Some 2016 Bash!
Let n be any odd positive integer. Then the expression
1 n + 2 n + ⋯ + 2 0 1 6 n
is divisible by which of the following numbers?
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1 solution
What's the remainder when you divide by 2016?
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We can do the same thing right? For n > 1 we have
1 n + 2 n + ⋯ + 2 0 1 5 n + 2 0 1 6 n ( m o d 2 0 1 6 ) ≡ 1 n + 2 n + 3 n + ⋯ 1 0 0 8 n ⋯ + ( − 3 ) n + ( − 2 ) n + ( − 1 ) n + 0 n ( m o d 2 0 1 6 ) ≡ 1 0 0 8 n ≡ 0 ( m o d 2 0 1 6 )
Thus the sum is divisible by 2016 too!
Yes sir it is divisible by 2016 too (my bad!!) but initially i was looking for making negative pairs and with 2017 it seemed easy and eventually i came up with a result that 2017 is a divisor of above summation without looking 2016.
I think the question should ask which odd integer is divisible by the above summation..
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Quite right. But note that when n = 1 the expression is not divisible by 2 0 1 6 .
1 0 0 8 n ≡ 0 ( m o d 2 0 1 6 ) ∀ n ≥ 2 ≡ 1 0 0 8 ( m o d 2 0 1 6 ) when n = 1
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@Sravanth C. – Yes it is for n >= 2 and i think question should be changed to which positive odd integer is a divisor of above summation
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@Ankit Nigam – That will not work either: Since the sum is divisible by 2016, it is also divisible by the odd integers 3, 7, 9, etc., all the odd factors of 2016 (and many more).
What about my idea of making n any odd positive integer, allowing n = 1 ?
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@Otto Bretscher – Yes sir i think it will work because if we put n = 1 sum will be equal to 1 0 0 8 ⋅ 2 0 1 7 and 1 0 0 8 ⋅ 2 0 1 7 ( m o d 2 0 1 6 ) ≡ 1 0 0 8 ( m o d 2 0 1 6 ) .
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@Ankit Nigam – Exactly! Maybe you can add a remark to your solution explaining why the sum fails to be divisible by 2015, 2016, and 2018 for all odd n .
Great solution, +1!
Bonus Question: What will happen if n is even?
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I would expect 48 to be the largest divisor that works for all even n , but things get a bit messy
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Sir Could you please explain Why 48?
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@Mr Yovan – Just a hunch... somebody should work it out!
1 n + 2 n + ⋯ + 2 0 1 5 n + 2 0 1 6 n ( m o d 2 0 1 7 ) ≡ 1 n + 2 n + 3 n + ⋯ ( − 3 ) n + ( − 2 ) n + ( − 1 ) n ( m o d 2 0 1 7 ) ≡ 0 ( m o d 2 0 1 7 ) I generated negative pairs and since n is odd all pairs will be equal to 0 .
∴ 1 n + 2 n + ⋯ + 2 0 1 5 n + 2 0 1 6 n is divisible by 2 0 1 7