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If n is the product of the 5 different prime numbers, how ma

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by Max@Math Revolution » Tue Jun 28, 2016 5:41 pm
If n is the product of the 5 different prime numbers, how many factors n have except 1 and n?
A. 20
B. 22
C. 24
D. 30
E. 32

*An answer will be posted in 2 days.
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by Danny@GMATAcademy » Wed Jun 29, 2016 4:51 am
Here's a video that develops a method for quickly determining how many factors a number has. It leads up to a formula that can make quick work of questions such as this.

https://www.youtube.com/watch?v=njePP0ZK2bY
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by 800_or_bust » Wed Jun 29, 2016 6:24 am
Max@Math Revolution wrote:If n is the product of the 5 different prime numbers, how many factors n have except 1 and n?
A. 20
B. 22
C. 24
D. 30
E. 32

*An answer will be posted in 2 days.
The total number of factors can be calculated from the prime factorization. Add one to the exponent of each prime factor and multiply each of these numbers together. This yields the total number of factors. Here, since we have five different prime numbers, the exponent of each is one. Hence, the product is 2x2x2x2x2 = 2^5 = 32. Now, this is the TOTAL number of different factors. Since we are told to toss 1 and the number itself, the final answer will be two less than this. Hence, the answer is D.
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by regor60 » Thu Jun 30, 2016 7:22 am
How many factors comprising one of the primes: 5!/1!4! = 5

" two " 5!/2!3! = 10

" three " 5!/3!2! = 10

" four " 5!/4!1! = 5

Total of 30
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by Max@Math Revolution » Fri Jul 01, 2016 1:17 am
(1+1)(1+1)(1+1)(1+1)(1+1)=32. Hence, from 32-2=30, we can see that D is the correct answer.
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