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Updated on: 05 Feb 2018, 07:11
Originally posted by GMATinsight on 05 Feb 2018, 03:44.
Last edited by GMATinsight on 05 Feb 2018, 07:11, edited 3 times in total.
Last edited by GMATinsight on 05 Feb 2018, 07:11, edited 3 times in total.
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B
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Difficulty:
65%
(hard)
Question Stats:
47% (01:42) correct
53%
(02:00)
wrong
based on 74
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Is Integer x divisible by 25?
(1) \(x^2\) is a multiple of 50
(2) \(x^3\) is a multiple of 625
Source: https://www.GMATinsight.com
(1) \(x^2\) is a multiple of 50
(2) \(x^3\) is a multiple of 625
Source: https://www.GMATinsight.com
05 Feb 2018, 05:59
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Answer c
1) let us assume x as 10
10^2=100 which is multiple of 50. But x which is 10 is not divisible by 25
2) let us assume x as 5
5^3=625
Again 5 is not divisible by 25
Together least value should be 1250, which is divisible by 25...
Sent from my Redmi Note 3 using GMAT Club Forum mobile app
1) let us assume x as 10
10^2=100 which is multiple of 50. But x which is 10 is not divisible by 25
2) let us assume x as 5
5^3=625
Again 5 is not divisible by 25
Together least value should be 1250, which is divisible by 25...
Sent from my Redmi Note 3 using GMAT Club Forum mobile app
05 Feb 2018, 08:13
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GMATinsight
If x is an integer, then its square and cube will also be integers.
(1) x^2 is a multiple of 50. Now 50 = 2 * 5^2
This tells us that prime factorisation of x^2 has at least one 2 and at least two 5's.
If x^2 has at least two 5's, then x will have at least one 5. So x will be divisible by 5, but we cannot be sure if it will be divisible by 25.
Eg., if x^2 = 100, then x = 10, not divisible by 25. But if x^2 = 2500, then x = 50, which is divisible by 25.
So not sufficient.
(2) x^3 is a multiple of 625. Now 625 = 5^4
So this tells us that cube of x has at least four 5's. So x also must have '5' in its prime factorisation.
But if x has only one 5 in its prime factorisation, then its cube will have only three 5's, and this given condition will not be satisfied. Case rejected
Only if x has at least two 5's in its prime factorisation, will then its cube have at least six 5's, and then only will its cube be divisible by 625.
This tells us that x must have at least two 5's in its prime factorisation, making it divisible by 25. So sufficient.
Hence B answer
