N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f

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)

CrackverbalGMAT · Accepted Answer

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!

shridhar786 · Answer

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

MathRevolution · Answer

=>

Since  \frac{N}{2}, \frac{N}{3}  and  \frac{N}{5}  are integers,  N  should have prime factors of  2, 3,  and  5.  Thus, the minimum value of  N  has the form  (2^a)(3^b)(5^c)  for some positive integers,  a, b  and  c.

Since  \frac{N}{2} = (2^{a-1})(3^b)(5^c)  is a square,  b  and  c  are even numbers and  a  is an odd number.

Since  \frac{N}{3} = (2^a)(3^{b-1})(5^c)  is a cube,  a  and  c  are multiples of  3  and  b  has remainder  1  when it is divided by  3.

Since  \frac{N}{5} = (2^a)(3^b)(5^{c-1})  is a fifth power,  a  and  b  are multiples of  5  and  c  has remainder  1  when it is divided by  5.

Since  a  is a multiple of  3  and  5  and an odd number, the smallest positive value of  a  is  15.

Since  b  is a multiple of  2  and  5  and has remainder  1  when it is divided by  3 , the smallest positive value of  b  is  10.

Since  c  is a multiple of  2  and  3  and has remainder  1  when it is divided by  5 , the smallest positive value of  c  is  6.

Thus, the minimum value of  N  is  N = (2^{15})(3^{10})(5^6).

Therefore, C is the answer.
Answer: C

sampriya · Answer

so basically in this question :

we are taking the LCM of the powers because :

N/2- is a square---- but N must be a cube and a 5th power
to be a square; a cube and a fifth power we take the LCM of 5 and 3- > 15

and the same goes for the other two :

given N/3= a cube then the minimum value would be 3^4---- but when we divide N by 3 we get 3^3 which is neither a square nor a 5th power therefore when we take the LCM of 2 and 5 which is 10-----> therefore 3^10 is a square and a fifth power!

Ahmed9955 · Answer

I apply this shortcut while solving such questions, it's fast and easy

>For a no. to be square-- the power of 2 has to be a multiple of two
in the above case of N/2 - 2^11/2 = 2^10

>For a no. to be a cube -- The power of 3 has to be a multiple of 3
in the above case of N/3 - 3^10/3 = 3^9

and the power fifth can be narrowed by the option given..
Hence, Option C is correct

Kudos, if you find this helpful.

bumpbot · Answer

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N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f

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N is a positive integer. N/2, N/3 and N/5 are a square, a cube and a f

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