Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or </math>5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require 2 or 5 to the fifth power two times. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>. | Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or </math>5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require 2 or 5 to the fifth power two times. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>. | ||
| − | -scrabbler94 | + | -scrabbler94 (edited by mshell214) |
==See Also== | ==See Also== | ||
Revision as of 08:34, 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 
.
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)
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 
5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require 2 or 5 to the fifth power two times. 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. 
