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Source: — Data Sufficiency |
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by GmatMathPro » Tue Oct 25, 2011 9:48 am
What is the gcd of a, b, and 3a+23b?

1) a=4. plug this in to get 4, b, 12+23b; If b=1, the gcd(4,1,35)=1. If b=4, the gcd(4,4,12+23(4))=4. INSUFFICIENT.

2) The greatest common divisor of a, b, and 23a+3b is 4. This implies that gcd(a,b)=4. If there were some divisor greater than 4, that a and b had in common, then it would also divide 23a+3b. But this would contradict the fact that gcd(a,b,23a+3b)=4. Hence, gcd(a,b)=4.

Back to the original question, the fact that gcd(a,b)=4 implies that 4 is a common divisor of a, b, and 3a+23b. But it also must be the GREATEST common divisor of a, b, 3a+23b, because if there were some common divisor greater than 4, it would contradict the fact that gcd(a,b)=4. SUFFICIENT

Ans: B
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by vaibhavgupta » Sat Oct 29, 2011 4:02 pm
GmatMathPro wrote:What is the gcd of a, b, and 3a+23b?

1) a=4. plug this in to get 4, b, 12+23b; If b=1, the gcd(4,1,35)=1. If b=4, the gcd(4,4,12+23(4))=4. INSUFFICIENT.

2) The greatest common divisor of a, b, and 23a+3b is 4. This implies that gcd(a,b)=4. If there were some divisor greater than 4, that a and b had in common, then it would also divide 23a+3b. But this would contradict the fact that gcd(a,b,23a+3b)=4. Hence, gcd(a,b)=4.

Back to the original question, the fact that gcd(a,b)=4 implies that 4 is a common divisor of a, b, and 3a+23b. But it also must be the GREATEST common divisor of a, b, 3a+23b, because if there were some common divisor greater than 4, it would contradict the fact that gcd(a,b)=4. SUFFICIENT

Ans: B
What if 3a+23b is not a factor of 4.. ?
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by GmatMathPro » Sat Oct 29, 2011 4:22 pm
3a+23b is not a factor of 4, it's a multiple of 4. And this is because from statement 2, you know that a and b are both multiples of 4. That is a=4x for some integer x, and b=4y for some integer y. So, 3a+23b=3(4x)+23(4y)=4(3x+23y)

In words, if 4 is a factor of BOTH a and b, then it must be a factor of all linear combinations of a and b because we can definitely factor out a 4 from every term.
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by vaibhavgupta » Sat Oct 29, 2011 10:49 pm
GmatMathPro wrote:3a+23b is not a factor of 4, it's a multiple of 4. And this is because from statement 2, you know that a and b are both multiples of 4. That is a=4x for some integer x, and b=4y for some integer y. So, 3a+23b=3(4x)+23(4y)=4(3x+23y)

In words, if 4 is a factor of BOTH a and b, then it must be a factor of all linear combinations of a and b because we can definitely factor out a 4 from every term.
Now, i get it! Thanks! :)
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