Sum of n n Consecutive Positive Integers Divided by n n and ( n + 1 ) (n+1)

Number Theory Level pending
  • Case-A : 1 1 is divisible by 1 \color{#20A900} 1 , but not \color{#D61F06} \text{not} by 2 = ( 1 + 1 ) 2=({\color{#20A900} 1}+1) .

  • Case-B : 2 + 3 2+3 is neither \color{#D61F06} \text{neither} divisible by 2 \color{#20A900} 2 , nor \color{#D61F06} \text{nor} by 3 = ( 2 + 1 ) 3=({\color{#20A900} 2}+1) .

  • Case-C : 3 + 4 + 5 3+4+5 is divisible by both 3 \color{#20A900} 3 and 4 = ( 3 + 1 ) 4=({\color{#20A900} 3}+1) .

Case-A describes a representative sum of n n consecutive positive integers that is divisible by n n , but not \color{#D61F06} \text{not} by ( n + 1 ) (n+1) .

Case-B describes a representative sum of n n consecutive positive integers that is neither \color{#D61F06} \text{neither} divisible by n n , nor \color{#D61F06} \text{nor} by ( n + 1 ) (n+1) .

Case-C describes a representative sum of n n consecutive positive integers that is divisible by both n n and ( n + 1 ) (n+1) .

Now, Let S n S_n be a sequence of n n consecutive positive integers and p p be its least element. Furthermore, the sum of integers in S n S_n is not \color{#D61F06} \text{not} divisible by n n , but divisible by ( n + 1 ) (n+1) . Also, let F ( n ) F(n) denote the number of such sequence(s) S n S_n with p 2017 p \leq 2017 .

Find F ( 2 ) + F ( 31 ) F(2)+F(31) .


The answer is 673.

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1 solution

Patrick Corn
Dec 19, 2017

If n n is odd, then F ( n ) = 0 , F(n)=0, because any sum of n n consecutive positive integers will be divisible by n n : indeed, a sum of n n consecutive positive integers equals n n times the middle one.

So F ( 31 ) = 0. F(31) = 0. Now a sum of two consecutive integers p , p + 1 p,p+1 is 2 p + 1. 2p+1. This is never divisible by 2 , 2, and is divisible by 3 3 if and only if p 1 p \equiv 1 mod 3. 3. So we want to count the number of p p in the sequence 1 , 4 , 7 , , 2017. 1,4,7,\ldots,2017. There are exactly 2017 + 2 3 = 673 \frac{2017+2}3 = 673 elements in this sequence.

So F ( 2 ) + F ( 31 ) = 673 + 0 = 673 . F(2) + F(31) = 673+0 = \fbox{673}.

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