Smooth

Algebra Level 5

We define a positive integer $n$ to be smooth if there exist a representation of real numbers $a_1,a_2,\ldots,a_{2n}$ such that $2n = a_1+a_2+\ldots+a_{2n} = a_1 \times a_2 \times \ldots \times a_{2n}$ . Find the sum of first 2015 smallest smooth number.

The answer is 2033135.

1 solution

Daniel Branscombe
Jul 30, 2015

smallest possible n is 1,

if n=1 we need to solve

x+y=2

xy=2

x(2-x)=2

2x-x^2=2

x^2-2x+2=0

discriminant is 4-8=-4

thus there are no real solutions

If n>1 then let

a1=2 a2=n and all remaining terms equal 1

then their sum is

2+n+(n-2) = 2n

and their product is

2*n*(1)^(n-2) = 2n

thus for any n>1 there exists a solutions

thus our sum is simply

2*(1+2+...+2016)-2

2*2016*2017/2-2

2016*2017-2

4066270

You logic of possible n is totally correct. However, from the problem statement, we are supposed to find the sum of first 2015 n, not the first 2015 2n. So the answer should be 2+3+...+2016 = 2033135, half the value of stated answer.

Wei Chen - 4 years, 9 months ago

Smooth

The answer is 2033135.

1 solution

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