j_shreyans 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

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 -- since 2 is the least possible prime factor -- without exceeding the threshold of 1000.
(A fuller explanation of this line of reasoning appears below.)

Let x = 2� = 512.
Length of x = 9.

Substituting x = 512 into x+3y < 1000, we get:
512 + 3y < 1000
3y < 488
y < 163 (approx).
Thus, y = 2� = 128.
Length of y = 7.

Total length = 9+7 = 16.

The correct answer is D.

Here's why the prime-factorization of x+3y must include as many 2's as possible:

Let's say the constraint is that x < 40.
Thus, the maximum possible value of x is 39.

If the prime-factorization of x is composed solely of 2's, then the greatest possible value of x is 32:
2*2*2*2*2 = 32.
No more prime factors can be included without exceeding the threshold of 39.
Here, since x is composed of 5 prime factors, the length of x is 5.

If the prime-factorization of x is composed solely of 5's, then the greatest possible value of x is 25:
5*5 = 25.
No more prime factors can be included without exceeding the threshold of 39.
Here, since x is composed of 2 prime factors, the length of x is 2.

To MAXIMIZE the length of x, we must MINIMIZE the value of each prime factor.
Thus, we want the prime-factorization of x to include as many 2's as possible.

x max = 2^9 = 512 (because 2^10 = 1024 > 1000)
1000-512=488
So y max < 488/3 = 162.66666...
y = 2^7 = 128 (because 2^8 = 256 > 162.66666...)

Adding indices yields: 9+7 = 16