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01 Aug 2019, 00:08
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C
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:
69% (01:59) correct
31%
(02:04)
wrong
based on 229
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[GMAT math practice question]
\(N\) is a positive integer. \(\frac{N}{2}, \frac{N}{3}\) and \(\frac{N}{5}\) are a square, a cube and a fifth power, respectively. What is the minimum possible value of \(N\)?
\(A. (2^{11})(3^8)(5^8)\)
\(B. (2^{13})(3^8)(5^6)\)
\(C. (2^{15})(3^{10})(5^6)\)
\(D. (2^{17})(3^{12})(5^6)\)
\(E. (2^{19})(3^{13})(5^6)\)
\(N\) is a positive integer. \(\frac{N}{2}, \frac{N}{3}\) and \(\frac{N}{5}\) are a square, a cube and a fifth power, respectively. What is the minimum possible value of \(N\)?
\(A. (2^{11})(3^8)(5^8)\)
\(B. (2^{13})(3^8)(5^6)\)
\(C. (2^{15})(3^{10})(5^6)\)
\(D. (2^{17})(3^{12})(5^6)\)
\(E. (2^{19})(3^{13})(5^6)\)
01 Aug 2019, 04:38
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This is a slightly different question compared to the usual ones on the concept of squares and cubes.
A perfect square is the square of an integer. A perfect square always has an even power for all its prime factors.
A perfect cube is the cube of an integer. A perfect cube always has a multiple of 3 as the power for all its prime factors.
Similarly, the fifth power should have the powers of its prime factors as multiples of 5.
With this information, let’s look at the question. The best approach to solve this question in the shortest time is to eliminate options, taking them one by one.
For option A, N = \(2^{11} * 3^8 * 5^8\). So, \(\frac{N}{2}\) = \(2^{10} * 3^8 * 5^8\) which is a perfect square. \(\frac{N}{3}\) = \(2^{11} * 3^7 * 5^8\) which is not a perfect cube.
Option A can be eliminated.
Following the same procedure for option B, we see that although \(\frac{N}{2}\) is a perfect square, \(\frac{N}{3}\) is not a perfect cube and so option B can also be eliminated.
When we come to option C, we see that \(\frac{N}{2}\) is a perfect square, \(\frac{N}{3}\) is a perfect cube and \(\frac{N}{5}\) is a fifth power. Option C has to be the answer.
Hope this helps!
A perfect square is the square of an integer. A perfect square always has an even power for all its prime factors.
A perfect cube is the cube of an integer. A perfect cube always has a multiple of 3 as the power for all its prime factors.
Similarly, the fifth power should have the powers of its prime factors as multiples of 5.
With this information, let’s look at the question. The best approach to solve this question in the shortest time is to eliminate options, taking them one by one.
For option A, N = \(2^{11} * 3^8 * 5^8\). So, \(\frac{N}{2}\) = \(2^{10} * 3^8 * 5^8\) which is a perfect square. \(\frac{N}{3}\) = \(2^{11} * 3^7 * 5^8\) which is not a perfect cube.
Option A can be eliminated.
Following the same procedure for option B, we see that although \(\frac{N}{2}\) is a perfect square, \(\frac{N}{3}\) is not a perfect cube and so option B can also be eliminated.
When we come to option C, we see that \(\frac{N}{2}\) is a perfect square, \(\frac{N}{3}\) is a perfect cube and \(\frac{N}{5}\) is a fifth power. Option C has to be the answer.
Hope this helps!
General Discussion
01 Aug 2019, 01:35
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N is a positive integer and is divisible by 2,3 and 5
so N is of form = \(2^a*3^b*5^c\)
now given \(\frac{N}{2}\) is a square (that means powers of N are multiple of 2
\(\frac{N}{2}\) = 2^(a-1)*3^b*5^c (when a-1 ,b and c divided by 2 result is integer)
a = 2I+1, b = 2I, c = 2I
a= 3,5,7...., b = 2,4,6...., c = 2,4,6.....
similarly, \(\frac{N}{3}\) is a cube (that means powers of N are multiple of 3)
\(\frac{N}{3}\) = 2^a*3^(b-1)*5^c (when a ,b-1 and c divided by 3 result is integer)
a = 3I, b = 3I+1 , c = 3I
a = 3,6,9..., b = 4,7,10..., c = 3,6,9...
\(\frac{N}{5}\) is a fifth power (that means powers of N are multiple of 5)
\(\frac{N}{5}\) = 2^a*3^b*5^(c-1) (when a ,b and c-1 divided by 5 result is integer)
a = 5I, b = 5I+1 , c = 5I
a= 5,10,15.. b = 6,11,16... c= 5,10,15
taking the LCM of all the 3 cases of value of a, b and c (to satisfy \(\frac{N}{2}\) square, \(\frac{N}{3}\) cube,\(\frac{N}{5}\) fifth power)
a = 15,30,45...., b= 10,20,30...., c = 6,12,18
minimum values of a,b, and c = 15,10, and 6
so minimum value of N = 2^15*3^10*5^6
so N is of form = \(2^a*3^b*5^c\)
now given \(\frac{N}{2}\) is a square (that means powers of N are multiple of 2
\(\frac{N}{2}\) = 2^(a-1)*3^b*5^c (when a-1 ,b and c divided by 2 result is integer)
a = 2I+1, b = 2I, c = 2I
a= 3,5,7...., b = 2,4,6...., c = 2,4,6.....
similarly, \(\frac{N}{3}\) is a cube (that means powers of N are multiple of 3)
\(\frac{N}{3}\) = 2^a*3^(b-1)*5^c (when a ,b-1 and c divided by 3 result is integer)
a = 3I, b = 3I+1 , c = 3I
a = 3,6,9..., b = 4,7,10..., c = 3,6,9...
\(\frac{N}{5}\) is a fifth power (that means powers of N are multiple of 5)
\(\frac{N}{5}\) = 2^a*3^b*5^(c-1) (when a ,b and c-1 divided by 5 result is integer)
a = 5I, b = 5I+1 , c = 5I
a= 5,10,15.. b = 6,11,16... c= 5,10,15
taking the LCM of all the 3 cases of value of a, b and c (to satisfy \(\frac{N}{2}\) square, \(\frac{N}{3}\) cube,\(\frac{N}{5}\) fifth power)
a = 15,30,45...., b= 10,20,30...., c = 6,12,18
minimum values of a,b, and c = 15,10, and 6
so minimum value of N = 2^15*3^10*5^6
