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If a, b, and c are positive integers such that x/(2^a3^b5^c) is an int 19 Jun 2020, 05:20
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D
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95% (01:03) correct 5% (01:43) wrong based on 60 sessions

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If a, b, and c are positive integers such that \(\frac{x}{2^a3^b5^c}\) is an integer, which of the following could be the value of x ?


A. 216

B. 225

C. 232

D. 240

E. 250
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D
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If a, b, and c are positive integers such that x/(2^a3^b5^c) is an int 19 Jun 2020, 05:37
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Given : X/(2^a * 3^b * 5^c) = An integer
To find : Value of X

Solution: X is atleast multiple of 2*3*5 = 30
Among all option, only D is a said multiple of 30.

Answer should be D

ALTERNATIVELY,

Factorizing options,

A. 216 = 2*2*2*3*3*3 (5 is not a factor)

B. 225 = 5*5*3*3 (2 is not a factor)

C. 232 = 2*2*2*29 (3&5 both are not a factor)

D. 240 = 2*2*2*2*3*5 (Satisfying)

E. 250 = 5*5*5*2 (3 is not a factor)

Answer is D

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If a, b, and c are positive integers such that x/(2^a3^b5^c) is an int 19 Jun 2020, 05:43
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you can either factor the numbers as the must be in a form of 2 (something)*3 (somthing)* 5 something

225 (not even-not factor of 2), 232 (no factor 5 ), 250 (not divisible by 3) are off since they dont have factors needed.

216 is not having 5

240 is even+ digits divisible by 3 + 0/5 divisible by 5.

checking- 240=2^4*5^1*3^1 (F) so 240 / F is the integer 1
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If a, b, and c are positive integers such that x/(2^a3^b5^c) is an int 19 Jun 2020, 06:08
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Bunuel
If a, b, and c are positive integers such that \(\frac{x}{2^a3^b5^c}\) is an integer, which of the following could be the value of x ?


A. 216

B. 225

C. 232

D. 240

E. 250
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Key: "Could be" means we have to find possibility in one option because other 4 options will be IMPOSSIBLE values of x

\(\frac{x}{2^a3^b5^c}\) since a, b and c are Positive Integers so x must have atleast 1 power of 2, 3 and 5 all




A. \(216 = 6^3 = 2^3*3^3\) NOT Possible because 216 is not a multiple fo 5

B. \(225 = 15^2 = 3^2*5^2\) NOT Possible because 225 is not a multiple fo 2

C.\( 232 = 4*58 = 2^3*29\) NOT Possible because 232 is not a multiple fo 3 and 5

D. \(240 = 2^4*3*5\) POSSIBLE

E. 250 NOT Possible because 250 is not a multiple of 3

Answer: Option D
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If a, b, and c are positive integers such that x/(2^a3^b5^c) is an int 19 Jun 2020, 09:17
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A number divisible by any positive power of both 2 and 5 must end with 0.

This eliminates option A, B and C.

250 is not divible by 3.

Correct answer is 240
Option D

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