Difference between revisions of "2020 AMC 12A Problems/Problem 21"
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There can be anywhere between <math>3</math> and <math>8</math> <math>2</math>'s and <math>1</math> to <math>4</math> <math>3</math>'s. However, since <math>n</math> is a multiple of <math>5</math>, and we multiply the <math>\text{gcd}</math> by <math>5</math>, there can only be <math>3</math> <math>5</math>'s in <math>n</math>'s prime factorization. Finally, there can either <math>0</math> or <math>1</math> <math>7</math>'s. | There can be anywhere between <math>3</math> and <math>8</math> <math>2</math>'s and <math>1</math> to <math>4</math> <math>3</math>'s. However, since <math>n</math> is a multiple of <math>5</math>, and we multiply the <math>\text{gcd}</math> by <math>5</math>, there can only be <math>3</math> <math>5</math>'s in <math>n</math>'s prime factorization. Finally, there can either <math>0</math> or <math>1</math> <math>7</math>'s. | ||
| − | Thus, we can multiply the total possibilities of <math>n</math>'s factorization to determine the number of integers <math>n</math> which satisfy the equation, giving us <math>6 \times 4 \times 1 \times 2 = \boxed{\textbf{(D) } 48}</math>. ~ciceronii | + | Thus, we can multiply the total possibilities of <math>n</math>'s factorization to determine the number of integers <math>n</math> which satisfy the equation, giving us <math>6 \times 4 \times 1 \times 2 = \boxed{\textbf{(D) } 48}</math>. ~ciceronii osas |
{{AMC12 box|year=2020|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2020|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 03:28, 20 March 2020
Problem 21
How many positive integers 
are there such that is a multiple of 
, and the least common multiple of 
and equals times the greatest common divisor of 
and 
Solution
We set up the following equation as the problem states:
![\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\]](http://latex.artofproblemsolving.com/e/4/1/e4182bee46f971119822c1847de1c6dd5e26a37f.png)
Breaking each number into its prime factorization, we see that the equation becomes
![\[\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}.\]](http://latex.artofproblemsolving.com/3/5/6/356f8d9a66c1cdd0ad995e7193a7bed51528f404.png)
We can now determine the prime factorization of . We know that its prime factors belong to the set 
, as no factor of has 
in its prime factorization, nor anything greater. Next, we must find exactly how many different possibilities exist for each.
There can be anywhere between 
and 
's and 
to 
's. However, since is a multiple of , and we multiply the 
by , there can only be 's in 's prime factorization. Finally, there can either 
or 
's.
Thus, we can multiply the total possibilities of 's factorization to determine the number of integers which satisfy the equation, giving us 
. ~ciceronii osas
| 2020 AMC 12A (Problems • Answer Key • Resources) | |
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Followed by Problem 22 |
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