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kiran120680
Moderator - Masters Forum
Joined: 18 Feb 2019
Last visit: 27 Jul 2022
Posts: 706
Given Kudos: 276
Location: India
Schools: AGSM '20 EDHEC Sept 2019 Intake Great Lakes '21 IE
GMAT 1: 460 Q42 V13
GPA: 3.6
09 Jun 2019, 10:16
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
57% (01:06) correct
43%
(01:05)
wrong
based on 533
sessions
History
Date
Time
Result
Not Attempted Yet
If n is the product of all odd prime numbers less than 16, how many factors does n have?
A. 5
B. 6
C. 16
D. 32
E. 64
A. 5
B. 6
C. 16
D. 32
E. 64
10 Jun 2019, 02:51
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There is an important formula to find the factors of a number 'N'
if \(N=(a^x)*(b^y)*(c^z)\)... (where a,b,c,d.. are prime factors of N)
so, number of all factors of a \(N = (x+1)*(y+1)*(z+1)\)....
Back to the question
N = (Product of all odd prime numbers less than 16) = \((3^1)*(5^1)*(7^1)*(11^1)*(13^1)\)
number of all factors = \((1+1)*(1+1)*(1+1)*(1+1)*(1+1)=2*2*2*2*2=32\)
if \(N=(a^x)*(b^y)*(c^z)\)... (where a,b,c,d.. are prime factors of N)
so, number of all factors of a \(N = (x+1)*(y+1)*(z+1)\)....
Back to the question
N = (Product of all odd prime numbers less than 16) = \((3^1)*(5^1)*(7^1)*(11^1)*(13^1)\)
number of all factors = \((1+1)*(1+1)*(1+1)*(1+1)*(1+1)=2*2*2*2*2=32\)
General Discussion
27 Oct 2020, 09:09
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kiran120680
Solution:
Recall that to determine the number of positive factors that a number n has, we first prime factorize n and then add 1 to each prime’s exponent. We then calculate the product of those “adjusted” exponents.
First, we prime factorize n:
n = 3^1 x 5^1 x 7^1 x 11^1 x 13^1
We add 1 to each exponent and then find the product of the adjusted exponents. Therefore, n has (1 + 1) x (1 + 1) x (1 + 1) x (1 + 1) x (1 + 1) = 2^5 = 32 factors.
Answer: D
