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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.

Mathematics
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f(124^24,99) = 37 f(173^24,99) = 37 37 - 37 = 0
wolframalpha gives that as a mixed fraction.
I changed the problem statement to avoid the obvious electronic aided solution(s) :)

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Other answers:

arggh...
I think we have to search for a pattern here, and we must have to use a calculator here to some extent.
^ calculator is only as smart as the idiot pushing the buttons ^_^
Does it have anything to do with the fact that 124^24mod99 = 25^24mod99 = 31^12 mod 99 and so on so forth?
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:\[124^{24}|99-173^{24}|99=25^{24}|99-25^{24}|99=0\]
That's how I did it @asnaseer. Well done!
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.
I should have written "173" in that first line.
That's seems like a valid approach.

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