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 "2009 AIME I Problems/Problem 9"
(→‎Video Solution)
 
(One intermediate revision by the same user not shown)
Line 2: Line 2:
 
A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from <math>\$ 1</math> to <math>\$ 9999</math> inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were <math>1, 1, 1, 1, 3, 3, 3</math>. Find the total number of possible guesses for all three prizes consistent with the hint.
 
A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from <math>\$ 1</math> to <math>\$ 9999</math> inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were <math>1, 1, 1, 1, 3, 3, 3</math>. Find the total number of possible guesses for all three prizes consistent with the hint.
  
−
== Solution ==
+
== Solution 1 ==
 
[Clarification: You are supposed to find the number of all possible tuples of prices, <math>(A, B, C)</math>, that could have been on that day.]
 
[Clarification: You are supposed to find the number of all possible tuples of prices, <math>(A, B, C)</math>, that could have been on that day.]
  
Line 24: Line 24:
  
 
Thus, each arrangement has <cmath>\binom{6}{2}-3=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways.
 
Thus, each arrangement has <cmath>\binom{6}{2}-3=12</cmath> ways per arrangement, and there are <math>12\times35=\boxed{420}</math> ways.
 +
 +
== Solution 1a (Casework)==
 +
 +
Follow Solution 1 until the partitioning of all the possible strings into 3 groups. Another way to partition the strings is by casework.
 +
 +
Case 1: one of [A, B, or C] has one digit, another has two digits, and the last has four digits.
 +
There are <math>3!=6</math> ways this can happen.
 +
 +
Case 2: one of [A, B, or C] has two digit, another has two digits, and the last has three digits.
 +
There are <math>3!/2=3</math> ways this can happen.
 +
 +
Case 3: one of [A, B, or C] has one digit, another has three digits, and the last has three digits.
 +
There are <math>3!/2=3</math> ways this can happen.
 +
 +
The total numbers of ways per arrangement is <math>6+3+3=12</math> ways. Following Solution 1, there are <math>12\times35=\boxed{420}</math> total ways.
 +
 +
-unhappyfarmer
  
 
==Video Solution==
 
==Video Solution==
−
https://youtu.be/VhyLeQufKr8
+
https://youtu.be/VhyLeQufKr8 (unavailable)
  
 
== See also ==
 
== See also ==

Latest revision as of 12:42, 13 September 2025

Problem

A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $$ 1$ to $$ 9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$. Find the total number of possible guesses for all three prizes consistent with the hint.

Solution 1

[Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$, that could have been on that day.]

Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$, then the string is

\[1131331.\]

Since the strings have seven digits and three threes, there are $\binom{7}{3}=35$ arrangements of all such strings.

In order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.

Let's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to

\[x+y+z=7, x,y,z>0.\]

This gives us

\[\binom{6}{2}=15\]

ways by stars and bars. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$.

Thus, each arrangement has \[\binom{6}{2}-3=12\] ways per arrangement, and there are $12\times35=\boxed{420}$ ways.

Solution 1a (Casework)

Follow Solution 1 until the partitioning of all the possible strings into 3 groups. Another way to partition the strings is by casework.

Case 1: one of [A, B, or C] has one digit, another has two digits, and the last has four digits. There are $3!=6$ ways this can happen.

Case 2: one of [A, B, or C] has two digit, another has two digits, and the last has three digits. There are $3!/2=3$ ways this can happen.

Case 3: one of [A, B, or C] has one digit, another has three digits, and the last has three digits. There are $3!/2=3$ ways this can happen.

The total numbers of ways per arrangement is $6+3+3=12$ ways. Following Solution 1, there are $12\times35=\boxed{420}$ total ways.

-unhappyfarmer

Video Solution

https://youtu.be/VhyLeQufKr8 (unavailable)

See also

2009 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_9&oldid=260537"