Find the number n

Number Theory Level 3

What is the maximum positive integer n n such that 2013 can be written as a sum of n n consecutive positive integers?


The answer is 61.

Vijay Simha by Vijay Simha , 22, USA

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1 solution

k + ( k + 1 ) + ⋯ + ( k + ( n − 1 ) ) = n k + ( n − 1 ) n 2 = 2013 ⟹ n ( 2 k + n − 1 ) = 2.3.11.61 k+(k+1)+\dots + (k+(n-1)) = nk+\frac{(n-1)n}{2}=2013 \implies n(2k+n-1)=2.3.11.61

knowing that n ≤ 2 k + n − 1 n\leq 2k+n-1 , for k ≥ 1 k\geq 1 , we need to put 2.3.11.61 2.3.11.61 in two groups such that they have the least difference. One can offer ( n , 2 k + n − 1 ) = ( 61 , 66 ) (n,2k+n-1)=(61,66) . This cannot be beaten and n = 61 n=61 is the answer.

2 Helpful 1 Interesting 2 Brilliant 0 Confused

The source is PRMO 2013 right?????

Mohammed Imran - 1 year, 2 months ago

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