1987 AJHSME Problems/Problem 9

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Problem

When finding the sum $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}$, the least common denominator used is

$\text{(A)}\ 120 \qquad \text{(B)}\ 210 \qquad \text{(C)}\ 420 \qquad \text{(D)}\ 840 \qquad \text{(E)}\ 5040$

Solution

We want the least common multiple of $2,3,4,5,6,7$, which is $420$, or choice $\boxed{\text{C}}$.

Solution 2

Since we want to find the LCM (least common multiple) of 2,3,4,5,6,7 we would be able to do this simply by finding the prime factorization of each of these numbers. First, we know that since 2 and 3 are already prime numbers, it would be completely factored. But, since 4 is a composite number, we would need to find the prime factorization of this, which would simply be 2 x 2. Afterward, we would basically just do this for all the numbers that are not prime (composite). Once we find the prime factorization of all of these numbers we would multiple the greatest prime factors.

2 --> 2 3 --> 3 4 --> 2^2 5 --> 5 6 --> 2 x 3 7 --> 7

Since 2^2 is the greatest number of 2's we can have we would simply multiply this by the greatest number of 3's, 5's, and 7's we could have in being the factors of each number. So,

2^2 x 3 x 5 x 7 = 420

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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