2016 AMC 8 Problems/Problem 9: Difference between revisions

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Problem

What is the sum of the distinct prime integer divisors of 735 ?

$\textbf{(A) }9\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

Solutions

Solution 1

The prime factorization is $2016=2^5\times3^2\times7$ . Since the problem is only asking us for the distinct prime factors, we have $2,3,7$ . Their desired sum is then $\boxed{\textbf{(B) }12}$ .

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/GNFnta9MF9E

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Video Solution

https://youtu.be/1KN7OTG3k-0

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2016 AMC 8 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

@@ Line 10: / Line 10: @@
 ===Solution 1===
-The prime factorization is <math>735=2^5\times3^2\times7</math>.  Since the problem is only asking us for the distinct prime factors, we have <math>2,3,7</math>.  Their desired sum is then <math>\boxed{\textbf{(B) }12}</math>.
+The prime factorization is <math>2016=2^5\times3^2\times7</math>.  Since the problem is only asking us for the distinct prime factors, we have <math>2,3,7</math>.  Their desired sum is then <math>\boxed{\textbf{(B) }12}</math>.
 ==Video Solution (CREATIVE THINKING!!!)==

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