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MathRevolution
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A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 childre 29 Oct 2018, 00:58
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E
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85% (00:44) correct 15% (00:58) wrong based on 112 sessions

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[Math Revolution GMAT math practice question]

A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 children with no pencils left over. What is the smallest possible number of pencils in the set?

A. 6
B. 12
C. 15
D. 30
E. 60
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E
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shashaankbhat
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A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 childre 29 Oct 2018, 01:00
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IMO E. LCM is the smallest possible number. So LCM of given no.s is 60

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A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 childre 29 Oct 2018, 05:30
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shashaankbhat
IMO E. LCM is the smallest possible number. So LCM of given no.s is 60

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Can you explain process?
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A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 childre 29 Oct 2018, 07:58
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The smallest number which is evenly divisible by numbers 2, 3, 4, 5 and 6 is 60 which is the LCM. LCM can be found by using prime factorization method Ritace
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A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 childre 29 Oct 2018, 08:53
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Ritace

Alternatively check using the option

(A) Not divisible by 4 & 5
(B) Not divisible by 5
(C) Not divisible by 2 , 4 & 6
(D) Not divisible by 4
(E) Divisible by all the numbers...

Hence, Answer must be (E)
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A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 childre 30 Oct 2018, 19:10
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MathRevolution
[Math Revolution GMAT math practice question]

A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 children with no pencils left over. What is the smallest possible number of pencils in the set?

A. 6
B. 12
C. 15
D. 30
E. 60
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We need to determine the LCM of 2, 3, 4, 5 and 6. Breaking each number into its prime factors, we have:

2 = 2
3 = 3
4 = 2^2
5 = 5
6 = 2 x 3

Thus, the LCM is 2^2 x 3 x 5 = 60.

Answer: E
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A set of pencils can be evenly shared between 2, 3, 4, 5 and 6 childre 31 Oct 2018, 02:35
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=>

The smallest possible number of pencils must be the least common multiple of \(2, 3, 4, 5\)and \(6\).
\(2 = 2^1, 3 = 3^1, 4 = 2^2, 5 = 5^1\) and \(6 = 2^1*3^1.\)
We multiply the maximum powers of each base to obtain the least common multiple \(2^2*3^1*5^1 = 60.\)

Therefore, the answer is E.
Answer: E
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