Question

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.

Answer 1

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

Answer 2

wolframalpha gives that as a mixed fraction.

Answer 3

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

Answer 4

arggh...

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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:\[124^{24}|99-173^{24}|99=25^{24}|99-25^{24}|99=0\]

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

That's seems like a valid approach.