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What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is

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Joined: 24 Jun 2008
Posts: 1346

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12 Aug 2008, 09:55
x2suresh wrote:
IanStewart wrote:
arjtryarjtry wrote:
What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is divided by 58?
57
1
30
0
28

If you know modular arithmetic:

13^7 + 14^7 + 15^7 + 16^7 is clearly even --> divisible by 2

If it's also divisible by 29, it's divisible by 58.

Notice
16 ~ -13 mod 29
15 ~ -14 mod 29

(~ means 'congruent to')
13^7 + 14^7 + 15^7 + 16^7 ~ 13^7 + 14^7 + (-14)^7 + (-13)^7 mod 29
~ 0 mod 29

So the expression is divisible by 2 and 29, and thus by 58.

I couldn't understand the highlighted part.

Could you elaborate

I was using modular arithmetic, which isn't needed on the GMAT. I can attempt a brief summary, but this would normally be a chapter in a book. When you learn modular arithmetic, you learn in which arithmetic situations you can replace one number with another number that gives the same remainder. Here, we are dividing by 29; the remainder you get when you divide 16 by 29 is the same as the remainder you get when you divide 45 by 29, or 74 by 29, or -13 by 29 (adding or subtracting multiples of 29 from any number will not affect the remainder you get when you divide by 29). We say that "16 is congruent to 45 mod 29", which means "16 has the same remainder as 45 when divided by 29". This ensures that 16^7 will have the same remainder as 45^7 when you divide either by 29 (a fact not hard to prove). 16^7 will also have the same remainder as (-13)^7 when divided by 29, and that's what I used in the solution quoted above. Indeed, 16^7 has the same remainder as (-13)^7, 15^7 has the same remainder as (-14)^7, and 13^7 + 14^7 + 15^7 + 16^7 must have the same remainder as 13^7 + 14^7 + (-14)^7 + (-13)^7 when divided by 29. Clearly 13^7 + 14^7 + (-14)^7 + (-13)^7 is equal to zero; thus 13^7 + 14^7 + 15^7 + 16^7 is divisible by 29.
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12 Aug 2008, 10:37
If a^n + b^n + c^n + d^n is divisible by a + b + c + d as long as n is odd, then in our case, wouldn't the remainder be 0?
Re: funny remainder...   [#permalink] 12 Aug 2008, 10:37

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What will be the remainder when 13^7 + 14^7 + 15^7 + 16^7 is

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