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Sum of exponents in the prime breakup of s

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Sum of exponents in the prime breakup of s

by gmattesttaker2 » Wed Feb 19, 2014 3:27 pm
Hello,

Can you please tell me how to solve this:

Function f(s) equals the sum of the exponents in the prime break-up of s.
For example: f(120) = 5, since 120 = (23)(31)(51), and 3 + 1 + 1 = 5. If
f(m) = 4 and f(n) = 5, what is the value of f(mn)?

(A) 4
(B) 5
(C) 9
(D) 18
(E) 20

OA: C


Thanks,
Sri
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by Matt@VeritasPrep » Wed Feb 19, 2014 5:13 pm
Let's start with one of the functions to get a sense of what's happening.

If f(m) = 4, then m is either x�, x³ * y, or x² * y², where x and y are distinct prime factors of m.

(For instance, f(16) = 4, since 16 = 2�, f(24) = 4, since 24 = 2³ * 3, etc.)

In any case, as we see, the sum is 4. f(n) is similar, but with partitions of 5 instead of 4 (e.g. x³ * y², etc.) ("Partitions" here means "ways you can add positive integers to obtain a sum of 5".)

When we multiply the two numbers together, our possibilities include

m * n = x� * (x³ * y²)
or
m * n = (x² * y²) * (x� * y)

Notice that in these cases (and any others) the sum is 9, so you're set!

The easiest way to do this on the test is probably to pick small numbers, such as m = 16 and n = 32, that fit the functions, then to see what your result is. Since the question implies the answer will be the same for ALL integers m and n, any two you pick should suffice.
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by [email protected] » Thu Feb 20, 2014 12:25 am
Hi Sri,

What is the source of this question? I ask because this question essentially asks you to "prime factor" a number AND consider duplicates, which is something that GMAT usually does NOT ask for.

Since f(m) has 4 prime factors (including duplicates) and f(n) has 5 prime factors (including duplicates), the f(mn) must have 9 prime factors (including duplicates); the 4 from the first number and the 5 from the second number. This doesn't match the "design" of GMAT questions that cover this concept, so you're not gong to end up seeing this on Test Day.

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by GMATGuruNY » Thu Feb 20, 2014 4:25 am
The portion in red reflects the intent of the problem.
gmattesttaker2 wrote: Function f(s) equals the sum of the exponents in the prime break-up of s.
For example: f(120) = 5, since 120 = (2³)(3¹)(5¹), and 3 + 1 + 1 = 5. If
f(m) = 4 and f(n) = 5, what is the value of f(mn)?

(A) 4
(B) 5
(C) 9
(D) 18
(E) 20

OA: C
f(x) = the sum of the exponents when x is prime-factored.
To illustrate:
If x = 2³3�, then f(2³3�) = 3+4 = 7.
If x = 5²11�, then f(5²11�) = 2+8 = 10.
But there is no requirement that the prime-factorization of x be composed of more than one exponent.
If the prime-factorization of x is composed of only ONE exponent, then that ONE exponent is also the SUM of the exponents:
If x = 2³, then f(2³) = 3.

Since f(m) = 4, let m = 2�.
Since f(n) = 5, let n = 2�.
Result:
f(mn) = f(2�2�) = f(2�) = 9.

The correct answer is C.
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