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Number Theory Level 3

Among all pairs of positive integers that sum to $2n,$ the one that gives the greatest product is $(n, n),$ with $n\times n = n^{2}.$ Now, if the sum were $2n+1$ instead, there may be multiple ways of getting a higher product than $n^{2}.$

For example, if $n=8,$ then $2n=16$ and we have $8^2=64.$ However, $2n+1=17$ for $n=8$ and we have $9\times 8=72,\quad 10 \times 7=70,\quad 11\times 6=66,$ that are all greater than $64.$

What is the minimum $n$ for which there are $2018$ distinct ways of getting a greater product than $n^{2}$ with integers that add to $2n+1?$

The answer is 4070307.

1 solution

Jeremy Galvagni
Apr 25, 2018

We seek each of the following $2018$ products to be greater than $n^{2}$

$n(n+1) > n^{2}$

$(n-1)(n+2) > n^{2}$

$(n-2)(n+3) > n^{2}$

...

$(n-2017)(n+2018) > n^{2}$

Of these products, the last of these is smallest. Solving:

$n^{2} + n - 4070306 > n^{2}$

$n > 4070306$

Therefore the minimum $n=\boxed{4070307}$

There's a typo in your solution (3070307 should read 4070307)

David Vreken - 3 years, 1 month ago

Fixed it. Thanks!

Jeremy Galvagni - 3 years, 1 month ago

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The answer is 4070307.

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