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anonymous
 4 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.
anonymous
 4 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.

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BTaylor
 4 years ago
Best ResponseYou've already chosen the best response.0f(124^24,99) = 37 f(173^24,99) = 37 37  37 = 0

BTaylor
 4 years ago
Best ResponseYou've already chosen the best response.0wolframalpha gives that as a mixed fraction.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I changed the problem statement to avoid the obvious electronic aided solution(s) :)

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I think we have to search for a pattern here, and we must have to use a calculator here to some extent.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^ calculator is only as smart as the idiot pushing the buttons ^_^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does it have anything to do with the fact that 124^24mod99 = 25^24mod99 = 31^12 mod 99 and so on so forth?

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.2is 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:\[\begin{align} 173&=2*9925\\ \therefore173^{24}&=(2*9925)^{24}\\ \therefore173^{24}99&=(2*9925)^{24}99=25^{24}99 \end{align}\]therefore:\[124^{24}99173^{24}99=25^{24}9925^{24}99=0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's how I did it @asnaseer. Well done!

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1How 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 \(74\equiv25\pmod{99}\), 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.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1I should have written "173" in that first line.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's seems like a valid approach.
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