Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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Notice that this is similar to choosing a divisor of <math>100,000^2 = 2^{10}5^{10}</math>, which has <math>(10+1)(10+1) = 121</math> divisors. 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 cannot be so written because the maximum factor of <math>100,000</math> containing only <math>2</math>s or <math>5</math>s (and not both) is only <math>2^5</math> or <math>5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. The first two would require <math>1 \cdot 1</math> and <math>2^{5}5^{5} \cdot 2^{5}5^{5}</math>, respectively. 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 <math>0</math> or <math>10</math>, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>. | Notice that this is similar to choosing a divisor of <math>100,000^2 = 2^{10}5^{10}</math>, which has <math>(10+1)(10+1) = 121</math> divisors. 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 cannot be so written because the maximum factor of <math>100,000</math> containing only <math>2</math>s or <math>5</math>s (and not both) is only <math>2^5</math> or <math>5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. The first two would require <math>1 \cdot 1</math> and <math>2^{5}5^{5} \cdot 2^{5}5^{5}</math>, respectively. 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 <math>0</math> or <math>10</math>, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>. | ||
− | + | ==Video Solution== | |
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https://youtu.be/ZhAZ1oPe5Ds?t=3975 | https://youtu.be/ZhAZ1oPe5Ds?t=3975 | ||
Revision as of 08:25, 25 October 2022
- 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
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 not both or , can be written as a product of two distinct elements in . Hence the answer is .
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=3975
~ pi_is_3.14
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.