Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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==Problem== | ==Problem== | ||
| − | Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math> | + | @Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math>@ |
<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> | ||
Revision as of 11:55, 19 March 2019
- The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.
Problem
@Letbe the set of all positive integer divisors of
How many numbers are the product of two distinct elements of
@
How many numbers are the product of two distinct elements of 
@
Solution
The prime factorization of 
is 
. Thus, we choose two numbers 
and 
where 
and 
, whose product is 
, where 
and 
.
Notice that this is analogous 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 
. This gives 
candidate numbers. It is not too hard to show that every number of the form 
, where 
, and 
are not both 
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. 
