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Man Cat 3 #29-Lengthy Problem

Expert replies
by joyseychow » Sat May 16, 2009 6:55 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?

(A)5
(B)6
(C)15
(D)16
(E)18

Any shortcut to this? Thanks!
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Source: — Problem Solving |

Re: Man Cat 3 #29-Lengthy Problem

by dumb.doofus » Sat May 16, 2009 7:34 am
joyseychow 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?

(A)5
(B)6
(C)15
(D)16
(E)18

Any shortcut to this? Thanks!
I think the answer is D

I approached this problem with the following things in mind:
1. The length of any number would be highest if the number is made up of the first couple of prime numbers.. i.e. 2, 3 or 5..
2. We need to mostly focus on 2 here..

With the above in mind.. 2^9 = 512.. so let x = 512..

x + 3y < 1000 => y < 163 (approx).. Focusing more on even numbers as we want to increase the count of 2, we know 2^7 = 128

So the maximum possible length can be 9 +7 = 16..

this is option D
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by joyseychow » Thu May 21, 2009 7:10 pm
Thanks!
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