Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math> | <math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math> | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/ZhAZ1oPe5Ds?t=3975 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==Solution== | ==Solution== |
Revision as of 21:28, 17 January 2021
- The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.
Contents
[hide]Problem
Let be the set of all positive integer divisors of
How many numbers are the product of two distinct elements of
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=3975
~ pi_is_3.14
Solution
The prime factorization of is
. Thus, we choose two numbers
and
where
and
, whose product is
, where
and
.
Notice that this is similar to choosing a divisor of , which has
divisors. However, some of the divisors of
cannot be written as a product of two distinct divisors of
, namely:
,
,
, and
. The last two cannot be so written because the maximum factor of
containing only
s or
s (and not both) is only
or
. Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require
or
. The first two would require
and
, respectively. This gives
candidate numbers. It is not too hard to show that every number of the form
, where
, and
are both not
or
, can be written as a product of two distinct elements in
. Hence the answer is
.
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AMC 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.