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- Thread starter Wilmer
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- Jan 26, 2012

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You have the problem: given \(m\) solve:p = product of 2 consecutive integers n-1 and n.

s = sum of m consecutive integers, the first being n+1.

s = p

Example (n = 12, m = 8):

p = 11 * 12 = 132

s = 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 = 132

If m = 4,541,160 then what's n ?

\( n \times m+ \frac{m(m+1)}{2}=n^2-n\)

or:

\( n^2 -(m+1)n-\frac{m(m+1)}{2}=0\)

and you want the positive root of this.

This does involve arithmetic with nice long integers by Dr Wolfram's Alpha can handle it

CB

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