Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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<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== |
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>. | ||
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~hashbrown2009 (revised and fully updated) | ~hashbrown2009 (revised and fully updated) | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Set \( S \) has the number of factors in 100,000. The prime factorization of 100,000 is | ||
| + | <cmath> | ||
| + | 100,000 = 2^5 \times 5^5 | ||
| + | </cmath> | ||
| + | Therefore set \( S \) has 36 elements. | ||
| + | |||
| + | Now, notice how \( 5^n \) will always end in 5, and \( 2^n \) will end in 2, 4, 8, or 6. We also include \( 2^0 \), and we see the unique units digits include 0, 1, 2, 4, 5, 6, and 8. However, most of the multiples of these numbers cannot divide 100,000, so we cancel out most of them and see that most units digits will end in 0, 1, and 5. | ||
| + | |||
| + | So we multiply the elements in \( S \) by 3, to get \( 36 \times 3 = 108 \). | ||
| + | |||
| + | We have 3 units digits that we can distribute to either the factors \( 2^0 \) or \( 2^n \). We can't distribute anything to \( 5^n \) because no matter what we multiply the units digits by, we always result in a units digits of 5, which we already considered. Therefore the number of ways to do this is \( 3^2 = 9 \). | ||
| + | |||
| + | We add 108 and 9 to get \( 108 + 9 =\) <math>\boxed{\textbf{(C) } 117}</math>. | ||
| + | |||
| + | This problem was much harder than anticipated. | ||
| + | |||
| + | ~Pinotation | ||
==Video Solution by OmegaLearn== | ==Video Solution by OmegaLearn== | ||
Latest revision as of 17:09, 3 September 2025
- 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 1
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 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 
.
~hashbrown2009 (revised and fully updated)
Solution 2
Set \( S \) has the number of factors in 100,000. The prime factorization of 100,000 is
![\[100,000 = 2^5 \times 5^5\]](http://latex.artofproblemsolving.com/c/0/3/c032ae693471928b14e52440afa72c28d1d4bcc3.png)
Therefore set \( S \) has 36 elements.
Now, notice how \( 5^n \) will always end in 5, and \( 2^n \) will end in 2, 4, 8, or 6. We also include \( 2^0 \), and we see the unique units digits include 0, 1, 2, 4, 5, 6, and 8. However, most of the multiples of these numbers cannot divide 100,000, so we cancel out most of them and see that most units digits will end in 0, 1, and 5.
So we multiply the elements in \( S \) by 3, to get \( 36 \times 3 = 108 \).
We have 3 units digits that we can distribute to either the factors \( 2^0 \) or \( 2^n \). We can't distribute anything to \( 5^n \) because no matter what we multiply the units digits by, we always result in a units digits of 5, which we already considered. Therefore the number of ways to do this is \( 3^2 = 9 \).
We add 108 and 9 to get \( 108 + 9 =\) .
This problem was much harder than anticipated.
~Pinotation
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
