Find all pairs of positive integers (a, n) such that n ≥ 2 and

a + (a+1) + (a+2) + ... + (a+n-1) = 100.

noobieatmath Jan 17, 2019

#2**+1 **

18 + 19 + 20 + 21 + 22 =

18 + 22 + 19 + 21 + 20 =

40 + 40 + 20 =

100

So ( a, n) = (18, 5)

CPhill Jan 17, 2019

#5

#6**+3 **

We can write:

a + (a+1) + (a+2) + ... + (a+n-1) = na + (n-1)n/2

Hence we have: na + (n-1)n/2 = 100, which we can rewite as: n^{2} + (2a-1)n - 200 = 0

Solve this for n in terms of a: \(n=\frac{\sqrt{4a^2-4a+801}-2a+1}{2}\) (the other solution results in negative values for n).

Let a run from 1 to 100 to find the only integer solutions are: (a,n) = (9,8) and (18,5) as found by Chris and Guest.

(There is also a = 100 and n = 1, but the question requires n>=2).

(Alternatively, solve for a in terms of n and run through n = 2 to 100, and look for integer solutions for a)

Alan
Jan 18, 2019