## FoolForMath 3 years ago Fool's problem of the day, If $$f(x,y)$$ is a function that gives the remainder when $$x$$ divided by $$y$$. Prove that the value of $$f(124^{24},99) - f(173^{24},99) =0$$. Problems credits: IMS india.

1. BTaylor

f(124^24,99) = 37 f(173^24,99) = 37 37 - 37 = 0

2. BTaylor

wolframalpha gives that as a mixed fraction.

3. FoolForMath

I changed the problem statement to avoid the obvious electronic aided solution(s) :)

4. BTaylor

arggh...

5. ParthKohli

I think we have to search for a pattern here, and we must have to use a calculator here to some extent.

6. harharf

^ calculator is only as smart as the idiot pushing the buttons ^_^

7. m_charron2

Does it have anything to do with the fact that 124^24mod99 = 25^24mod99 = 31^12 mod 99 and so on so forth?

8. asnaseer

is this a valid method of solving this?\begin{align} 124&=99+25\\ \therefore124^{24}&=(99+25)^{24}\\ \therefore124^{24}|99&=(99+25)^{24}|99=25^{24}|99\\ \end{align}similarly:\begin{align} 173&=2*99-25\\ \therefore173^{24}&=(2*99-25)^{24}\\ \therefore173^{24}|99&=(2*99-25)^{24}|99=25^{24}|99 \end{align}therefore:$124^{24}|99-173^{24}|99=25^{24}|99-25^{24}|99=0$

9. FoolForMath

That's how I did it @asnaseer. Well done!

10. KingGeorge

How about... Notice that this function is the same as the mod function We want to show $124^{24}\equiv174^{24}\pmod{99}$Reduce, and get that this is equivalent to asking $25^{24}\equiv74^{24}\pmod{99}$Since $$74\equiv-25\pmod{99}$$, we can simplify to $25^{24}\overset{?}\equiv(-25)^{24}\pmod{99}$Since we have an even exponent, this is clearly true. This is basically the same thing asnaseer did.

11. KingGeorge

I should have written "173" in that first line.

12. FoolForMath

That's seems like a valid approach.