Difference between revisions of "2019 AMC 10B Problems/Problem 19"
m (→Solution) |
|||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}} | ||
| + | |||
==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> | ||
| Line 4: | Line 6: | ||
<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> | ||
| − | ==Solution== | + | ==Solution 1== |
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>S</math> and factor it. Factoring, we find <math>S^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>S^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>S^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>S</math> and factor it. Factoring, we find <math>S^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>S^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>S^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</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) | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</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>. 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 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}} | ||
| + | {{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:03, 14 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 . 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
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. 
