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Question
James, Jeremy, and John are placing 105, 112, and 119 traffic cones on different construction sites. If they each want to place equal number of traffic cones on each site, what is the greatest number of traffic cones each site can have?
James, Jeremy, and John are placing 105, 112, and 119 traffic cones on different construction sites. If they each want to place equal number of traffic cones on each site, what is the greatest number of traffic cones each site can have?
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Solution
The correct option is A 7
To solve this problem, first we have to break down both the numbers into prime factors.
The prime factorization of 105 is: 3 × 5 × 7
The prime factorization of 112 is: 2 × 2 × 2 × 2 × 7
The prime factorization of 119 is: 7 × 17
Thus, the greatest common divisor of 105, 112, and 119 is 7.
Hence, each site can have 7 traffic cones.
To solve this problem, first we have to break down both the numbers into prime factors.
The prime factorization of 105 is: 3 × 5 × 7
The prime factorization of 112 is: 2 × 2 × 2 × 2 × 7
The prime factorization of 119 is: 7 × 17
Thus, the greatest common divisor of 105, 112, and 119 is 7.
Hence, each site can have 7 traffic cones.
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