Math Olympiad Question | A Trick Everyone Should Know!

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

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.

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!!!

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😄.

This type question cames in jee mains. In advanced more tough.

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?.

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

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

Please solve IMO shortlist thanks)

Thanks

UR the best 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.

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.

Hi Ellie, what olympiad is this question from? thx!

It is also can be solved by binomial theorem

what tablet and software do you use for these videos?

Hi ELLIE can u solve ISI bstat entrance exam paper's UGB part ? That is tougher than Jee advance of INDIA❤❤❤

Binomial