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Is the natural number n a multiple of 40? (1) 20 is a factor of \(n^2\). (2) \(\frac{n^3}{128}\) is an integer.

Hi, Is n= 40x where x is an integer.. lets see the statement.. (1) 20 is a factor of \(n^2\). if 20 is a factor of \(n^2\), n has to have 2*5 as its factor... since 20=2*2*5.. so n can be 10,30,50,80.. ans can be yes if n is 80, or NO if n is 10... insuff

(2) \(\frac{n^3}{128}\) is an integer. this means n^3 has 128 as its factor.. 128=2^7.. therefore n has to have 2^3 as its factor... insuff..

combined we know n is a multiple of 10 and also a multiple of 8.. so n has to be a multiple of 40 (LCM of 10 and 8).. suff ans C
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Re: Is the natural number n a multiple of 40? [#permalink]

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05 Jun 2016, 01:42

For n to be a multiple of 40, it should have at least three 2’s and one 5 in it (i.e. 23 and 5).

Statement 1) 20 is a factor of n2 i.e. n2 has 22 and a 5 in it. As n is an integer n should have at least a 2 and a 5 in it. Thus the minimum value of n = 2 x 5 = 10. But we want to check whether n is a multiple of 40 or not. Being multiple of 10 doesn’t really tell us about it being a multiple of 40. Thus statement (1) is insufficient alone.

Statement 2) Given n3/128 is an integer, it implies n3 has 128 i.e. 27 in it. When n3 has seven 2’s in it, n should have at least three 2’s in it. Thus n is a multiple of 8 but it is not sufficient as we want to know whether n is a multiple of 40 or not. As a 5 may not necessarily be there in n, the statement remains insufficient.

Combining both the statements: we know n is a multiple of 10 (using statement 1) and a multiple of 8 (using statement 2). Thus we can say n should be a lowest common multiple of 10 and 8 it implies that n should be a multiple of 40. Thus combining both the statements is sufficient to answer the question.

Is the natural number n a multiple of 40? [#permalink]

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29 Nov 2017, 11:36

Is the natural number n a multiple of 40? (1) 20 is a factor of \(n^2\). (2)\(n^3/128\) is an integer.

Statement 1: if \(n^2\) = 20, then n is not an integer, it can't be multiple of 40 if n is integer, then \(n^2 = 2^2 * 5^2 * m^2\) (where m is an integer), n must have 2 and 5 as factor. Not Sufficient

Statement 2: \(n^3/128\) is an integer if \(n^3 = 128\), then n is not integer, it can't be multiple of 40 if n is integer, then \(n^3\) is divisible by 128, \(n^3 = 2^7 * 2^2 * x^3\) (where x is integer), n must be \(2^3 x\), so n has 8 as factor.

No Sufficient.

(1)+(2), since \(n^3/128\) is integer => \(n^3\) must be integer since 20 is factor of \(n^2\)=> \(n^2\) must be integer since both \(n^2\) and \(n^3\) is integer, \(n\) must be integer. From statement 1, n has 5 as factor, from statement 2, n has 8 as factor, combining, n has both 8 and 5, so n must be multiple of 40

Re: Is the natural number n a multiple of 40? [#permalink]

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29 Nov 2017, 22:30

n is given to be a natural number. So n^2, n^3 both are also integers. Also we should know that in the prime factorisation, n^2 will have all powers of prime numbers as multiples of 2, and n^3 will have all powers of prime numbers as multiples of 3. Also 40 = 2^3 * 5, so for a number to be a multiple of 40, it MUST have at least three 2's and one 5, in its prime factorisation.

(1) n^2 is a multiple of 20. Now 20 = 2^2 * 5 , and since n^2 will have all powers as multiples of 2, the LEAST value of n^2 can be 2^2 * 5^2, so the least value of n will be 2*5 = 10, OR we can say that n MUST have at least one 2 and one 5. But we cant be sure whether it will be a multiple of 40 or not. Insufficient.

(2) n^3 is a multiple of 128. now 128 = 2^7, and since n^3 will have all powers as mutiples of 3, the LEAST value of n^3 can be 2^9, so the least value of n will be 2^3 = 8, OR we can say that m MUST have at least three 2's. But we cant be sure whether it will be a multiple of 40 or not. Insufficient.

Combining the two statements, n MUST have at least three 2's and one 5. So it has to be a multiple of 40. Sufficient.