[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
A set of pencils can be evenly shared between 2, 3, 4, 5 and
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- Max@Math Revolution
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For the set of pencils to be shared evenly, the total number of pencils must be divisible by the total number of children.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
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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- fskilnik@GMATH
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$$? = LCM\left( {2,3,4,5,6} \right) = LCM\left( {6,4,5} \right) = LCM\left( {12,5} \right) = 60$$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
The correct answer is therefore (E).
This solution follows the notations and rationale taught in the GMATH method.
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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: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
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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- Max@Math Revolution
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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
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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