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Difficult Math Question #29 - Arithmetic

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by rajs.kumar » Wed Oct 18, 2006 1:09 am
Answer is E.

It is the only number that has a prime factor greater than 13.

99 - 11 x 3 x 3

84 - 4 x 3 x 7

72 - 3 x 3 x 2 x 4

65 - 13 x 5

57 - 51 x 1
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Re: Difficult Math Question #29 - Arithmetic

by gmat_enthus » Wed Oct 18, 2006 5:27 pm
800guy wrote:If P represents the product of the first 15 positive integers, then P is not a multiple of:

a) 99 b) 84 c) 72 d) 65 e) 57
P just represent 15! .

a = 11*9 both are present in 15!
b = 12*7 same as above
c= 12*6 same
d = 13*5 same
e = 19*3, 3 is present in 15! but 19 doesnt. Hence P is not a multiple of 57.

Ans E
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Re: Difficult Math Question #29 - Arithmetic

by anandsebastin » Wed Oct 18, 2006 6:24 pm
800guy wrote:If P represents the product of the first 15 positive integers, then P is not a multiple of:

a) 99 b) 84 c) 72 d) 65 e) 57
E

Divide the options into multiples of 1st 15 positive integers

a. 11*9
b. 12*7
c. 9*8
d. 13*5
e. odd man out
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here's the OA

by 800guy » Fri Oct 20, 2006 10:36 am
OA:

If P represents the product of the first 15 integers, P would consist of the prime numbers that are below 15.

2,3,5,7,11,13

Any value that has a prime higher than 13 would not be a value of P.

57 = 3, 19

19 is a prime greater than 13, so the answer is E.
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by fr743 » Tue Jun 07, 2011 8:44 am
Somebody, help me understand this question.
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by Frankenstein » Tue Jun 07, 2011 9:06 am
fr743 wrote:Somebody, help me understand this question.
Hi,
P = 1*2*3*...15 = 15!
For a number not to be factor of P, it should a prime number that is greater than 15 as a factor.
a)99 = 3.3.11 (all <15)
b)84 = 2.2.3.7 (all <15)
c)72 = 2^3.3^2 (all <15)
d)65 = 5.13 (all <15)
e)57 = 3.19 (19(prime) > 15).
None of the numbers in P is a multiple of 19.

Hence, E
Cheers!

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by fr743 » Tue Jun 07, 2011 10:17 am
Thanks.
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by cans » Tue Jun 07, 2011 9:03 pm
If P represents the product of the first 15 positive integers, then P is not a multiple of:

a) 99 b) 84 c) 72 d) 65 e) 57
p=15!
99=9*11. multiple
84 = 12*7 multiple
72=9*8 multiple
65=5*13 multiple
57=3*19 not
IMO E
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by Brian@VeritasPrep » Wed Jun 08, 2011 9:45 am
Hey all,

Just proposing a quick follow-up on this question:

Did anyone (correctly) pick E but because they (incorrectly) saw 57 as a prime number at first glance?

In my initial 10-second read of this one, my first inclination before ever picking up a pencil was that A, B, C, and C were all pretty easy to recognize as non-primes, so 57 seemed like an obvious right answer here...but for the wrong reason. To thoroughly and correctly do this problem, you need to make sure that EACH prime factor of each answer choice is one of the first 15 integers. Otherwise, a number like 95 (5 * 19) might look okay, but 19 is NOT accounted for in 15! so it is NOT a multiple of 15!.

I'd say that something like 95 would be a terrific hard-to-get right answer on this one since it might look at first glance like "oh, it's divisible by 5 so clearly it's okay" to someone who hasn't thought all the way through that they need to break down ALL prime factors.


Anyway...as I've mentioned on here a few times I think that one way to become excellent at this test is to start to think about potential traps and tweaks that the test COULD throw at you, and this one jumped out at me as something that could easily catch me if I'm being lazy (like, when I'm browsing this forum casually on my first cup of coffee and trying to solve problems without writing anything down).
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

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