Hi, I found this problem in the Kaplan Premier book, can someone please tell me what is the correct approach to solve it?
thanks!
if both 5^2 and 3^3 are factors of n x 2^5 x 6^2 x 7^3, what is the smallest possible positive value of n?
a. 25
b. 27
c. 45
d. 75
e. 125
d
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prime factorization
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Solution:
n * 2^5 * 6^2 * 7^3 = n * 2^5 * 2^2 * 3^2 * 7^3 = n * 2^7 * 3^2 * 7^3.
Since two 3's are there in the above expression, we need one 3 more to get 3^3 as factor.
Also since 5 is not there in the expression, we need two 5's as factors.
So, the minimum value of n is 5^2 * 3 = 75.
The correct answer is hence d.
n * 2^5 * 6^2 * 7^3 = n * 2^5 * 2^2 * 3^2 * 7^3 = n * 2^7 * 3^2 * 7^3.
Since two 3's are there in the above expression, we need one 3 more to get 3^3 as factor.
Also since 5 is not there in the expression, we need two 5's as factors.
So, the minimum value of n is 5^2 * 3 = 75.
The correct answer is hence d.
Rahul Lakhani
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Gurome, Inc.
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Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
