Is the positive integer n a multiple of 20?

Question

(1) n^3 is a multiple of 60.
(2) 10 is a factor of square root on n.

MahmoudFawzy · Accepted Answer

Question:   does  n = 2^2*5^1*k?

Statement 1:  
 n^3 = 60m 
 n^3 = 2^2*5^1*3^1*m 
since n is a positive integer,  n = 2^1*5^1*3^1*q  , so n is for sure a multiple of 30.
We don't know whether q is an even number, so insufficient.

Statement 2:  
 \sqrt{n} = 10p 
 n = 100p^2  , so n is divisible by 20 regardless of the value of p --> sufficient

B

4d · Answer

Statement 1 :
 n^3=1000 
n=10
Or 
 n^3=4000 , n=20 
Not sufficient
Statement 2 :
 \sqrt{n}=10k 
 n=100k^2 
Sufficient 
B

EgmatQuantExpert · Answer

Solution

Given:  
 • The number n is a positive integer

To find:  
 • Whether n is a multiple of 20 or not

Approach and Working:  
If we express 20 in terms of its prime factors, we get 20 = 2^2 x 5^1
 • Therefore, if n is a multiple of 20, then n must have 2^2 and 5 as its factors.

Analysing Statement 1  
As per the information given in statement 1, n^3 is a multiple of 60.
 • Or, n^3 = 2^2 x 3 x 5 x k, where k is a positive integer
• From this, we can say that k must have at least 2, 3^2 and 5^2.
• However, we cannot say whether n is a multiple of 20 or not.

Hence, statement 1 is not sufficient to answer the question.

Analysing Statement 2  
As per the information given in statement 2, 10 is a factor of square root of n.
 • Therefore, we can say that 100 should be a factor of n.
• As 100 is a multiple of 20, we can say that n is also a multiple of 20.

Hence, statement 2 is sufficient to answer the question.

Hence, the correct answer is option B.

Answer: B

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-NP-e1533292144892.png

Siddharthjain07 · Answer

Statement 1. 
 N^3/60 = integer (stated)
So least value of n for which it can be divided by 60 is n= 30 which is not divisible by 20 .
And if n=60 which is divisible by 20 . 
Therefore 2 values and insufficient.

Statement 2. 
Rootn/10 =integer
Squaring n/100 = integer 
Factors of 100 are also factors of n therefore sufficient. 
B. Is the answer

Posted from my mobile device

kunalbean · Answer

Hi! Can you please explain that how why did you take all the prime factors power as 1 ?

MahmoudFawzy · Answer

lets try a simplified one first,

if  n^3  is a multiple of 2,
then  n^3 = 2a  , where a is an integer.

remember that n is also an integer, and  n = \sqrt {2a} 
this constrain forces us to conclude that  a = 2^2b  , where  b  is an integer.
why we concluded that? because otherwise  \sqrt {2}  will produce an irrational number.

back to our question,
 n^3 = 2^2*5^1*3^1*m 
so m must be equal to  2^15^23^2p , where p is an integer
and n must have 2,5,3 as primes of it

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Is the positive integer n a multiple of 20?

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