If positive integer x is a multiple of 6 and positive integer y is a multiple of 14,is xy a multiple of 105?
1.x is a multiple of 9.
2.y is a multiple of 25.
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105 = 3*5*7
Statement 1) x is a multiple of 9. But we know that x is a multiple of 6 and y a multiple of 14.
So, xy has got 2,3^2,7 for sure but may or may have 5. if x = 90, then xy is a multiple of 105, but if x = 54, then xy is not a multiple of 105.
Not Sufficient!!!
Statement 2) y is a multiple of 25. But we know that x is a multiple of 6 and y a multiple of 14.
So, xy has got 2,3,5^2,7 i.e. all multiples of 105. Sufficient!!!
Answer B.
Statement 1) x is a multiple of 9. But we know that x is a multiple of 6 and y a multiple of 14.
So, xy has got 2,3^2,7 for sure but may or may have 5. if x = 90, then xy is a multiple of 105, but if x = 54, then xy is not a multiple of 105.
Not Sufficient!!!
Statement 2) y is a multiple of 25. But we know that x is a multiple of 6 and y a multiple of 14.
So, xy has got 2,3,5^2,7 i.e. all multiples of 105. Sufficient!!!
Answer B.
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Target question: Is xy a multiple of 105?theachiever wrote:If positive integer x is a multiple of 6 and positive integer y is a multiple of 14,is xy a multiple of 105?
1.x is a multiple of 9.
2.y is a multiple of 25.
Important stuff:
First, If N is a multiple of k, then N is divisible by k.
Second, a lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7
Since 105 = (3)(5)(7), then we can rewrite the target question as . . .
Rephrased target question: Is there a 3, a 5 and a 7 hiding in the prime factorization of xy?
Given: x is a multiple of 6
In other words, x = (2)(3)(other possible prime numbers)
Given: y is a multiple of 14
In other words,y = (2)(7)(other possible prime numbers)
Combine both of the above to see that xy = (2)(2)(3)(7)(other possible prime numbers)
So, the given information tells us that we ALREADY have a 3 and a 7 hiding in the prime factorization of xy. The only piece missing is the 5.
So, we can rephrase our target question one last time. . .
Rephrased target question: Is there a 5 hiding in the prime factorization of xy?
Statement 1: x is a multiple of 9.
Since 9 = (3)(3), all this tells us is that there are two 3's hiding in the prime factorization of xy.
So, there may or may not be a 5 hiding in the prime factorization of xy.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 1: y is a multiple of 25.
Since 25 = (5)(5), this tells us is that there is definitely a 5 hiding in the prime factorization of xy.
Since we can answer the target question with certainty, statement 2 is SUFFICIENT
Answer = B
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Brent
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If x=6a and y=14b, from the problem statement, we have is 84ab a multiple of 105? or is ab a multiple of 5?theachiever wrote:If positive integer x is a multiple of 6 and positive integer y is a multiple of 14,is xy a multiple of 105?
1.x is a multiple of 9.
2.y is a multiple of 25.
I does not help. II does.
B is the right choice.