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Difficulty:
45%
(medium)
Question Stats:
70% (01:48) correct
30%
(02:00)
wrong
based on 317
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What is the remainder when the difference of \(3191^{2020}\) and \(3159^{2020}\) is divided by 16 ?
A. 0
B. 3
C. 4
D. 6
E. 7
A. 0
B. 3
C. 4
D. 6
E. 7
19 Nov 2022, 20:21
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Afroditee
IanStewart follows a very apt logical approach.
But if you are looking for a quick answer, then remember these rules/facts/deductions
\(a^n-b^n\) is always divisible by a-b
So, \(3191^{2020}\) - \(3159^{2020}\) is divisible by 3191-3159 or 32, and so by 16 also.
Thus, remainder is 0.
Other points that one could remember are
\(a^n-b^n\) is divisible by a+b if n is even
\(a^n+b^n\) is divisible by a+b if n is even
A
General Discussion
12 Nov 2022, 10:38
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The numbers don't really matter here; all that matters is that the bases are odd, and the exponents are multiples of 4. So we really just have odd^4 - odd^4 = (odd^2 + odd^2)(odd^2 - odd^2) = (odd^2 + odd^2)(odd + odd)(odd - odd), and all three of those factors are even, so their product will be a multiple of 8. A multiple of 8 can only have a remainder of 0 or 8 when you divide by 16, and 0 is the only possibility among the choices.
It's not hard to prove that odd^2 - odd^2 is actually always a multiple of 8 (which is why the answer is zero here), but we don't need to bother doing that for this question with these answer choices.
It's not hard to prove that odd^2 - odd^2 is actually always a multiple of 8 (which is why the answer is zero here), but we don't need to bother doing that for this question with these answer choices.
