Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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==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> | |
<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> | ||
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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>. | 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 | + | 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 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 by OmegaLearn== | ||
+ | https://youtu.be/ZhAZ1oPe5Ds?t=3975 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Latest revision as of 20:43, 25 June 2023
- 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 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 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 by OmegaLearn
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.