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 "1987 AIME Problems/Problem 7"

(solution)
m (→‎Solution 1)
 
(19 intermediate revisions by 9 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Let <math>\displaystyle [r,s]</math> denote the [[least common multiple]] of [[positive]] [[integer]]s <math>\displaystyle r</math> and <math>\displaystyle s</math>.  Find the number of ordered triples <math>\displaystyle (a,b,c)</math> of positive integers for which <math>\displaystyle [a,b] = 1000</math>, <math>\displaystyle [b,c] = 2000</math>, and <math>\displaystyle [c,a] = 2000</math>.
+
Let <math>[r,s]</math> denote the [[least common multiple]] of [[positive integer]]s <math>r</math> and <math>s</math>.  Find the number of [[ordered tuple | ordered triples]] <math>(a,b,c)</math> of positive integers for which <math>[a,b] = 1000</math>, <math>[b,c] = 2000</math>, and <math>[c,a] = 2000</math>.
  
−
== Solution ==
+
__TOC__
−
<math>\displaystyle 1000 = 2^35^3</math> and <math>2000 = 2^45^3</math>. By [[LCM#Using prime factorization|looking at the prime factorization]] of <math>2000</math>, <math>c</math> must have a [[factor]] of <math>2^4</math>. If <math>c</math> has a factor of <math>5^3</math>, then there are two [[casework|cases]]: either (1) <math>a</math> or <math>b = 5^32^3</math>, or (2) one of <math>a</math> and <math>b</math> has a factor of <math>5^3</math> and the other a factor of <math>2^3</math>. For case 1, the other number will be in the form of <math>2^x5^y</math>, so there are <math>4 \cdot 4 = 16</math> possible such numbers; since this can be either <math>a</math> or <math>b</math> there are a total of <math>2(16)-1=31</math> possibilities. For case 2, <math>a</math> and <math>b</math> are in the form of <math>2^35^x</math> and <math>2^y5^3</math>, with <math>x < 3</math> and <math>y < 3</math> (if they were equal to 3, it would overlap with case 1). Thus, there are <math>2(3 \cdot 3) = 18</math> cases.
+
==Solution 1==
 +
It's clear that we must have <math>a = 2^j5^k</math>, <math>b = 2^m 5^n</math> and <math>c = 2^p5^q</math> for some [[nonnegative]] [[integer]]s <math>j, k, m, n, p, q</math>.  Dealing first with the powers of 2: from the given conditions, <math>\max(j, m) = 3</math>, <math>\max(m, p) = \max(p, j) = 4</math>. Thus we must have <math>p = 4</math> and at least one of <math>m, j</math> equal to 3.  This gives 7 possible triples <math>(j, m, p)</math>: <math>(0, 3, 4), (1, 3, 4), (2, 3, 4), (3, 3, 4), (3, 2, 4), (3, 1, 4)</math> and <math>(3, 0, 4)</math>.
  
−
If <math>c</math> does not have a factor of <math>5^3</math>, then at least one of <math>a</math> and <math>b</math> must be <math>2^35^3</math>, and both must have a factor of <math>5^3</math>. Then, there are <math>4</math> solutions possible just considering <math>a = 2^35^3</math>, and a total of <math>4 \cdot 2 - 1 = 7</math> possibilities. Multiplying by three, as <math>0 \le c \le 2</math>, there are <math>7 \cdot 3 = 21</math>. Together, that makes <math>31 + 18 + 24 = 70</math> solutions for <math>(a, b, c)</math>.
+
Now, for the powers of 5: we have <math>\max(k, n) = \max(n, q) = \max(q, k) = 3</math>.  Thus, at least two of <math>k, n, q</math> must be equal to 3, and the other can take any value between 0 and 3.  This gives us a total of 10 possible triples: <math>(3, 3, 3)</math> and three possibilities of each of the forms <math>(3, 3, n)</math>, <math>(3, n, 3)</math> and <math>(n, 3, 3)</math>.
 +
 
 +
Since the [[exponent]]s of 2 and 5 must satisfy these conditions independently, we have a total of <math>7 \cdot 10 = 70</math> possible valid triples.
 +
 
 +
==Solution 2==
 +
 
 +
<math>1000 = 2^35^3</math> and <math>2000 = 2^45^3</math>. By [[LCM#Using prime factorization|looking at the prime factorization]] of <math>2000</math>, <math>c</math> must have a [[factor]] of <math>2^4</math>. If <math>c</math> has a factor of <math>5^3</math>, then there are two [[casework|cases]]: either (1) <math>a</math> or <math>b = 5^32^3</math>, or (2) one of <math>a</math> and <math>b</math> has a factor of <math>5^3</math> and the other a factor of <math>2^3</math>. For case 1, the other number will be in the form of <math>2^x5^y</math>, so there are <math>4 \cdot 4 = 16</math> possible such numbers; since this can be either <math>a</math> or <math>b</math> there are a total of <math>2(16)-1=31</math> possibilities. For case 2, <math>a</math> and <math>b</math> are in the form of <math>2^35^x</math> and <math>2^y5^3</math>, with <math>x < 3</math> and <math>y < 3</math> (if they were equal to 3, it would overlap with case 1). Thus, there are <math>2(3 \cdot 3) = 18</math> cases.
 +
 
 +
If <math>c</math> does not have a factor of <math>5^3</math>, then at least one of <math>a</math> and <math>b</math> must be <math>2^35^3</math>, and both must have a factor of <math>5^3</math>. Then, there are <math>4</math> solutions possible just considering <math>a = 2^35^3</math>, and a total of <math>4 \cdot 2 - 1 = 7</math> possibilities. Multiplying by three, as <math>0 \le c \le 2</math>, there are <math>7 \cdot 3 = 21</math>. Together, that makes <math>31 + 18 + 21 = 70</math> solutions for <math>(a, b, c)</math>.<!--
 +
{| border="1px" style="width:25%" align=center
 +
|-
 +
! c || a || b || solutions
 +
|-
 +
| rowspan = 2 | <math>5^32^4</math>
 +
| <math>2^35^3</math>
 +
| <math>5^x2^y</math>
 +
| 31
 +
|-
 +
| <math>2^x5^3</math> || <math>2^35^y</math> || 18
 +
|-
 +
| rowspan = 1 | <math>5^x2^4</math> || <math>2^35^3</math> || <math>5^y2^z</math> || 24
 +
|}-->
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1987|num-b=6|num-a=8}}
 
{{AIME box|year=1987|num-b=6|num-a=8}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 +
[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 09:00, 18 June 2024

Problem

Let $[r,s]$ denote the least common multiple of positive integers $r$ and $s$. Find the number of ordered triples $(a,b,c)$ of positive integers for which $[a,b] = 1000$, $[b,c] = 2000$, and $[c,a] = 2000$.

Solution 1

It's clear that we must have $a = 2^j5^k$, $b = 2^m 5^n$ and $c = 2^p5^q$ for some nonnegative integers $j, k, m, n, p, q$. Dealing first with the powers of 2: from the given conditions, $\max(j, m) = 3$, $\max(m, p) = \max(p, j) = 4$. Thus we must have $p = 4$ and at least one of $m, j$ equal to 3. This gives 7 possible triples $(j, m, p)$: $(0, 3, 4), (1, 3, 4), (2, 3, 4), (3, 3, 4), (3, 2, 4), (3, 1, 4)$ and $(3, 0, 4)$.

Now, for the powers of 5: we have $\max(k, n) = \max(n, q) = \max(q, k) = 3$. Thus, at least two of $k, n, q$ must be equal to 3, and the other can take any value between 0 and 3. This gives us a total of 10 possible triples: $(3, 3, 3)$ and three possibilities of each of the forms $(3, 3, n)$, $(3, n, 3)$ and $(n, 3, 3)$.

Since the exponents of 2 and 5 must satisfy these conditions independently, we have a total of $7 \cdot 10 = 70$ possible valid triples.

Solution 2

$1000 = 2^35^3$ and $2000 = 2^45^3$. By looking at the prime factorization of $2000$, $c$ must have a factor of $2^4$. If $c$ has a factor of $5^3$, then there are two cases: either (1) $a$ or $b = 5^32^3$, or (2) one of $a$ and $b$ has a factor of $5^3$ and the other a factor of $2^3$. For case 1, the other number will be in the form of $2^x5^y$, so there are $4 \cdot 4 = 16$ possible such numbers; since this can be either $a$ or $b$ there are a total of $2(16)-1=31$ possibilities. For case 2, $a$ and $b$ are in the form of $2^35^x$ and $2^y5^3$, with $x < 3$ and $y < 3$ (if they were equal to 3, it would overlap with case 1). Thus, there are $2(3 \cdot 3) = 18$ cases.

If $c$ does not have a factor of $5^3$, then at least one of $a$ and $b$ must be $2^35^3$, and both must have a factor of $5^3$. Then, there are $4$ solutions possible just considering $a = 2^35^3$, and a total of $4 \cdot 2 - 1 = 7$ possibilities. Multiplying by three, as $0 \le c \le 2$, there are $7 \cdot 3 = 21$. Together, that makes $31 + 18 + 21 = 70$ solutions for $(a, b, c)$.

See also

1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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=1987_AIME_Problems/Problem_7&oldid=220722"