Length of an integer - MGMAT CAT 3

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Length of an integer - MGMAT CAT 3

by josh80 » Mon Dec 09, 2013 4:37 pm
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?


5

6

15

16

18
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by Brent@GMATPrepNow » Mon Dec 09, 2013 5:02 pm
josh80 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
Notice that powers of 2 will maximize the length of an integer.
For example, 16 (aka 2^4) has length 4, whereas 21 (which is greater than 16) has a length of only 2.

Let's recall a few of the higher powers of 2.
2^7 = 128 (length is 7)
2^8 = 256 (length is 8)
2^9 = 512 (length is 9)

At this point, we might fiddle with some possible values of x and y and test them.
If x = 512 and y = 128, then x + 3y < 1000
Here, the length of x is 9, and the length of y is 7.
So, the sum of the length of x and the length of y = 9 + 7 = 16

So, the correct answer might be D, or it might be E

The correct answer can't be E. ,
We know this because 2^10 is greater than 1000, which means neither x nor y can be 2^10 or greater.
Also, 2^9 = 512, so x and y cannot both equal 2^9 (since x + 3y would be GREATER THAN 1000)
These two facts rule out the possibility that the lengths of x and y add to 18.

This means the correct answer must be D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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