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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote A set of pencils can be evenly shared between 2, 3, 4, 5 and tagged by: Max@Math Revolution This topic has 4 expert replies and 0 member replies GMAT/MBA Expert A set of pencils can be evenly shared between 2, 3, 4, 5 and Timer 00:00 Your Answer A B C D E Global Stats Difficult [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 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only Worldâ€™s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course
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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.

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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.

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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.

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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.

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