MHB Find "n" Given p=s and m=4,541,160 | Integer Solution

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The discussion centers on finding the integer n given p as the product of two consecutive integers (n-1 and n) and s as the sum of m consecutive integers starting from (n+1). For m = 4,541,160, the equation derived is n^2 - (m+1)n - (m(m+1)/2) = 0, which can be solved using the quadratic formula. The positive root calculated is n = 6,203,341, while a negative solution of n = -1,662,180 is also valid under certain conditions. The example provided illustrates how these calculations work with smaller values of n and m. The thread emphasizes the arithmetic involved in solving for n using large integers.
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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 ?
 
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Wilmer said:
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 ?

You have the problem: given \(m\) solve:

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

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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Agree. n = [m + 1 +- SQRT(3m^2 + 4m + 1)] / 2

So n = 6,203,341 or -1,662,180

The negative solution is also valid; using m = 8, then n = -3:
-4 * -3 = 12
-2 + -1 + 0 + 1 + 2 + 3 + 4 + 5 = 12
 
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