Question 1175017: Find a formula for the nth triangular number Sn = 1 + 2 + 3 + · · · + n.
For what values of n is Sn: i: divisible by 5, ii: even?
Found 2 solutions by ikleyn, math_helper:
Answer by ikleyn(52787)   (Show Source):
Answer by math_helper(2461)  (Show Source):
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One way to get a formula is to add 1+2+...+n to itself, but write the 2nd line in reverse, then add them:
Sn = 1 + 2 + 3 + ... + (n-1) + n
Sn = ( n + (n-1) + (n-2) + ... + 2 + 1 )
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2Sn = (n+1) + (n+1) + (n+1) + ... + (n+1) + (n+1)
The RHS is just n terms of n+1, thus it is n(n+1)
2Sn = n(n+1)
(i)
As a hint for divisibility by 5:
If n is even, n+1 is odd
If n is odd, n+1 is even
Thus, a factor of 2 is always present in n(n+1), and the (1/2) in the  formula will divide that out ("cancels" it).
So there is just the consideration about n or n+1 being divisible by 5. Numbers ending with 5 or 0 meet this condition, and as long as EITHER n or n+1 ends with 5 or 0 you will find to be divisible by 5:
Examples:
n=10 >>> n*(n+1)/2 = 110/2 = 55, that works.
n=9 >>> n*(n+1)/2 = 9*10/2 = 45, that is divisible by 5
n=8 >>> n*(n+1)/2 = 8*9/2 = 36, NOT divisible by 5 (neither n nor n+1 is divisble by 5)
n=15 >>> n*(n+1)/2 = 15*16/2 = 120, divisible by 5
n=14 >>> n*(n+1)/2 = 14*15/2 = 105, divisible by 5
A subtle note about the above: notice how the factor of 5 comes from 'n' or 'n+1' in a mutually exclusive way. That is, if n has a factor of 5, n+1 does not; if n+1 has a factor of 5, n does not. You may need this observation for part(ii) when considering the factor 4.
(ii)
is even: you give it some thought. Remember, you always get one factor of 2 automatically in the product n(n+1) and that will cancel with the 1/2 of the formula, so the question is, for what values of n(n+1) do you get n(n+1) divisible by 4?
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