A factorization question

Let N N be a positive integer ending in 9 such that N < 50000 N < 50000 .

Let a a and b b be positive integers such that a × b = N a \times b = N and a > b.

Let c c and d d be positive integers such that c × d = 2 N c \times d = 2N and c > d c > d .

Is it possible to find a value of N N such that it meets the specifications of a b < 1.25 \dfrac ab < 1.25 and c d < 1.02 \dfrac cd < 1.02 ?

No Yes

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1 solution

Denton Young
Feb 24, 2016

Let N = 43659

a = 231, b = 189

c= 297, d = 294

How did you get these numbers?

Pi Han Goh - 5 years, 3 months ago

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N = 9 x 11 x 21 x 21

a= 9 x 21

b = 11 x 21

2N = 2 x 9 x 11 x 21 x 21

c= 9 x 11 x 3

d= 2 x 3 x 7 x 7

Denton Young - 5 years, 3 months ago

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I mean, how did you find these number to begin with?

Pi Han Goh - 5 years, 3 months ago

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@Pi Han Goh Worked them out in my head. You can't use even numbers or a multiple of 5 as any of the factors of N, which restricts the possibilities. In order to have 2N also be factored into two close numbers, your factorization of N has to be one where you can shift by a factor of close to 2. 81/49 came to mind, so that means you need a factor of 3 and a factor of 7 in both a and b. The rest was simple.

Denton Young - 5 years, 3 months ago

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