Difference between revisions of "2009 Grade 8 CEMC Gauss Problems/Problem 21"

Latest revision as of 11:43, 1 October 2025

This page has been proposed for deletion. Reason: being moved Note to sysops: Before deleting, please review: • What links here • Discussion page • Edit history

@@ Line 1: / Line 1: @@
-==Problem==
+{{delete|being moved}}
-The product of four ''different'' positive integers is <math>360</math>. What is the maximum possible sum of these four integers?
-<math> \text{ (A) }\ 68 \qquad\text{ (B) }\ 66 \qquad\text{ (C) }\ 52 \qquad\text{ (D) }\ 39 \qquad\text{ (E) }\ 24 </math>
-==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 <math>1</math> and the last <math>360</math>.
-We can find the prime factorization of <math>360</math>, which is <math>2^{3} \times 3^{2} \times 5</math>.
-We can now see that the three smallest integers are <math>1</math>, <math>2</math>, and <math>3</math>, which multiply to <math>6</math>. This means the last number must be <math>\frac{360}{6} = 60</math>.
-Summing all of our numbers together, we get <math>1 + 2 + 3 + 60 = \boxed {\textbf {(B) } 66}</math>
-~anabel.disher