Difference between revisions of "2016 AMC 8 Problems/Problem 9"
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We notice that <math>9 \mid 2016</math>, since <math>2+0+1+6 = 9</math>, and <math>9 \mid 9</math>. We can divide <math>2016</math> by <math>9</math> to get <math>224</math>. This is divisible by <math>4</math>, as <math>4 \mid 24</math>. Dividing <math>224</math> by <math>4</math>, we have <math>56</math>. This is clearly divisible by <math>7</math>, leaving <math>8</math>. We have <math>2016 = 9\cdot 4\cdot 7\cdot 8</math>. We know that <math>4</math> and <math>8</math> are both multiples of <math>2</math>, <math>9</math> is <math>3^2</math>, and <math>7</math> is prime. This means that the distinct prime factors are <math>2,3,</math> and <math>7</math>. Their sum is <math>\boxed{\textbf{(B) }12}</math>. | We notice that <math>9 \mid 2016</math>, since <math>2+0+1+6 = 9</math>, and <math>9 \mid 9</math>. We can divide <math>2016</math> by <math>9</math> to get <math>224</math>. This is divisible by <math>4</math>, as <math>4 \mid 24</math>. Dividing <math>224</math> by <math>4</math>, we have <math>56</math>. This is clearly divisible by <math>7</math>, leaving <math>8</math>. We have <math>2016 = 9\cdot 4\cdot 7\cdot 8</math>. We know that <math>4</math> and <math>8</math> are both multiples of <math>2</math>, <math>9</math> is <math>3^2</math>, and <math>7</math> is prime. This means that the distinct prime factors are <math>2,3,</math> and <math>7</math>. Their sum is <math>\boxed{\textbf{(B) }12}</math>. | ||
+ | ==Video Solution== | ||
+ | https://youtu.be/1KN7OTG3k-0 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2016|num-b=8|num-a=10}} | {{AMC8 box|year=2016|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:28, 21 March 2022
Problem
What is the sum of the distinct prime integer divisors of ?
Solutions
Solution 1
The prime factorization is . Since the problem is only asking us for the distinct prime factors, we have
. Their desired sum is then
.
Solution 2
We notice that , since
, and
. We can divide
by
to get
. This is divisible by
, as
. Dividing
by
, we have
. This is clearly divisible by
, leaving
. We have
. We know that
and
are both multiples of
,
is
, and
is prime. This means that the distinct prime factors are
and
. Their sum is
.
Video Solution
~savannahsolver
See Also
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.