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MGMAT CAT Question Help! EXPERT Advice Needed!

Problem Solving — algebra and arithmetic
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Junior | Next Rank: 30 Posts
Posts: 17
Joined: Tue Aug 03, 2010 12:13 pm
GMAT Score:660
by Purusha » Fri Aug 13, 2010 5:31 pm
Hello,

I got this question on my last MGMAT CAT that I took:


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

The answer is D.

I reviewed their answer a few times and kind of understand it but I would like to know how to go about setting such a problem up, figuring out which values to use, and then solving them all in TWO MINUTES!!

Please explain how one would solve this problem thoroughly.

Thanks.
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Source: — Quantitative Reasoning |
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by roh00kan » Fri Aug 13, 2010 8:30 pm
I approached the problem like this:
Max length of an integer is possible if the number has max numbers of 2 as prime factors.

In this case, x + 3y is less than 1000. That means, max length of x is possible if x has the highest possible powers of 2 (but less than 1000).
So the value of x can go up to 2^9 = 512 (= 2*2*2*2*2*2*2*2*2).

Now let's find out the max value 3y, such that x + 3y is less than 1000. In other words, 3y should be less than 486 (1000-512). Again, in order to get the max length of the integer y, find an integer which has max number of prime factors of 2.

2^7 (128= 2*2*2*2*2*2*2). Multiply this with 3 will give a value of 384, and x + 3y = 512 + 384 = 898.
So in short we can write, 512 = 2^9 and 384 = 3*2^7

Maximum possible sum of the length of x and the length of y is: 9 + 7 = 16

I took less than 2 minutes to get the idea of the problem. Then, under 40 secs I found out the prime factors. I know powers of 2 up to 2^10 and powers of 3 up to 3^6 and these info helped me to solve this problem.

If there is any short cuts, pls explain.

Thank you!
Ren
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by sgmuthukumar » Tue Sep 21, 2010 9:00 am
roh00kan wrote:I approached the problem like this:
Max length of an integer is possible if the number has max numbers of 2 as prime factors.

In this case, x + 3y is less than 1000. That means, max length of x is possible if x has the highest possible powers of 2 (but less than 1000).
So the value of x can go up to 2^9 = 512 (= 2*2*2*2*2*2*2*2*2).

Now let's find out the max value 3y, such that x + 3y is less than 1000. In other words, 3y should be less than 486 (1000-512). Again, in order to get the max length of the integer y, find an integer which has max number of prime factors of 2.

2^7 (128= 2*2*2*2*2*2*2). Multiply this with 3 will give a value of 384, and x + 3y = 512 + 384 = 898.
So in short we can write, 512 = 2^9 and 384 = 3*2^7

Maximum possible sum of the length of x and the length of y is: 9 + 7 = 16

I took less than 2 minutes to get the idea of the problem. Then, under 40 secs I found out the prime factors. I know powers of 2 up to 2^10 and powers of 3 up to 3^6 and these info helped me to solve this problem.

If there is any short cuts, pls explain.

Thank you!
Ren
This looks to me the best way to solve the problem... Cheers
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