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Number Theory; Difficulty: Hard

Expert replies
by GMATGuruNY » Tue Jul 26, 2016 2:43 am
Alphonsaj wrote:In how many ways can 480 be expressed as a product of two positive integers that are relatively prime (co-prime) to each other?

A) 24
B) 12
C) 4
D) 8
E) 2
Co-primes are integers that have no factors in common other than 1.
The prompt should make clear that xy and yx are to be considered the same product.

480 = (2�)(3)(5).

From the prime-factorization above, we can form the following factor pairs that have a product of 480 and are co-primes:
(1)(2�3*5)
(2�)(3*5)
(3)(2�5)
(5)(2�3).
Total ways = 4.

The correct answer is C.
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Source: — Problem Solving |

by OptimusPrep » Tue Jul 26, 2016 7:03 pm
Alphonsaj wrote:In how many ways can 480 be expressed as a product of two positive integers that are relatively prime (co-prime) to each other?

A) 24
B) 12
C) 4
D) 8
E) 2


Correct answer:
C
Co-prime numbers: If two numbers have just a factor "1" in common, then the numbers are called co-prime.

480 = 2^5*3*5
The following will be the co-prime factors:
1 and 2^5*3*5
3 and 2^5*5
5 and 2^5*3
2^5 and 3*5
Total 4 ways.

These are the only possible ways.

Correct Option: C
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by ceilidh.erickson » Fri Jul 29, 2016 5:59 am
Alphonsaj wrote:In how many ways can 480 be expressed as a product of two positive integers that are relatively prime (co-prime) to each other?

A) 24
B) 12
C) 4
D) 8
E) 2


Correct answer:
C
What is the source of this question? I have never seen an official question that asks about co-primes. The GMAT would almost certainly never ask about such a specific mathematical definition without also defining it.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
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by Matt@VeritasPrep » Thu Aug 04, 2016 9:03 pm
480 =

2 * 2 * 2 * 2 * 2 * 3 * 5

If a * b = 480, and a and b are coprime, then all the factors of 2 must be contained in a, or all the factors of 2 must be contained in b. Since they have to go together, we could simplify the problem by pretending that we only have one 2, or 2 * 3 * 5. Then, considering this, we have

1 * (2 * 3 * 5)
or
2 * (3 * 5)
or
3 * (2 * 5)
or
5 * (2 * 3)

So there are four pairs, and we're set!
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