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maximum possible sum

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by GmatKiss » Sat Oct 22, 2011 1:59 am
For any integer k > 1, the term "length of an integer" refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

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Source: — Problem Solving |

by shankar.ashwin » Sat Oct 22, 2011 2:11 am
x+3y<1000

2^9 + 3(2^7) < 1000

9+7 = 16. D IMO
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by GMATGuruNY » Sat Oct 22, 2011 2:20 am
GmatKiss wrote:For any integer k > 1, the term "length of an integer" refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

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Length = the number of prime factors.
Thus, MAXIMUM length = the MAXIMUM number of prime factors.
To maximize the number of prime factors of x+3y, we must include as many 2's as possible without exceeding the threshold of 1000.

Let x = 2^9 = 512.
Length of x = 9.

3y < 1000-512
y < 163 (approx)
Thus, y = 2^7 = 128.
Length of y = 7.

Total length = 9+7 = 16.

The correct answer is D.
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