Someone help me figure this one out please.
The minimum case is 1 + 2 + 3 + … + n (because any other combination will be greater than this).
Now, knowing that
1 + 2 + 3 + ... + n = \frac{1}{2} n (n + 1)
all you need to do is find the largest value of n such that
\frac{1}{2} n (n + 1) < 100
and you’re done.
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Hi, I would do this using Greg’s BD strategy.
But first, N = # of distinct positive integers that have the sum of less than 100. We want the greatest possible value of N which means we will have to start at the least possible value and increasing by the smallest amount possible. Since only positive integers allowed, the first number will be 1 and the second will be 1+1, 1+2, 1+3… which is an arithmetic progression. Hence we can use the equation Sum of N = n/2 * (first term + last term)
BD strategy:
Plug in B - 11/2 * 12 = 66 – the value is too small
Plug in D - 13/2 * 14 = 91
From N=13 to N=14 we add + 14, 91 + 14 is greater than 100. Answer D.