Difference between revisions of "2016 AMC 8 Problems/Problem 9"

Revision as of 03:49, 16 January 2021

Problem

What is the sum of the distinct prime integer divisors of $2016$ ?

$\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63$

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 $\boxed{\textbf{(B) }12}$ .

Solution 2

We notice that $9 \mid 2016$ , since $2+0+1+6 = 9$ , and $9 \mid 9$ . We can divide $2016$ by $9$ to get $224$ . This is divisible by $4$ , as $4 \mid 24$ . Dividing $224$ by $4$ , we have $56$ . This is clearly divisible by $7$ , leaving $8$ . We have $2016 = 9\cdot 4\cdot 7\cdot 8$ . We know that $4$ and $8$ are both multiples of $2$ , $9$ is $3^2$ , and $7$ is prime. This means that the distinct prime factors are $2,3,$ and $7$ . Their sum is $\boxed{\textbf{(B) }12}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 23:43, 1 November 2020 (view source) Ab2024 (talk \| contribs) (→‎Solution 2) ← Older edit		Revision as of 03:49, 16 January 2021 (view source) Hashtagmath (talk \| contribs) Newer edit →
Line 1:		Line 1:
		+	== Problem ==
		+
	What is the sum of the distinct prime integer divisors of <math>2016</math>?		What is the sum of the distinct prime integer divisors of <math>2016</math>?

Page

Toolbox

Search

Difference between revisions of "2016 AMC 8 Problems/Problem 9"

Revision as of 03:49, 16 January 2021

Problem

Solution 1

Solution 2