I recall it being covered in HS, but it was one of those short, 1 week, never gets used again things, which is a shame
@@oxylepy2 I was in high school in the 90's. It's possible they taught it, but I wouldn't remember if they did.
I had seen people do modular arithmetic on Quora, but they never explained the work in such detail for me to understand what they were doing, this video helped me understand a lot of it, thanks!
Modulus math... stuff I actually learned for programming games, web sites, and other random projects. Most of the functions you feature I have no clue about, but I got this one down pat. It's super useful in so many different programming situations.
this one’s quite easy with a little logic! since the number is clearly even, can’t be the case that it’s divisible by 3 and not 6, and vice versa, so neither of those are the answer-
then you can see that it’s divisible by 4 when you see that the first added clearly is and the second added is a power of an even number, so 26^3 will have 2^3 in it’s prime factorization, so that’s also divisible by 4!
so the sum is divisible by 4, leaving the only possibility to be 5- heckin’ brilliant :3
It's been 33 years since I had to deal with mod. Took me back for sure!
Process of elimination.
Four is easy since 4^8 and 26^3 are both 0 (mod 4) therefore divisible by 4.
Three is also easy since 6 is 0 (mod 3) AND 0 (mod 2). If not divisible by 3 then it's also not divisible by 6 and only looking for single answer SOOO divisible by 3.
That makes six easy because we've already shown divisible by 3 and 4 which are coprime. That means divisible by 12 which is divisible by 6.
That leaves five as the odd one out.
This is how I did it too! There are some "dirty tricks" you can do with multiple choice questions like this that can cut down on a lot of unnecessary computation.
They never taught me anything about modular arithmatic when I was in school and I had no idea it even existed until just now, but now I definitely want to learn it. How cool.
A great lecture. I never knew much arithmetic in remainders and mods to solve something in addition that way. It was a fresh new teaching for me! 👏
I love learning things that I don't know and have no need to know. Knowledge is power. I love your videos.
I have taken a lot of math. I've not seen this solution before. Very cool.
Perfectly clear, excellent explanation.
Ty
I noticed that 4 and 26 are both even numbers, so no matter what the exponent is (as long as it is at least 2) then they will both be multiples of 4. A multiple of four added to another multiple of four results in a multiple of four.
Of course a multiple of four is even, so the result will also be a multiple of six as long as the result is also a multiple of three. This means that, if the result is not a multiple of three, it will not be a multiple of three OR six. Since we cannot select two correct answers in a multiple choice question, I made the assumption that the result must be a multiple of three (and therefore six as well).
By process of elimination, the result is not a multiple of five.
You must try solving IOQM(Indian Olympiad Qualifier in Mathematics) problems and INMO problems ....they are really good ...
I tried a different logic for as effective solving as possible.
1) Notice how we can factor 4 from the expression. So it will be divisible by 4.
2) If 3 was the answer meaning that 3 will not be a factor will imply 6 to also be an answer. As only one option is correct we are left with 5
4 to any odd power ends in 4, and even powers in 6.
6 to any power ends in 6.
As it is an even power of 4, it ends in 6, the two added together ends in 2, thus not a multiple of 5.
My thought exactly.
Great minds think alike.....
Issue is that you would have to prove the 4/6 shortcut, you can't just assume that.
You can exclude the 4 based on 26 = 13*2 and 26^3 = (13*2)^3 = (13^3 * 2^3) = (13^3 * 8) = (13^3 * 2 * 4), meaning both parts of the addition are divisible by 4.
3 would require modular math to exclude, which combined with 4's exclusion means 6 is excluded. (3*4 = 12 = 2*6)
You could stop there, as the question does not have an all-of-the-above, so 5 would have to be the answer unless the question is wrong.
But if you HAVE to prove it as well, 5 would still need its own modular math, as there is no factorial of 3, 4 and 6 that can get you n*5.
Que vídeo massa!!! Não sabia esse forma de encarar esse tipo de problema. Parabéns🎉🎉🎉
Guessing strat:
It can’t be 3 because if it isn’t divisible by 3 it’s not divisible by 6, this leads to 2 answers which isn’t the case here. 4^8 is obv divisible by 4 and 26^3 = (2*13)^3 = 2^3*13^3=8*13^3 which is also divisible by 4. If it is divisible by 4 and 3 then since 4=2*2 it is divisible by 2*3 I.e. 6. Hence ans is 5.
Damn, you're smart, I didn't think of how you could exclude 2 numbers right away
I used the exact same method and, when I saw that a different method was used in the video, I made my own comment about how I answered the question. I wouldn't have written it out if I had seen this comment earlier...
I like how you excluded the one answer by deduction. I used a different approach, as it's immediately obvious that the number doesn't end in a 5 or 0 (because you're adding two numbers that end in a 6), so it has to 5 (doesn't require calculation).
@@Misteribel how would you know 4^8 ends in a 6 without calculating it?
Nvm i realised 4^(any even number) ends in a 6 and odd numbers end in a 4
Man, I remember doing this such a long time ago, but I completely forgot how to do it until re-watching this video. Same with your polynomial factors for quadratic equation video. I love re-learning this kind of stuff, and I actually think I now understand it even better than when I first learned it.
I also remember a few neat tricks for some numbers...
Everything is divisible by 1.
Every even number is divisible by 2.
If you add up the digits of a number and it's a multiple of three, the original number is a multiple of three. (E.G., 216 is 2+1+6=9, 9 is divisible by 3, so 216 is.)
If the LAST TWO digits are divisible by 4, the number is divisible by 4.
If the LAST digit is 5 or 0, the number is divisible by 5.
6 of course follows the 2 and 3 rules.
I don't know one for 7.
If the LAST THREE digits are divisible by 8, the number is divisible by 8.
If the digits add up to a multiple of 9, is divisible by 9. (E.G., 1701, 9999, 4230)
If last digit is zero, divisible by 10.
I don't know one for 11.
12 is of course 3 and 4.
First I tried to work this out by looking at the prime factorization, which was only useful for eliminating Four as the solution because a 4 cleanly factors out of both 2^16 and (2*13)^3, the prime factorization of both terms. Then, since I didn't remember the rules of modulo arithmetic, I did a few examples in my head for the mod 3 case. After convincing myself that I can do operations on their remainder, three was quickly found to be a divisor of the expression. Six was then eliminated immediately because anything divisible by 3 and also even (which this clearly will be) is divisible by 6. That only left Five as a potential answer, but I worked that out with modulo arithmetic also, though I didn't use that -1 trick so it took a little more effort.
Testing:
3: 4^8+26^3≡1^8+(-1)^3=1-1=0
4: 4^8 and 26^3 are both obvious multiples of 4.
6: given due to the previous ones.
5: 4^8+26^3≡(-1)^8+1^3=1+1=2
So 5 is the answer.
General
4^8+26^3=2^3*3*3463
Yes this factorisation was done with Wolfram Alpha.
This is new concept for me and this is so cool
Modular arithmetic is interesting but ultimately useless in a modern world. Any math question that has to specify "Do not use a calculator" is like a flashing sign that says "this is a concept from decades ago that no one uses anymore."
@@Phyrre56 i see, I suppose that's why it's not listed on my hs curriculum then
@Phyrre56 modulo arithmetic is why RSA exists but ok
4 to some power end in 4, 6, 4, 6, 4, etc, with 6 for the even powers. 6 (or 26) to some power always ends in 6. 6+6=12, that is, the resulting number cannot end in 0 or 5, so it's not divisible by 5. (this is similar to the modular approach, but it's trivial to do in your head).
While this is true, I wonder if this explanation would be accepted as correct in a math class. Whenever simplifications like this are used, teachers like to reply with "Well imagine instead of 4 it was 17 to the 8th power, now your simplification doesn't work anymore and you can't do it in your head, so it wasn't the right way to do it."
@@Phyrre56 Actually, it's still works. With 17 you will get numbers that end in ...7 ...9 ...3 ...1 ...7 etc. So if you add this with a power of 26 that always ends in 6, it still won't be divisible by 5 (nor by 4 or 6 or any even numbers).
In an MC, this kind of thinking is often the simplest way.
@@Phyrre56Exactly. It's cool you memorized a rule that's useful in a singular case, but the point of this is to develop a skill that's useful across the board.
@Phyrre56 it works with any number and any power, as shown above for 17. No need to memorize any rules. Several math videos, including ones from this channel, use this approach. It's just a simplification of modular math, and it's sound.
To explain why the modular arithmetic works this way, consider 4^2 (mod 3): 4x4 = (3+1)x(3+1) = 3x3 + 2(3x1) + 1x1. Any integer multiplied by N is 0 (mod N) so the first and second parts cancel out, leaving only 4^2 = 1^2 (mod 3).
speaking of power, the mitochondria is the powerhouse of the cell
We can cross out the option B obviously. Then 4^n is always 1 mod 3 & 26^n is 2 mod 3 if n is odd, meaning the total is divisible by 3. Since it's divisible by 3 & 4, so do 6. So option C is the answer
I love prime factorization, so that's what I used. It is not nearly as efficient, but it's one of my favorite techniques.
First, I'll do the prime factorization of both terms
4^8=2^16 , 26^3 = (13^3)*(2^3)
Next, I'll isolate 2^3 from each, because that is what's common. I've already ruled out 4 because of the common 2^3.
2^16+(13^3)*(2^3) = 2^3(2^13+13^3)
Any term ending with 3 taken to a third power will end with 7. Powers of 2 end with '2, 4, 8, 6, 2, 4, 8, 6, ...' in that order, assuming the exponent is greater than 0.
To factor by 5, the 2^13 term must end in 8, because anything perfectly divisible by 5 ends with either 5 or 0. I can group 3 and 6 together, since I have an 8 outside, so as long as it's divisible by 3, it's divisible by 6. This one I have to admit, I wasn't confident about, so I'm shooting a bit blindly here.
When dividing 13, I get 3R1, meaning it ends with 2, which looks like this: "2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6, *2*", and 7+2=9, not 10 or 5, telling me that 5 is not divisible.
Using negative modular numbers is indeed a useful trick (which was great to point out), but you could also do:
4^8 = (4^2)^4 = 16^4
16 ≡ 1 (mod 5)
(4)^8 ≡ (16)^4 ≡ (1)^4 ≡ 1 (mod 5)
However, it's also interesting to note that if the exponent on the 4 was an odd number, then the result _would_ be divisible by 5, for example:
4^9 + 26^3 ≡ (-1)^9 + (1)^3 ≡ -1 + 1 ≡ 0 (mod 5)
I figured that after we find that 3 is a factor of the number, we can easily show through prime factorization that 2 is a factor at least twice, then use the proof that says if a|b and a|c then a|b+c to show that choices a,b,d are all possible, so we only need to show that c is impossible
Just see that 4=5-1, 26=25+1, then you immediately get that the remainder is 2. (mod 5)
4⁸ is obviously divisible by 4. And since 26³ = 8 · 13³, this as well. So, the sum of two numbers which are divisible by 4 is divisible by 4 as well.
The number is even. So, if it is divisible by 3, it is divisible by 6 as well.
Thus, by ruling out the other options, the solution must be (C).
There is a fast way to logic this out.
4^8+26^3.
A. 3
B. 4
C. 5
D. 6
The easiest one to knockout is B. Both 4^8 and 26^3 would be divisible by 4. 4^8 because you are multiplying 4s to get it and 26^3 because you can simplify to that to (13*2)^3. Since you are multiply 2 in there at least twice, you know it's divisible by 4.
By using logic, you can rule out A. If it isn't divisible by 3, then it can't be divisible by 6 either since it must be divisible by both 3 and 2 to also be divisible 6. In other words, if the answer was A, then there'd be multiple correct answers.
The last step uses the information we gathered from the 2 previous steps. In order to be divisible by 4 it must be divisible by 2 and we know it's divisible by 3 from the previous step.
The only possibility you are left with is 5.
I'm curious to see how he does it.
26 isn't a nice number but we can still factot it into 2^3(2^13+13^3) and based on this it is divisible by 4 so 4 is ecluded but we need to calculate the parantese to see if it's divisible by 3(and 6) or if it's divisible by 5
I got there with:
Both have multiple 2 factors, therefore sum will be divisible by 4. For 3 and 6, since I already know there is a 2 factor at the end, it can't be one without being the other, so discount these two. Therefore (C).
Here’s how you can do it with minimal math: process of elimination. It can’t be B (4) because the base is even and the exponent is 2 or more. 4 = 2*2. It can’t be A (3) and D (6). We know that the sum is even and therefore divisible 2. The common factor of A and D is 3. If the answer is A, then it is also D. Since you can’t select both, it can’t be these two. There’s 4 choices. If it can’t be A,B, and D, then it has to be C.
The powers of 4’s last digits alternate between 6 (16, 256, 4096…) and 4 (64, 1024, 16,384) and even is 6. The 8th power is even so 4^8’s last digit is a 6.
26^3 for the last digit is always 6 (6, 36, 216), so the last digit of this term is a 6
6+6=12, so the sum of both terms will have a 2 as it’s last digit. Any number ending in 2 is not divisible by 5.
Without using mod arithmetic, my reasoning was:
1. All integer powers (greater than 1...) of an even number are divisible by 4. As the result is the sum of a power of 4 and a number divisible by 4, B is wrong.
2. If A is right, considering B is already proven wrong, D must also be right. So both must be wrong for the question be answerable.
3. This leaves C, assuming no errors from the examiner.
I think this kind of question is more about quick thinking and identifying and extrapolating the little bits of info given than formal knowledge.
I only thought about mod arithmetics when the explanation started!
Solution:
First off, you can rewrite the term to easily calculate the result without a calculator:
4⁸ + 26³
= (2²)⁸ + (2 * 13)³
= 2¹⁶ + 2³ * 13³
= 2³ * (2¹³ + 13³)
= 8 * (2³ * 2¹⁰ + 13 * 13²)
= 8 * (8 * 1024 + 13 * 169)
= 8 * (8192 + 2197)
= 8 * 10389
= 83112
You can immediately see, that it can't be divisible by 5, but let's check the others to be sure:
To be divisible by 3, the cross sum has to be divisible by 3: 8 + 3 + 1 + 1 + 2 = 15 → divisible by 3
To be divisible by 4, the last two digits have to be divisible by 4: 12 → divisible by 4
To be divisible by 6, it has to be even and divisible by 3: both are true → divisible by 6
Minimum calculations:
4⁸ ends in 6, because 4*4 is 16 and any number ending in 6 to any power ends in 6.
The other number already ends in 6.
So the final answer will end in a 2.
One way of doing this is elimination:
If the number is *not* divisible by three, then it's *not* divisible by six, giving two answers. Thus D can't be the correct answer.
The number *is* divisible by two (they're both even numbers), meaning if it's *not* divisible by six, it must *not* be divisible by three, giving two answers. Thus A can't be the correct answer.
The number is the sum of two even numbers to powers greater than one, thus both are divisible by four, meaning B is *not* the answer.
By elimination, C is the answer.
4⁸ + 26³ = 2¹⁶+ 2³13³ = 2³(2¹³ + 13³)
Apply Eugene's theorem to find the factors of 2¹³ + 13³ since the first base is equal to the second exponent, 4⁸ + 26³ is thus 2³3¹3463.
It is hence, obviously C.
Here's what I did (i know this CANNOT be a good explanation without the options 💀)
After rewriting 4 and 26, using exponent rules and factoring, we get:
(2^3)(2^13+13^3),
after which we can rule out option B, four.
When we take a closer look at option D (six), it is really just asking us whether the number is divisible by two and three. We know it is divisible by two, so three is the only check required. However option A (three) also asks whether the number is divisible by three. So there are two options asking whether it is divisible by three, which is redundant, so it's not any of them, leaving us with option C (five)
Yes, I know this is stupid.
The divisibility check for 5 requires only the last digit, 4 needs the last two, and 3 and 6 require all the digits, so I'll answer (C) 5 and move on to the next question.
This is easy. Every multiple of 5 ends in either 5 or 0. 4^8 ends in 6. 26^anything ends in 6. The sum ends in 2
I love your videos❤
5 facts
1) 3 is a divisor of 6, so it can't be A just from looking at the answers because then both A and D would be correct
2) 6ˣ always ends in a 6
3) 2⁸ is 256 so 4⁸ which is 2⁸×2⁸ also ends in a 6
Alternately if you don't know the powers of 2
2²=4 so 2²×2² = 4² = 16
And (4²)⁴ = 4⁸ which will end in a 6
4) when you add the 2 numbers the last digit will be 2 since 6+6 = 12
5) no multiple of 5 ends in 2
The answer is C and there is no need to know to the actual value of 4⁸ + 26³
Assuming only 1 right answer:
1) if dividible by 3 then also by 6 (it's obviously even), so these answers are out
2) 26^3 = (24 + 2)^3 which leaves a sum of multiples of 24 and 2^3 which is 8 and so is divisible by 4
3) 5 is the only answer left
Edit: Also 4^8 = 16^4 = (12 + 4)^4 which is a sum of multiples of 12 plus 4^4 which is 256, or if you don't know that it is also 4^4 = 16^2 = (12 + 4)^2 which is a sum of multiples of 12 plus 4^2 which is 16, and 16 + 8 = 24 is divisible by 6 and thus also 3, as is 256 + 8 = 264 (the 8 comes from point 2), it's the only part that is not divisible by 6).
Not something I learned, but I do know 2^10 is 1024, so the first part is 1024 x 64, and you can just take the ones digit of 4x4 + 6x6x6 = 2.
4⁸ is equal to 4×4×4×4×4×4×4×4 which in simpler terms, 16×16×16×16
When we multiply 6 by 6, then the last digit is still 6.
4⁸ is equal to ___6
26³ is equal to ____6
___6 + ____6, just add the last digits
= ____2
Therefore, answer is c. as it cannot be divisible by 5 because it doesn't end with either 0 or 5
This was cool!
Process of elimination: if D were correct, the number would be divisible by 3 and by 4, but not by 6 which is impossible. If A were correct the number would be divisible by 6, but not by 3 which is impossible. Both 4^8 and 26^3=2^3*13^3 are divisible by 4, so their sum is which eliminates B. This leaves only answer choice C.
This is very easy actually. Whenever you have high powers and want to know the units digit divide the power by 4 and the remainder is the new power you need to use. Here for example- 4^8 here 8 would be divided by 4 and leave remainder 0. (4th power of an even number is always 6 and of odd number except 5 is 1 so here in case of 0 it would be 6) So we get 6 as a units digit of 4. In 26^3, 6 always has of any power units digit 6.
So adding both of their units digit the answer is 12. So 12 is not divisible by 5. This is the first approach you should use always. Seems long but it's really easy and can be done in 2 seconds.
I've got an easiest way to prove that 4^8 + 26^3 isn't dividable by 5 :
1) 4^8 = 2^8 * 2^8 = 256 * 256, we don't care about the whole result, we just look at the last digit : 6 * 6 = 36, so the result of 4^8 ends with a 6.
2) Same goes for 26^3, we just look at the last digit : 6 * 6 * 6 ends with 6.
3) The sum of a number ending with 6 plus a number ending with a 6 ends with a 2. 2 is neither 0 nor 5, then it's not dividable by 5.
Three is really the only one you need to do mod calculations for. Any even number that is divisible by 3 must also be divisible by 6, and it's clear that 4⁸ + 26³ is going to be even, so once we know 3 is divisible, we know 6 is divisible. And any even number raised to the power of 2 or higher is divisible by 4 (since it must include 2²⁺ as part of its prime factors), so the sum of any two such numbers will also be divisible by 4.
Assume from the format of the question that only one answer is true.
Eliminate a) 3, because anything divisible by 6 will also be divisible by 3.
Eliminate d) 6, because anything divisible by both 3 and 4 will always be divisible by 6. (Technically, anything divisible by both 2 and 3, but 4 is divisible by 2, so...)
4^x will always be divisible by 4. 26 may not be divisible by 4, but if you double it, it is - so multiplying it by any even number will always make it divisible by 4. Eliminate b) 4
That which remains is the correct answer.
I need a bit of help with a formula (not related to this video). I can't remember the formula, and it's making me nuts. I am retired so I haven't used it in years.
As a nurse, if I have 0.9% Normal Saline in a bag for an IV and I have 5% Saline in a vial, what is the formula if I want to make 1.3% Saline IV solution? I used to do this ALL the time, and now I can't remember the formula. Any help out there? Thanks!!!
What I did is only keep the last digit, So 4^8=6, 6^3=6, and 6+6=2, immediately answer is 5.
Of course if the answer is not five, I can also eliminate it, and 3 can be eliminated also because 6, a multiple of 3, is also an option. So it would be either 4 or 6 if 5 is not the answer
An even number multiplied by an even number will always be even.
Therefore 4^n and 26^n will always be even.
An even number plus an even number will always be even.
To be divisible by 5 it must end with a 5, which is odd, which is impossible, therefore the answer is c) 5.
Is there any numerical method available to calculate gamma function of non analytical numbers like (1/3), (1/5) ?
26 single digit is always 6 not matter the power, the 4 is the tricky one because it alternated between 4 and 6 with 4 making the number divisible by 5, we can see that even power of 4 is 6 whilst odd is 4( i.e. 4^2 is 16, 4^4 is 256 etc.., 4^1 is 4, 4^3 is 64, 4^5 is 1024, etc...) therofore the single digit for 4^8 is 6 making for 6+6 which the single digit is 2 which is not divisiable by 5.
Even without modular arithmetic, if a sharp student know that the product of two even numbers is even, they can reason their way to the answer solely based on the fact that it is MC.
Assume exactly one option is correct.
We can immediately eliminate A. If it not divisible by 3, surely it is not divisible by 6.
The number is a sum of two evens and hence even. We eliminated A, so it an even number divisible by 3, hence divisible by 6.
Clearly 4^8 is divisible by 4. 26^3 = (2*13)(2*13)(2*13) so clearly it is divisible by 4.
C is the only remaining.
I was lazy and used process of elimination. Both terms are clearly divisible by 4, so (B) is out, and since the number is even, (A) and (D) are either both correct or both incorrect. Since there's only one right answer, it must be C.
I normally calculate the last digit first so that it can check for even or odd, divisible for 5 or not. After that then just do normal maths 😅
Thought 1: If only one answer is right, it can't be 3. Operating with definitely even numbers disallow 6 as well.
Intermediate thought: 4 to the power of some positive integer is divisible by 4. 26² is also divisible by 4, so 26³ has to, as well. Same for the sum.
Thought 2: 4 to the power of some positive integer is always one over a multiple of 3.
26 to the power of an odd positive integer is one under a multiple of 3. So the sum matches.
Even numbers divisble by 2 and 3 are also divisible by 6.
Remains 5.
They dont teach us basic arithmetic like this in NY b4 they move you on to a higher level math!
Haven't watched the video, but my way of proving it is:
For a number to be divisible by 5, it needs a 5 or a 0 as the one digits.
26^3 has a one digits of 6 since time itself is always a 6 at the one digits. For it to be 5 or a 0 as the one digits, 26^3 must be added with either a number with a 9 or 4 as the one digits.
However, 4^8 = 16^4, which again has a one digits of 6 with the same reason. Since 6+6 = 12, it is not divisible by 5.
I didn't know you were Red Pen Blue Pen
Ok so 4⁸ + 26³ is:
Modulo 3:
1⁸ + (-1)³ = 1 - 1 = 0 therefore divisble by 3
Modulo 4:
0⁸ + 2³ = 8 = 0 therefore divisible by 4
Modulo 5:
(-1)⁸ + 1³ = 1 + 1 = 2 therefore not divisible by 5.
Divisibility by 4 and 3 automatically give you 6.
What is the proof for exponentiation on both sides being possible in modular arithmetic?
Another way i cant see in these comments but might be here and i missed it whike scanning: 4^8=16^4, which must end in a 6 as any number ending in 6 raised to an integer power must also end in 6 (last digit is determined by multiplying the ones of both numbers, which will always be 6×6, which will always end in 6), 26^3 must also end in a 6 fir the same reason. Add both together and they must end in 2, so it can't be divisible by 5 😊
B) can be dispatched almost immediately, modular arithmetic unnecessary. 4^8 + 26^3 = (2^2)^8 + (2 * 13)^3 = (2^2)^8 + (2^2)(2 * 13^3). 2^2 = 4 and can be factored out, so the original expression must be divisible by 4. As for A) and D), modular arithmetic also unnecessary there as you have 4 raised to an even power and a base ending in 6 raised to an odd power. Both results will end in 6, the sum will end in a 2 making divisibility by 3 and 6 possible. So the original problem can't be divisible by 5.
I solved it by checking the prime factors. Both 4⁸ and 26³ have at least two 2s in their factorization, making them divisible by 4. The sum of two multiples of a number is still a multiple of that number. 6 and 3 can be used to eliminate each other. The result of the addition has to be even, and any even number that is divisible by 3 is also divisible by 6. Since only one answer can fail to divide the addition, neither 3 nor 6 can be correct. The only answer remaining is 5.
How come I never learnt this in school?
My school taught me to break the numbers and use binomial theorem and group numbers to find the remainder
For the UK, modular arithmetic is usually early university level.
@@darranrowe174 yeah
But this modular arithmetic method is much easier than using binomial theorems to break the numbers in multiples
@@For_watching_yt I would guess the reasons are two fold.
First, modular arithmetic has limited use earlier on. Secondly, modular systems are tougher to reason about.
I would guess that mathematics education is always going to be a trade off between what is useful and what is practical.
You did skip over a proof of why (q*d+r)^x is congruent with r^x (mod d) with all variables integers.
Proof: if you expand the sum to the power x via binomial expansion, every term but the last one of the expansion (= r^x) is divisible by d.
Another easy way to solve it is that the number is a multiple of 8 (since both addends are even numbers raised to at least the third power), therefore a multiple of 4, and therefore even (so if 3 is false 6 is also false, therefore both 3 and 6 must be true).
1) Any even number squared is a multiple of 4, so adding a power of 4 yields another multiple of 4. That rules out 4.
2) It's already even, so it's either divisible by 3 AND 6 or neither one. Therefore 3 and 6 must be wrong.
That leaves 5.
Is this the first video you've done with number theory? I can't recall one.
Not sure how to solve it completely mathematically, but with those 4 options you can get the answer quite easily
4 is a factor of both 4^8 and 26^3 (26 is divisible by 2), so B is out.
Assume the answer is A, the number is divisible by 2, so if A is the answer it must also be divisible by 6. Since there can only be one correct answer, both A and D are out.
Thus, the answer must be C (assuming there ia exactly one correct answer among the options)
i was about to say that there was only one wrong answer, but then i realised that it was addition and not multiplication
yeas easy with minimum calculations doing in your head in 30 - 60 seconds.
It takes more to read the next solution than it is to do in the head.
Last digit is important in each of the 2 numbers added.
Powers of 6 always end up in 6
Powers of 4 end up in 4 and 6, 4 and 6, etc.. so at the 8th power it will be a 6
So we will have a ...6 + ...6 that will end in a 2. Which is not divisible by 5.
all of these comments are pointing out how easy it is to see that 5 works but if you don’t assume it’s only one correct answer choice it is pretty nice to work out the more difficult ones
26³ is 2³*13³ so it is divisible by 4.If it was not divisible by 3 or would not be divisible by 6 either. Since they ask one answer that is not possible so it must be 5.
Could also do it by eliminating 4 as it goes into both for and 26 on thier own. Then eliminate 3 and 6 as for even numbers being divisible by 3 is the same thing as being divisible by 6. This leaves 5 as the only possible answer.
*wrote for when I meant 4
You can eliminate 6 right off the bat. Divisble by 3 and 4 implies divisible by6
Pleeease! Clip this Clip-Mic to your shirt! It is MADE for that, it sounds BETTER that way and it doesn't look so freakin embarassing :c
It all about the last number
We have 26³ the last number is 6 and about the 4⁸ we realise a pattern 4¹ Last diget is 4, 4² last digite is 6 4³ last digit is 4 and so on, so 4⁸ last number is 6 and 6+6 = 12 and that's divisible by 3, 4 and 6 but not 5
You don't actually know that it's divisible by 12 just because the last digits are 12. 112 is not divisible by 6, for example.
4^8 and 26^3 are both divisible by 4 therefore 4^8 + 26^3 is divisible by 4. We know that 3 or 6 must be true, but one implies the other, therefore they are both true. That leaves you with 5.
3 does not imply 6 but 6 implies 3. However, you can figure this out by just adding 4 and 26 to get 30 which is divisible by 3 and 6.
Yeah *Social Media* was a mistake alright
This all seems way more complicated than just doing the calculations as presented. 26 cubed isn't that bad.
Well, we can eliminate B almost immediately. The answer will most definitely be even. Any even number multiple by itself any number of times will always be divisible by 4. And adding numbers divisible by 4, will result in a number divisible by 4. Now, to get to the correct answer without using too much math, we have to look at the question more so than the equation. It is implied that there is only 1 correct answer. If something is divisible by 4, but not by 3, then it also cant be divisible by 6. The logic is even simpler by stating that if the answer isnt divisible 6, then it also isnt divisible by 3, therefore, A and D arent the right answers. That leads us to the conclusion that C, is the correct answer.
In order to be divisible by 5 the answer has to end in 5 or 0. These are not possible with those exponents.
Use binomial theorem
I just got the ans by looking at the options idk its the right approach tho
Last digit of 4^(2n) is always 6 and 6^x is also 6 hence the last digit will be 2 so it won't be divisible by 5
I do not understand how and why this thing with powers is working.
Here's much simpler solution:
1. We add even number to even number, so it will be even, therefore divisible by 2.
2. Number is divisible by 6 only if it is divisible by 2 and 3. So now either 3 and 6 are both correct, or both incorrect. Since we should have only one answer, both of those are not the answers.
3. Being left with 4 and 5 we really want to know the last digit of this sum. And it's quite easy to get, we just need to calculate the last digits of both parts of sum and add them.
4. 4^8 = 256^2, 6*6 = 36, so last digit is 6. 26^3 = 26*26*26. 6*6 = 36, we know that, and multiplying 36*6 last digit will also be 6, obviously. 6+6 = 12, so last digit of the sum will be 2. So the number is not divisible by 5.
Also you can calculate this thing pretty easy.
256*256 = 256/4*1000 + 256*4 + 256*2 = 64*1000 + 1024 + 512 = 64000 + 1536 = 65536
26*26 = 26/4*100 + 26 = 6.5*100 + 26 = 650 + 26 = 676
676*26 = 676/4*100 + 676 = 169*100 + 676 = 16900 + 676 = 17576
65536+17576 = 83112
Basically, the wild thing about modular arithmetic is that everything you think is true about numbers is still true - addition and multiplication works the exact samew ay.
Proof that addition works:
bm+k+bn+g mod b= b(m+n)+k+g (the k and g parts are the only parts you see)
Proof that multiplication works:
(bm+k)(bn+g)=bmbn+bmg+bnk+kg, but mod b, only kg is left. So both addition and multiplication work *exactly* the way you assume they would.
Since we use nothing but those relations here, we know that we can freely replace 4 and 26 with their easier versions.
4^8+26^3 mod 5 is the exact same as (-1)^8+1^3, because 26 is 1 more than a multiple of five, and 4 is one less than a multiple of five. So you can immediately see that (-1)^8=1 and 1^3=1, so even if you know *nothing* else about the number, you know that it is 2 more than a multiple of five. So it's not divisible by five.
@@XCC23 So, I believe this doesn't always work for division or if some of numbers are not integers?
But I guess, I got the point. It's sad that those things weren't explained in the video.
@@F1r1at Division is a little scary in modular arithmetic, because it's much easier to accidentally divide by 0 in there.
You can do it by multiplying by the multiplicative inverse, so mod 5, you can see, say, 4/3 mod 5 and go "well, that's the same as 4*n, where n is some number whose product with 3 is congruent to 1. In this case, it's 2, as 2*3=6 which is congruent to 1 mod 5.)
But sometimes this number doesn't exist - if you're looking at mod 4, for example, you can't divide by 2, because 2 can't multiply with anything to give 1. 2*1=2, 2*2=4 (so 0), 2*3=2 again.
And yeah. I think it's an elegant solution here that can be done with mental arithmetic (that doesn't involve metagaming the options, because that's obviously easier), but he does kind of assume that people will be comfortable with modular arithmetic, and a lot of people are not, unless they've been sniffing at some number theory.