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A set of pencils can be evenly shared between 2, 3, 4, 5 and

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by Max@Math Revolution » Sun Oct 28, 2018 11:51 pm

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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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by GMATGuruNY » Mon Oct 29, 2018 2:25 am
Max@Math Revolution wrote:[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
For the set of pencils to be shared evenly, the total number of pencils must be divisible by the total number of children.
Thus, the correct answer must be divisible by 2, 3, 4, 5 and 6.
Of the five answer choices, only E is divisible by 2, 3, 4, 5 and 6.

The correct answer is E.
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by fskilnik@GMATH » Mon Oct 29, 2018 2:26 am
Max@Math Revolution wrote:[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
$$? = LCM\left( {2,3,4,5,6} \right) = LCM\left( {6,4,5} \right) = LCM\left( {12,5} \right) = 60$$

The correct answer is therefore (E).

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Scott@TargetTestPrep » Tue Oct 30, 2018 6:05 pm
Max@Math Revolution wrote:[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
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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by Max@Math Revolution » Wed Oct 31, 2018 1:26 am
=>

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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