2016 is coming!
What is the remainder when
1 2 0 1 5 + 2 2 0 1 5 + ⋯ + 2 0 1 5 2 0 1 5
is divided by 2016?
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1 solution
Yes Same Way.
Nicely Explained!
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Nice problem!
I got the first part 1 ≤ k ≤ 1 0 0 7 . I don't get for 1 0 0 8 ≤ k ≤ 2 0 1 5 . Can you explain?
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The second part is included in first part. See 2016 - k
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@Dev Sharma – I see, I thought 2 0 1 6 − k ranges from 1 to 1007 only. Stupid me.
Sir will you please elaborate ? How did you find k , and k 2 0 1 5 + ( 2 0 1 6 − k ) 2 0 1 5 = 0 (mod 2016 ) @Otto Bretscher
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Just use modular arithmetic: k 2 0 1 5 + ( 2 0 1 6 − k ) 2 0 1 5 ≡ k 2 0 1 5 + ( − k ) 2 0 1 5 = k 2 0 1 5 − k 2 0 1 5 = 0 ( m o d 2 0 1 6 )
If you expand it, we get:
k 2 0 1 5 + ( 2 0 1 6 − k ) 2 0 1 5 = k 2 0 1 5 + 2 0 1 6 2 0 1 5 − ( 2 0 1 5 1 ) 2 0 1 6 2 0 1 4 k + ( 2 0 1 5 2 ) 2 0 1 6 2 0 1 3 k 2 − ( 2 0 1 5 3 ) 2 0 1 6 2 0 1 2 k 3 + . . . − ( 2 0 1 5 2 0 1 3 ) 2 0 1 6 2 k 2 0 1 3 + ( 2 0 1 5 2 0 1 4 ) 2 0 1 6 k 2 0 1 4 − k 2 0 1 5 = 2 0 1 6 2 0 1 5 − ( 2 0 1 5 1 ) 2 0 1 6 2 0 1 4 k + ( 2 0 1 5 2 ) 2 0 1 6 2 0 1 3 k 2 − ( 2 0 1 5 3 ) 2 0 1 6 2 0 1 2 k 3 + . . . − ( 2 0 1 5 2 0 1 3 ) 2 0 1 6 2 k 2 0 1 3 + ( 2 0 1 5 2 0 1 4 ) 2 0 1 6 k 2 0 1 4
We note that every term above has 2 0 1 6 as a factor.
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But is there any method to find k 2 0 1 5 + ( 2 0 1 6 − k ) 2 0 1 5
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@A Former Brilliant Member – Aren't You aware of the rule that a^n+b^n is always divisible by a+b if n is an odd positive integer. It can be easily proven using factor theorem.
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@Kushagra Sahni – Thank you very much,,, Try my latest problem...
Maybe you can help me better understand. I was thinking the sum of 1 to an odd number is odd. Similarly this sum should be odd too, right? So how is it 0 mod 2016? Thanks in advance!
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The sum contains an even number of odd terms, ( 2 k − 1 ) 2 0 1 5 for k = 1 . . . 1 0 0 8 , so that the sum will be even.
For 1 ≤ k ≤ 1 0 0 7 we have k 2 0 1 5 + ( 2 0 1 6 − k ) 2 0 1 5 ≡ 0 ( m o d 2 0 1 6 ) . Also 1 0 0 8 2 0 1 5 ≡ 0 ( m o d 2 0 1 6 ) since 2 0 1 6 ∣ 1 0 0 8 2 . Thus the remainder is 0