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Manhattan GMAT challenge question

by gabriel » Mon Oct 29, 2007 8:15 am
Hi people, have been out of action for some days due to my exams, am done with them and so am ready to discuss some good GMAT questions. Here is one for everyone ..

It is this weeks Manhattan GMAT challenge question

q.) If the prime factorization of the integer q can be expressed as a^2x*b^x*c^(3x-1), where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?

(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer

Regards

PS: - I have no idea what the answer is ..
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by gabriel » Fri Nov 02, 2007 6:58 am
hmm ... no attempts so far ... come on guyz give this one a shot ..
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by samirpandeyit62 » Fri Nov 02, 2007 7:44 am
If the prime factorization of the integer q can be expressed as a^2x*b^x*c^(3x-1), where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?

so total factors will be

(2x +1) (x+1) (3x)

for odd value of x, X+1 will be even hence the product is even

for even value of x 3x will be even hence the product will be even

so we can say that the total nos of factors is an even nos

(A) 3j + 4, where j is a positive integer

this will be odd for odd vals & even for even vals of j

(B) 5k + 5, where k is a positive integer

again even -odd -even...
(C) 6l + 2, where l is a positive integer

always even
(D) 9m + 7, where m is a positive integer

even - odd -even.....
(E) 10n + 1, where n is a positive intege

always odd


so IMO it should be C 6l + 2
Regards
Samir
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by meetcarey » Fri Nov 02, 2007 1:29 pm
Picking numbers... a = 2, b =3, c=5 & x = 1,

q = 2 ^2 * 3^1 * 5^2
factors are : 1, 2, 3, 4, 5, 6, 10, 12, 25, 50, 75, 100, 150 and 300. (14 factors)

C) 14 can be expressed as 6l+2 where l =2 (positive integer). C) is the right answer.
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by samirpandeyit62 » Sat Nov 03, 2007 9:43 pm
Hi Guest,
if a nos n = a^x*b^y*c^z

where a,b,c are prime factors of n & x,y,z is some power of these factors starting from 1

then total nos of factors of the nos is given by

(x+1)(y+1)(z+1) i.e product of power of each prime + 1

this is a mathematical rule.
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Samir
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by beeparoo » Sun Nov 04, 2007 11:32 am
Samir,

Thanks very much for the explanation. Does the mathematical rule that you speak of (prime factors and their exponents) bear a name or title?

I'd like to google it to read about it further. Currently, when i search certain keywords, I get heaps of info on unwanted stuff.

Cheers!
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by gabriel » Sun Nov 04, 2007 1:21 pm
beeparoo wrote:Samir,

Thanks very much for the explanation. Does the mathematical rule that you speak of (prime factors and their exponents) bear a name or title?

I'd like to google it to read about it further. Currently, when i search certain keywords, I get heaps of info on unwanted stuff.

Cheers!
I don't think there is a lot to read about it .. it is basically a formula that tells you the number of factors any integer has ( it is derived using basic rules of permutation and combination ) ... the formula Samir mentioned should suffice .. but if you are still curious try searching for " Prime factorization of a number" or " Number of divisors (factors) of a given number" ... If you do find anything please post the link .. and if u don't find anything and still have doubts please do post on this thread.. I am sure a simple example will clear up all the doubts ..

Regards
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Re: Manhattan GMAT challenge question

by Ian Stewart » Thu Jul 31, 2008 5:10 am
gabriel wrote: It is this weeks Manhattan GMAT challenge question

q.) If the prime factorization of the integer q can be expressed as a^2x*b^x*c^(3x-1), where a, b, c, and x are distinct positive integers, which of the following could be the total number of factors of q?

(A) 3j + 4, where j is a positive integer
(B) 5k + 5, where k is a positive integer
(C) 6l + 2, where l is a positive integer
(D) 9m + 7, where m is a positive integer
(E) 10n + 1, where n is a positive integer
I was just trawling through the archives, and saw that this problem was never correctly solved. The question intends that a, b and c are distinct primes (though their phrasing is a bit awkward), and as is pointed out above, we can count the number of factors of q by adding one to each exponent and multiplying:

q has (2x + 1)*(x + 1)*(3x) divisors.

Now looking at this product:
-either x or x+1 must be even (they're consecutive): the product must be even, so the number of divisors of q is even.
-the number of divisors of q must also be divisible by 3 (there's a 3 in the product)

Of the answer choices:
-A will never be divisible by 3 (the remainder will always be 1)
-C will never be divisible by 3 (the remainder will always be 2)
-D will never be divisible by 3 (the remainder will always be 1)
-E will always be odd

B is the only answer choice that might be divisible by 2 and 3.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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