Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AIME I Problems/Problem 9"

(→‎Solution)
 
(13 intermediate revisions by 10 users not shown)
Line 3: Line 3:
 
Let <math>S</math> be the set of positive integer divisors of <math>20^9.</math> Three numbers are chosen independently and at random with replacement from the set <math>S</math> and labeled <math>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> divides <math>a_3</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m.</math>
 
Let <math>S</math> be the set of positive integer divisors of <math>20^9.</math> Three numbers are chosen independently and at random with replacement from the set <math>S</math> and labeled <math>a_1,a_2,</math> and <math>a_3</math> in the order they are chosen. The probability that both <math>a_1</math> divides <math>a_2</math> and <math>a_2</math> divides <math>a_3</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m.</math>
  
−
== Solution ==
+
== Solution 1 ==
 +
 
 +
<asy>
 +
size(12cm);
 +
for (int x = 1; x < 18; ++x) {
 +
    draw((x, 0) -- (x, 9), dotted);
 +
}
 +
for (int y = 1; y < 9; ++y) {
 +
    draw((0, y) -- (18, y), dotted);
 +
}
 +
 
 +
draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle);
 +
 
 +
pair b1, b2, b3;
 +
pair c1, c2, c3;
 +
pair a1, a2, a3;
 +
b1 = (3, 0); b2 = (12, 0); b3 = (16, 0);
 +
c1 = (0, 2); c2 = (0, 4); c3 = (0, 8);
 +
a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3;
 +
 
 +
draw(b1 -- a1 -- c1);
 +
draw(b2 -- a2 -- c2);
 +
draw(b3 -- a3 -- c3);
 +
 
 +
dot(a1); dot(a2); dot(a3);
 +
label("$a_1$", a1, NE);
 +
label("$a_2$", a2, NE);
 +
label("$a_3$", a3, NE);
 +
label("$b_1$", b1, S);
 +
label("$b_2$", b2, S);
 +
label("$b_3$", b3, S);
 +
label("$c_1$", c1, W);
 +
label("$c_2$", c2, W);
 +
label("$c_3$", c3, W);
 +
 
 +
</asy>
  
 
First, prime factorize <math>20^9</math> as <math>2^{18} \cdot 5^9</math>. Denote <math>a_1</math> as <math>2^{b_1} \cdot 5^{c_1}</math>, <math>a_2</math> as <math>2^{b_2} \cdot 5^{c_2}</math>, and <math>a_3</math> as <math>2^{b_3} \cdot 5^{c_3}</math>.
 
First, prime factorize <math>20^9</math> as <math>2^{18} \cdot 5^9</math>. Denote <math>a_1</math> as <math>2^{b_1} \cdot 5^{c_1}</math>, <math>a_2</math> as <math>2^{b_2} \cdot 5^{c_2}</math>, and <math>a_3</math> as <math>2^{b_3} \cdot 5^{c_3}</math>.
  
−
In order for <math>a_1</math> to divide <math>a_2</math>, and for <math>a_2</math> to divide <math>a_3</math>, <math>b_1\geb_2\geb_3</math>, and <math>c_1\egc_2\ge_3</math>. We will consider each case separately. Note that the total amount of possibilities is <math>190^3</math>, as there are <math>(18+1)(9+1)=190</math> choices for each factor.
+
In order for <math>a_1</math> to divide <math>a_2</math>, and for <math>a_2</math> to divide <math>a_3</math>, <math>b_1\le b_2\le b_3</math>, and <math>c_1\le c_2\le c_3</math>. We will consider each case separately. Note that the total amount of possibilities is <math>190^3</math>, as there are <math>(18+1)(9+1)=190</math> choices for each factor.
  
−
We notice that if we add <math>1</math> to <math>b_2</math> and <math>2</math> to <math>b_3</math>, then we can reach the stronger inequality <math>b_1<b_2<b_3</math>. Therefore, if we pick <math>3</math> integers from <math>0</math> to <math>20</math>, they will correspond to a unique solution, forming a 1-1 correspondence. The amount of solutions to this inequality is <math>\dbinom{21}{3}</math>.
+
We notice that if we add <math>1</math> to <math>b_2</math> and <math>2</math> to <math>b_3</math>, then we can reach the stronger inequality <math>0\le b_1<b_2+1<b_3+2\le 20</math>. Therefore, if we pick <math>3</math> integers from <math>0</math> to <math>20</math>, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers <math>b_1</math>, <math>b_2+1</math>, and <math>b_3+2</math>. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is <math>\dbinom{21}{3}</math>.
  
−
The case for <math>c_1</math>,<math>c_2</math>, and <math>c_3</math> proceeds similarly for a result of <math>\dbinom{12}{3}</math>. Therefore, the probability of choosing three such factors is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{77}</math>.
+
The case for <math>c_1</math>,<math>c_2</math>, and <math>c_3</math> proceeds similarly for a result of <math>\dbinom{12}{3}</math>. Therefore, the probability of choosing three such factors is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>.
  
 
-molocyxu
 
-molocyxu
 +
 +
== Solution 2==
 +
 +
Same as before, say the factors have powers of <math>b</math> and <math>c</math>. <math>b_1, b_2, b_3</math> can either be all distinct, all equal, or two of the three are equal. As well, we must have <math>b_1 \leq b_2 \leq b_3</math>. If they are all distinct, the number of cases is simply <math>{19 \choose 3}</math>. If they are all equal, there are only <math>19</math> cases for the general value. If we have a pair equal, then we have <math>2 \cdot {19\choose 2}</math>. We need to multiply by <math>2</math> because if we have two values <math>b_i < b_j</math>, we can have either <math>(b_i, b_i, b_j)</math> or <math>(b_i, b_j, b_j)</math>.
 +
 +
<cmath>{19 \choose 3} + 2 \cdot {19 \choose 2} + 19 = 1330</cmath>
 +
 +
Likewise for <math>c</math>, we get
 +
 +
<cmath>{10 \choose 3} + 2 \cdot {10 \choose 2} + 10 = 220</cmath>
 +
 +
The final probability is simply <math>\frac{1330 \cdot 220}{190^3}</math>. Simplification gives <math>\frac{77}{1805}</math>, and therefore the answer is <math>\boxed{077}</math>.
 +
 +
==Solution 3==
 +
 +
Similar to before, we calculate that there are <math>190^3</math> ways to choose <math>3</math> factors with replacement. Then, we figure out the number of triplets <math>{a,b,c}</math> and <math>{d,f,g}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> represent powers of <math>2</math> and <math>d</math>, <math>f</math>, and <math>g</math> represent powers of <math>5</math>, such that the triplets are in non-descending order. The maximum power of <math>2</math> is <math>18</math>, and the maximum power of <math>5</math> is <math>9</math>. Using the Hockey Stick identity, we figure out that there are <math>\dbinom{12}{3}</math> ways to choose <math>d</math>, <math>f</math> and <math>g</math>, and <math>\dbinom{21}{3}</math> ways to choose <math>a</math>, <math>b</math>, and <math>c</math>. Therefore, the probability of choosing <math>3</math> factors which satisfy the conditions is <cmath>\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.</cmath> This simplifies to <math>\frac{77}{1805}</math>, therefore <math>m =</math> <math>\boxed{077}</math>.
 +
 +
==Solution 4 (Official MAA)==
 +
Note that the prime factorization of <math>20^9</math> is <math>2^{18}\cdot5^{9}.</math> The problem reduces to selecting independently the powers of <math>2</math> and the powers of <math>5</math> in the numbers <math>a_1</math>, <math>a_2</math>, and <math>a_3</math>. This is equivalent to selecting <math>3</math> exponents for the powers of <math>2</math> and <math>3</math> exponents for the powers of <math>5</math> and determining in each case the probability that the 3 exponents are chosen in nondecreasing order. Given a positive integer <math>k</math>, the probability that three integers <math>a</math>, <math>b</math>, and <math>c</math> are chosen such that <math>0\le a \le b\le c \le k</math> is the probability that <math>a</math>, <math>b+1</math>, and <math>c+2</math> are chosen such that <math>0 \le a < b+1 < c+2 \le k+2</math>. There are <math>\binom {k+3}3</math> ways to choose <math>a</math>, <math>b+1</math>, and <math>c+2</math>, so the probability that the integers are chosen in order is
 +
<cmath>\frac{\binom{k+3}3}{(k+1)^3}.</cmath>Thus the probability that three chosen divisors of <math>20^9</math> satisfy the divisibility requirement is
 +
<cmath>\frac{\binom{12}3}{10^3}\cdot\frac{\binom{21}3}{19^3}=\frac{12\cdot11\cdot10}{6\cdot10\cdot10\cdot10}\cdot\frac{21\cdot20\cdot19}{6\cdot19\cdot19\cdot19}=\frac{77}{1805}.</cmath>The requested numerator is <math>77.</math>
  
 
==See Also==
 
==See Also==
  
 
{{AIME box|year=2020|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2020|n=I|num-b=8|num-a=10}}
 +
 +
[[Category: Intermediate Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:59, 24 February 2021

Problem

Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$

Solution 1

[asy] size(12cm); for (int x = 1; x < 18; ++x) { draw((x, 0) -- (x, 9), dotted); } for (int y = 1; y < 9; ++y) { draw((0, y) -- (18, y), dotted); } draw((0, 0) -- (18, 0) -- (18, 9) -- (0, 9) -- cycle); pair b1, b2, b3; pair c1, c2, c3; pair a1, a2, a3; b1 = (3, 0); b2 = (12, 0); b3 = (16, 0); c1 = (0, 2); c2 = (0, 4); c3 = (0, 8); a1 = b1 + c1; a2 = b2 + c2; a3 = b3 + c3; draw(b1 -- a1 -- c1); draw(b2 -- a2 -- c2); draw(b3 -- a3 -- c3); dot(a1); dot(a2); dot(a3); label("$a_1$", a1, NE); label("$a_2$", a2, NE); label("$a_3$", a3, NE); label("$b_1$", b1, S); label("$b_2$", b2, S); label("$b_3$", b3, S); label("$c_1$", c1, W); label("$c_2$", c2, W); label("$c_3$", c3, W); [/asy]

First, prime factorize $20^9$ as $2^{18} \cdot 5^9$. Denote $a_1$ as $2^{b_1} \cdot 5^{c_1}$, $a_2$ as $2^{b_2} \cdot 5^{c_2}$, and $a_3$ as $2^{b_3} \cdot 5^{c_3}$.

In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\le b_2\le b_3$, and $c_1\le c_2\le c_3$. We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor.

We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $0\le b_1<b_2+1<b_3+2\le 20$. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers $b_1$, $b_2+1$, and $b_3+2$. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is $\dbinom{21}{3}$.

The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\dbinom{12}{3}$. Therefore, the probability of choosing three such factors is \[\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.\] Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{077}$.

-molocyxu

Solution 2

Same as before, say the factors have powers of $b$ and $c$. $b_1, b_2, b_3$ can either be all distinct, all equal, or two of the three are equal. As well, we must have $b_1 \leq b_2 \leq b_3$. If they are all distinct, the number of cases is simply ${19 \choose 3}$. If they are all equal, there are only $19$ cases for the general value. If we have a pair equal, then we have $2 \cdot {19\choose 2}$. We need to multiply by $2$ because if we have two values $b_i < b_j$, we can have either $(b_i, b_i, b_j)$ or $(b_i, b_j, b_j)$.

\[{19 \choose 3} + 2 \cdot {19 \choose 2} + 19 = 1330\]

Likewise for $c$, we get

\[{10 \choose 3} + 2 \cdot {10 \choose 2} + 10 = 220\]

The final probability is simply $\frac{1330 \cdot 220}{190^3}$. Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{077}$.

Solution 3

Similar to before, we calculate that there are $190^3$ ways to choose $3$ factors with replacement. Then, we figure out the number of triplets ${a,b,c}$ and ${d,f,g}$, where $a$, $b$, and $c$ represent powers of $2$ and $d$, $f$, and $g$ represent powers of $5$, such that the triplets are in non-descending order. The maximum power of $2$ is $18$, and the maximum power of $5$ is $9$. Using the Hockey Stick identity, we figure out that there are $\dbinom{12}{3}$ ways to choose $d$, $f$ and $g$, and $\dbinom{21}{3}$ ways to choose $a$, $b$, and $c$. Therefore, the probability of choosing $3$ factors which satisfy the conditions is \[\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.\] This simplifies to $\frac{77}{1805}$, therefore $m =$ $\boxed{077}$.

Solution 4 (Official MAA)

Note that the prime factorization of $20^9$ is $2^{18}\cdot5^{9}.$ The problem reduces to selecting independently the powers of $2$ and the powers of $5$ in the numbers $a_1$, $a_2$, and $a_3$. This is equivalent to selecting $3$ exponents for the powers of $2$ and $3$ exponents for the powers of $5$ and determining in each case the probability that the 3 exponents are chosen in nondecreasing order. Given a positive integer $k$, the probability that three integers $a$, $b$, and $c$ are chosen such that $0\le a \le b\le c \le k$ is the probability that $a$, $b+1$, and $c+2$ are chosen such that $0 \le a < b+1 < c+2 \le k+2$. There are $\binom {k+3}3$ ways to choose $a$, $b+1$, and $c+2$, so the probability that the integers are chosen in order is \[\frac{\binom{k+3}3}{(k+1)^3}.\]Thus the probability that three chosen divisors of $20^9$ satisfy the divisibility requirement is \[\frac{\binom{12}3}{10^3}\cdot\frac{\binom{21}3}{19^3}=\frac{12\cdot11\cdot10}{6\cdot10\cdot10\cdot10}\cdot\frac{21\cdot20\cdot19}{6\cdot19\cdot19\cdot19}=\frac{77}{1805}.\]The requested numerator is $77.$

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_9&oldid=147906"