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prime p

by maihuna » Sun Jan 18, 2009 10:02 am
If x, y, and z are integers greater than 1, and (327)(510)(z) = (58)(914)(x^y), then what is the value of x?

(1) y is prime

(2) x is prime
Last edited by maihuna on Mon Jan 19, 2009 9:26 am, edited 1 time in total.
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Source: — Data Sufficiency |
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by DanaJ » Sun Jan 18, 2009 10:25 am
327 = 300 + 27 = 3 (100 + 9) = 3 * 109. Since no divisor of 109 pops up fast, we'll just leave it like this.
510 = 51 * 10 = 17 * 3 * 10
so you get 327*510*z = 2*3^2*5*17*109*z
58 = 29 * 2
914 = 2 * 457. Now, 457 is not divisible by either 3 (sum of digits is 16), 17 or 109. Its prime factorization doesn't matter for the moment.
so you get 58*914*xy = 2^2*29*457*xy.
then you have:

2*3^2*5*17*109*z = 2^2*29*457*xy, which is the equivalent of 3^2*5*17*109*z = 2*29*457*xy. Now you've got a lot of prime numbers to choose from!
Statement 1 is insufficient because y can be either 3, 5 or 17.
Statement 2 is insufficient because x can also be either 3, 5 or 17.
Both statements taken together are insufficient because x and y can be any combination of the three.

I'm pretty sure there is an easier way around this one. And please let me know if this is the OA, because I'm not confident that E is indeed the right answer.
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by thechamp » Sun Jan 18, 2009 3:24 pm
We need the info on y and z to calculate x.

St 1: No info on z
St 2: No info on y & z

Together, still no info on z, E is the answer
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by maihuna » Mon Jan 19, 2009 9:28 am
DanaJ wrote:327 = 300 + 27 = 3 (100 + 9) = 3 * 109. Since no divisor of 109 pops up fast, we'll just leave it like this.
510 = 51 * 10 = 17 * 3 * 10
so you get 327*510*z = 2*3^2*5*17*109*z
58 = 29 * 2
914 = 2 * 457. Now, 457 is not divisible by either 3 (sum of digits is 16), 17 or 109. Its prime factorization doesn't matter for the moment.
so you get 58*914*xy = 2^2*29*457*xy.
then you have:

2*3^2*5*17*109*z = 2^2*29*457*xy, which is the equivalent of 3^2*5*17*109*z = 2*29*457*xy. Now you've got a lot of prime numbers to choose from!
Statement 1 is insufficient because y can be either 3, 5 or 17.
Statement 2 is insufficient because x can also be either 3, 5 or 17.
Both statements taken together are insufficient because x and y can be any combination of the three.

I'm pretty sure there is an easier way around this one. And please let me know if this is the OA, because I'm not confident that E is indeed the right answer.
Sorry DanaJ,
The xy was a typo, its actually x^y.
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