Math Olympiad Question | A Trick Everyone Should Know!
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- čas přidán 23. 01. 2024
- Another beautiful Math Olympiad Question! Did you manage to solve it?
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I also just noticed that after 625 the 5^n series swtiches between ending in 125 and 625 so if you have a 5^n/1000 where n>=4 the remainder is 625 if n is even and 125 if n is odd
🤯
This is a cakewalk for a JEE aspirant (The toughest engineering enterance exam of India). We learn this remainder calculation concept as an application of binomial theorem. We further learn to apply this concept in finding the last two or three or even four digits of an ugly looking number with huge powers.
Make the number (25)¹⁰⁰²/100
Reduce it to (25)¹⁰⁰¹/4
Now, (24+1)¹⁰⁰¹/4
Now it is gonna look like an expanded binomial series and the compressed form is gonna look like : (24k + 1)/4 (where k is an integer, a very large whole number to be precise)
We are left with 1/4. Now everybody is gonna claim the remainder to be 1 but wait, we were asked to find the remainder when we divide the number by 100.
Multiply the number with 25/25, it looks like 25/100.
Boom! 25 is the answer!!!
You can also use modular arithmetic.
5 mod 100 = 5
5^2 mod 100 = 25
5^3 mod 100 = 25
5^4 mod 100 = 25
I think this is an easier way to see and prove that 5^n mod 100 = 25 for n > 1.
5^n/100 has a remainder of 25 for n≥2, it's easy to prove with induction.
Right now I am thinking about you Ellie and you posted this video😄.
Never thought about it but it makes so much sense that the 5^n series match the decimal series 2^-n because it's equivalent to 0.5^n. More of an artifact of the base 10 number system than anything else but for me an easy trick to remember both more easily
Can you pick-up some coordinate geometry questions for nxt video?.
This type question cames in jee mains. In advanced more tough.
Same
She already solved jee advanced 2016 paper
Here's how I got 1, what did I do wrong?
We have 100=5^2*4, so 5^2004/5^2*4=5^2002/4. Since, as you said, 5^2002 ends in 25, simply take 5^2002-1, which ends in 24 and is therefore divisible by 4, and then the remainder is 1, since we removed it at the previous step
11/2 has 1 as the remainder and 22/4 has 2 as the remainder although they're equal fractions.
Here, we multiply both the numerator and the denominator by 2 to go from 11/2 to 22/4, and the remainder goes from 1 to 2 (remainder also became twice)
So in the question when we go from (5^2004)/100 to (5^2002)/4, since we decreased both the numerator and denominator 25 times, the remainder (25) got reduced to its 1/25th, which is 1 (25×(1/25)).
Thank you so much!
@@Curious_george_3x1 I'm glad I could be of help :)
Please solve IMO shortlist thanks)
5 to the power of anything would give us 25 in unit and tens place so by dividing by 100 would give us 25 as remainder
Thanks
Hello Ellie. I have a really,really urgent question that I need to ask you. I’ve recently been required to read a book on harmonic analysis on finite groups by Silberstein. However, I’m still a high school student. I know it’ll take a miracle , but I’ll take my chances, so as of now I’m not sure what specific math topics do I really need to know in order to learn, and read the book on harmonic analysis on finite groups. Please tell me what math topics I need to know.
Hi Ellie, what olympiad is this question from? thx!
UR the best thanks)
what tablet and software do you use for these videos?
It is also can be solved by binomial theorem
Please explain how to do that?
This is just a 2 line question by using binomial theorem
Hi ELLIE can u solve ISI bstat entrance exam paper's UGB part ?
That is tougher than Jee advance of INDIA❤❤❤
I agree on this
Binomial
JEE aspirant🫂
@@mr.unknown1070bruh exam are coming 😂
Use binomial therem
❤
You can also use the identity: (a+b)(a-b) = a²-b².
Process:
5²⁰⁰⁴ ÷ 100
=(5²⁰⁰² × 25) ÷ (25 × 4)
= 5²⁰⁰² ÷ 4
As per identity (a+b)(a-b) = a² - b²
(5²⁰⁰² - 1²) should be divisible by (5-1) which is 4.
As the given number in question is one more than the number divisible by four.
Therefore, Remainder = 1
And I am from class 8.
Please pin this comment.
This easy
Hello ma'am...I'm vimal from India...good approach for teaching,ma'am I'm reading with you for jee advanced mathematics.if any other Chanel plz tell me....
ITNA BHEE MAT FEKO BHAI . ADVANCED KA TAIYARI AISE HOTI HAI ??😂😂😂😂
Hello hello ....ma'am learn math for Cambridge University...,this question is demo for advertisement....
@@itsinvalid07 Lagta hai Vimal ne Vimal khaya hai.....ladki ko impress karne ka naya tarika
Nice, but this must be one hell of a simple Olympiad if it's asking something this trivial...
I guess it's interesting to note the phenomenon, even if it's an inherently base 10 thing. That said, considering all even bases makes this much more interesting. Specifically, the powers of k in base 2k yield interesting patterns:
For bases that are multiples of 4, the powers of k behave a little too nicely e.g powers of 2 in base 4 are 2, 10, 20, 100, 200, 1000 etc
The others are bases 2, 6, 10, 14 etc.
Every other one of these, bases 2, 10, 18 etc exhibits the behaviour in the video: the residues mod "100" stay the same. This is because the "k" in this case happens to be 2n + 1 for an *even* n, I think. It relates to multiples of k in the base 2k being k, 10, 1k, 20 etc in their base k representation : the even ones are nice! and k^2 = "nk" in base 2k (just amounts to some algebra).
Then k^3 = "k*n0 + k^2" = "k*n*10 + nk". Here the k*n gives an extra factor of "10" when n is even, making "nk" still the remainder (adding "100" doesn't affect the "10"s digit)
e.g. in base 18, 9^2 = 81 = 4 * 18 + 9 = "49", then 9 * 49 = "9 * 40 + 9 * 9" = "(9*4)*10 + 49" = "200 + 49" = "249", and you always get 49 or "nk" as the remainder.
However, in bases 6, 14, 22 etc, the "n" we mentioned is odd, so while k^2 = "nk" still holds, k^3 = k*n*10 + k^2 = ("uk")*10 + "nk" = "uk0" + "nk" = "u, k+n, k"
Here the "10"s digit becomes even, since its k+n, where both are odd. From here, dropping the u digit (since "100"s aren't relevant) and multiplying by k again gives us "nk" as the remainder again, since the even "10"s digit "k+n" has the same effect as before.
e.g. in base 14, 7^2 = "nk" = "37", then 7^3 = "7*30 +7^2" = "(7*3)*10 + 37" = "170" + "37" = 1A7 (compare "u, k+n, k", note "A" is even, and 10 in base 10)
then dropping the 1 we have A7, then multiplying by 7 gives (7*A)*10 + 7^2 = 500 + 37 = 537 i.e. the even digit A gave us the multiple of "100", returning us to 37.
Long story short, the phenomenon in the video is specific to bases 2, 10, 18 etc so I guess the bases that are congruent to 2 mod 8.
In bases congruent to 0 or 4 mod 8, the phenomenon is even simpler.
In bases congruent to 6 mod 8, you get an alternating pattern "nk" -> "k+n, k" -> "nk" etc.
Other bases are odd so you can't really consider the "half" element.
Sorry for the rant, I found this kinda interesting so I figured I'd follow through - I'm sure I've overcomplicated it somehow haha.
You complicated it too much 😂 but nice
Solve (2023)^2023 / 35 .......ANS: 7
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