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30 Aug 2009, 11:20
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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OA Please explain
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OA Please explain
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30 Aug 2009, 11:42
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GCM of 16 and n is 4
=> n can have at max only 2 powers of 2 because if n is 8 or 16 ..GCM wil not be 4..
GCM of 45 and n is 3
=> n can have only 1 power of 3 and no power of 5 ..
Now, 210 = 2 * 3 * 5 * 7
and, n = 2 * 2 * 3 * k wkere k is any integer..
k can not be 2 or 3 or 5 as explained above.. or k can not be a multiple of 2,3 or 5.
We have 6 as a common factor in 210 and n without knowing the value of k
so, GCM should be a multiple of 6..
=> 3, 14 and 70 are out..
if GCM is 30 then k should be a multiple of 5 which is not possible..hence, c is out..
only 42 is left and hence, the answer..
if k = 7 GCM of 210 and n will be 42..
=> n can have at max only 2 powers of 2 because if n is 8 or 16 ..GCM wil not be 4..
GCM of 45 and n is 3
=> n can have only 1 power of 3 and no power of 5 ..
Now, 210 = 2 * 3 * 5 * 7
and, n = 2 * 2 * 3 * k wkere k is any integer..
k can not be 2 or 3 or 5 as explained above.. or k can not be a multiple of 2,3 or 5.
We have 6 as a common factor in 210 and n without knowing the value of k
so, GCM should be a multiple of 6..
=> 3, 14 and 70 are out..
if GCM is 30 then k should be a multiple of 5 which is not possible..hence, c is out..
only 42 is left and hence, the answer..
if k = 7 GCM of 210 and n will be 42..
30 Aug 2009, 11:47
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1st piece of info tells us that n has 2^2 in prime factorization.
1st piece of info tells us that n has 3^1 * 5^0 in its prime factorization.
n = 2 * 2 * 3 * 5^0 * ...
210 = 2 * 3 * 5 * 7
1. GCF = 2 * 3 * ... = multiple of 6
2. GCF cannot be multiple of 5 because n is not divisible by 5
only (D) satisfies both requirements
1st piece of info tells us that n has 3^1 * 5^0 in its prime factorization.
n = 2 * 2 * 3 * 5^0 * ...
210 = 2 * 3 * 5 * 7
1. GCF = 2 * 3 * ... = multiple of 6
2. GCF cannot be multiple of 5 because n is not divisible by 5
only (D) satisfies both requirements
