Prep Math

[BG]

If x and y are positive integers, what is the value of xy ?

(1) The greatest common factor of x and y is 10.

(2) The least common multiple of x and y is 180.

Answer: c

What is the best way to solve this kind of problem- greatest common factor, least common multiple which are common in GMAT? trying real numbers one by one is really time-consuming, but is easy to get the wrong results?

[Saurav]

If X and Y are two integers, and M is the GCD of X and Y, and N is the LCM of X and Y, then

X * Y = M * N

Test this for a certain values. Hence using 1 and 2 you can solve the problem.

[BG]

Thanks

[RonPurewal]

If X and Y are two integers, and M is the GCD of X and Y, and N is the LCM of X and Y, then

X * Y = M * N

Test this for a certain values. Hence using 1 and 2 you can solve the problem.

yes. well done.

i posted this here.
if i were you, i would know this fact, but i would avoid expressing it in the form promulgated above; i really, really, really don't want to have to probe into the meanings of FOUR different variables in order to understand a relatively simple (albeit obscure) relationship.

[RonPurewal]

you can also work this problem out explicitly, because there are only two numbers.

remember that the gcf contains the least powers of the common prime factors, and the lcm contains the greatest powers of all the prime factors in either factorization.

the gcf is 2 x 5.
the lcm is 2^2 x 3^3 x 5.

looking at the individual prime numbers gives the following results:
* either x or y contains 2^2; the other contains just 2 (i.e., 2^1).
* either x or y contains 3^2; the other contains no 3's at all.
* each of x and y contains 5 (i.e., 5^1).

in the first two cases, we don't know which of x and y contains which power, but we don't care because we're multiplying both quantities together anyway.
therefore, the product xy is (2^2)(2^1) x (3^3) x (5^1)(5^1).
sufficient.
no need to calculate.

[RonPurewal]

you can also work this problem out explicitly, because there are only two numbers.

remember that the gcf contains the least powers of the common prime factors, and the lcm contains the greatest powers of all the prime factors in either factorization.

the gcf is 2 x 5.
the lcm is 2^2 x 3^3 x 5.

looking at the individual prime numbers gives the following results:
* either x or y contains 2^2; the other contains just 2 (i.e., 2^1).
* either x or y contains 3^2; the other contains no 3's at all.
* each of x and y contains 5 (i.e., 5^1).

in the first two cases, we don't know which of x and y contains which power, but we don't care because we're multiplying both quantities together anyway.
therefore, the product xy is (2^2)(2^1) x (3^3) x (5^1)(5^1).
sufficient.
no need to calculate.

also note that, were there three numbers in the problem, you could no longer calculate their product.

for instance, if the gcf of x, y, and z contains 2^2 and their lcm contains 2^4, then you'd know at least one of them contains 2^2 and at least one of them contains 2^4. however, the third number could contain either 2^2, 2^3, or 2^4.
this ambiguity doesn't occur with only two numbers, because the gcf and the lcm give the 'floor' and the 'ceiling', respectively, on the powers of each prime. with more than two numbers, you can have values between the floor and the ceiling.

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my my, i do love this thread today, don't i.