Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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| − | To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. | + | To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <math>2^{10}</math> and <math>5^{10}</math> do not work since the factors chosen must be distinct, and those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. |
| − | Solution by greersc. (Edited by AZAZ12345 and then by greersc once again) | + | Solution by greersc. (Edited by AZAZ12345 and then by greersc once again and then added on by mshell214) |
==Solution 2== | ==Solution 2== | ||
Revision as of 08:41, 15 February 2019
- The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.
Contents
Problem
Let 
be the set of all positive integer divisors of 
How many numbers are the product of two distinct elements of 
Solution 1
To find the number of numbers that are the product of two distinct elements of , we first square 
and factor it. Factoring, we find 
. Therefore, 
has 
distinct factors. Each of these can be achieved by multiplying two factors of . However, the factors must be distinct, so we eliminate 
and , as well as 
and 
, so the answer is 
. and do not work since the factors chosen must be distinct, and those require 
or 
.
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again and then added on by mshell214)
Solution 2
The prime factorization of 100,000 is 
. Thus, we choose two numbers 
and 
where 
and 
, whose product is 
, where 
and 
.
Consider 
. The number of divisors is 
. However, some of the divisors of 
cannot be written as a product of two distinct divisors of , namely: 
, , , and . The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only 
or 
. Since the factors chosen must be distinct, the last two numbers cannot be created because those require or . This gives 
candidate numbers. It is not too hard to show that every number of the form 
where 
, and 
are not both 0 or 10, can be written as a product of two distinct elements in . Hence the answer is 
.
-scrabbler94 (edited by mshell214)
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. 
