MGMAT qurestion a tough one..pls advice a nice approach

alltimeacheiver 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?

ans choices

5
6
15
16
18

Hi!

Since length is related to the number of prime factors, and since we want to maximize the length, we want to use the smallest primes possible. So, let's make all the primes involved 2.

So, x and y will both be powers of 2. Let's let:

x = 2^a

and

y = 2^b

so that our condition becomes:

2^a + 3(2^b) < 1000

Now let's just write out the powers of 2 that are less than 1000:

2, 4, 8, 16, 32, 64, 128, 256, 512

Finally, let's play with possible values of x and y to max out the length:

If x=512, then the max possible value for y is 128 (since 3*128=384, giving us a sum less than 1000).

x = 2^9 and y = 2^7, giving us a length of 16, the second largest answer choice.

The only way to get a length of 18 (the largest choice) would be if each of x and y were 512; since that will violate our limitation, we can eliminate 18.

Choose (D)!