2009 Grade 8 CEMC Gauss Problems/Problem 21

Problem

The product of four different positive integers is $360$ . What is the maximum possible sum of these four integers?

$\text{ (A) }\ 68 \qquad\text{ (B) }\ 66 \qquad\text{ (C) }\ 52 \qquad\text{ (D) }\ 39 \qquad\text{ (E) }\ 24$

Solution

We want to make the first three integers the smallest that they can be, but make the fourth integer as large as it can be. However, since the integers are distinct, we cannot just make three of them $1$ and the last $360$ .

We can find the prime factorization of $360$ , which is $2^{3} \times 3^{2} \times 5$ .

We can now see that the three smallest integers are $1$ , $2$ , and $3$ , which multiply to $6$ . This means the last number must be $\frac{360}{6} = 60$ .

Summing all of our numbers together, we get $1 + 2 + 3 + 60 = \boxed {\textbf {(B) } 66}$

~anabel.disher

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2009 Grade 8 CEMC Gauss Problems/Problem 21

Problem

Solution