Not all 1

Number Theory Level 3

Out of 130 positive integers which satisfy

n 1 + n 2 + n 3 + … + n 130 > n 1 × n 2 × n 3 × … × n 130 , n_1 + n_2 + n_3 + \ldots + n_{130} > n_1 \times n_2 \times n_3 \times \ldots \times n_{130},

what is the minimum number of integers which are equal to 1?

123 122 125 129

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Calvin Lin by Calvin Lin

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1 solution

I think there are 'typo' word on the question because the answer for minimum number greater than 1 is 1.

I use all n that greater than 1 are 2, because we want to search the maximum integers greater than 1 so we must minimize the geometric.

The equation can be write as...

(2 n)+(130-n)>(2^n) (1^n)

as we can see the biggest for n is 7. And our answer 130-7=123 (where i think the right question is "Find the minimum number of one")

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Sorry for the confusion, I have updated the phrasing accordingly. I was debating between both versions, and ended up with the wrong question :(

Calvin Lin Staff - 6 years, 11 months ago

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