Difference between revisions of "2019 AMC 10B Problems/Problem 19"
Omerselim1 (talk | contribs) (→Solution) |
|||
(11 intermediate revisions by 6 users not shown) | |||
Line 6: | 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== |
− | |||
− | + | The prime factorization of <math>100,000</math> 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>. | |
− | + | Notice that this is analogous 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>. | |
− | |||
− | |||
==See Also== | ==See Also== | ||
Line 19: | Line 16: | ||
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− |
Latest revision as of 18:04, 14 July 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
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 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 . 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 .
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.