Case-A : is divisible by , but by .
Case-B : is divisible by , by .
Case-C : is divisible by both and .
Case-A describes a representative sum of consecutive positive integers that is divisible by , but by .
Case-B describes a representative sum of consecutive positive integers that is divisible by , by .
Case-C describes a representative sum of consecutive positive integers that is divisible by both and .
Now, Let be a sequence of consecutive positive integers and be its least element. Furthermore, the sum of integers in is divisible by , but divisible by . Also, let denote the number of such sequence(s) with .
Find .
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If n is odd, then F ( n ) = 0 , because any sum of n consecutive positive integers will be divisible by n : indeed, a sum of n consecutive positive integers equals n times the middle one.
So F ( 3 1 ) = 0 . Now a sum of two consecutive integers p , p + 1 is 2 p + 1 . This is never divisible by 2 , and is divisible by 3 if and only if p ≡ 1 mod 3 . So we want to count the number of p in the sequence 1 , 4 , 7 , … , 2 0 1 7 . There are exactly 3 2 0 1 7 + 2 = 6 7 3 elements in this sequence.
So F ( 2 ) + F ( 3 1 ) = 6 7 3 + 0 = 6 7 3 .