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 "2025 AIME I Problems/Problem 10"
Line 46: Line 46:
  
 
The number of different ways to fill such a grid can be written as <math>p^a \cdot q^b \cdot r^c \cdot s^d</math> where <math>p</math>, <math>q</math>, <math>r</math>, and <math>s</math> are distinct prime numbers and <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> are positive integers. Find <math>p \cdot a + q \cdot b + r \cdot c + s \cdot d</math>.
 
The number of different ways to fill such a grid can be written as <math>p^a \cdot q^b \cdot r^c \cdot s^d</math> where <math>p</math>, <math>q</math>, <math>r</math>, and <math>s</math> are distinct prime numbers and <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> are positive integers. Find <math>p \cdot a + q \cdot b + r \cdot c + s \cdot d</math>.
 +
 +
==Solution 1==
 +
 +
We will fill out the grid row by row. Note that there are <math>9!=2^7\cdot 3^4\cdot 5\cdot 7</math> ways to fill out the first row. For the second row, we will consider a little casework. WLOG let the first row be <math>123 | 456 | 789</math> (bars indicate between the <math>3\times 3</math> squares).
 +
 +
<u>Case 1:</u> Every number from the first square goes to the second square. Then there is <math>1</math> way along with <math>6^3</math> orderings giving us <math>6^3</math> cases.
 +
 +
<u>Case 2:</u> Two of the first numbers go in the second square and the last number goes in the third. This results in <math>3*3*3</math> ways to put the numbers along with <math>6^3</math> orderings. Thus there are <math>27\cdot 6^3</math> cases.
 +
 +
Note that we didn't consider <math>2</math> cases but these are symmetric so the number of ways to fill the second row is <math>56\cdot 6^3</math>.
 +
 +
There is <math>1</math> way to place the last numbers in with <math>6^3</math> orderings. Thus there are <math>2^7\cdot 3^4\cdot 5\cdot 7 \cdot 56\cdot 6^3 \cdot 6^3 = 2^{16}\cdot 3^{10}\cdot 5^1\cdot 7^2</math> ways to fill out our mini Sudoku puzzle. Thus the answer is <math>2\cdot 16+3\cdot 10 + 5\cdot 1+7\cdot 2 = \boxed{81}.</math>
 +
 +
~ [[User:zhenghua|zhenghua]]
  
 
==See also==
 
==See also==

Revision as of 21:02, 13 February 2025

Problem

The $27$ cells of a $3 \times 9$ grid are filled in using the numbers $1$ through $9$ so that each row contains $9$ different numbers, and each of the three $3 \times 3$ blocks heavily outlined in the example below contains $9$ different numbers, as in the first three rows of a Sudoku puzzle.

[asy] unitsize(20); add(grid(9,3)); draw((0,0)--(9,0)--(9,3)--(0,3)--cycle, linewidth(2)); draw((3,0)--(3,3), linewidth(2)); draw((6,0)--(6,3), linewidth(2)); real a = 0.5; label("5",(a,a)); label("6",(1+a,a)); label("1",(2+a,a)); label("8",(3+a,a)); label("4",(4+a,a)); label("7",(5+a,a)); label("9",(6+a,a)); label("2",(7+a,a)); label("3",(8+a,a)); label("3",(a,1+a)); label("7",(1+a,1+a)); label("9",(2+a,1+a)); label("5",(3+a,1+a)); label("2",(4+a,1+a)); label("1",(5+a,1+a)); label("6",(6+a,1+a)); label("8",(7+a,1+a)); label("4",(8+a,1+a)); label("4",(a,2+a)); label("2",(1+a,2+a)); label("8",(2+a,2+a)); label("9",(3+a,2+a)); label("6",(4+a,2+a)); label("3",(5+a,2+a)); label("1",(6+a,2+a)); label("7",(7+a,2+a)); label("5",(8+a,2+a)); [/asy]

The number of different ways to fill such a grid can be written as $p^a \cdot q^b \cdot r^c \cdot s^d$ where $p$, $q$, $r$, and $s$ are distinct prime numbers and $a$, $b$, $c$, $d$ are positive integers. Find $p \cdot a + q \cdot b + r \cdot c + s \cdot d$.

Solution 1

We will fill out the grid row by row. Note that there are $9!=2^7\cdot 3^4\cdot 5\cdot 7$ ways to fill out the first row. For the second row, we will consider a little casework. WLOG let the first row be $123 | 456 | 789$ (bars indicate between the $3\times 3$ squares).

Case 1: Every number from the first square goes to the second square. Then there is $1$ way along with $6^3$ orderings giving us $6^3$ cases.

Case 2: Two of the first numbers go in the second square and the last number goes in the third. This results in $3*3*3$ ways to put the numbers along with $6^3$ orderings. Thus there are $27\cdot 6^3$ cases.

Note that we didn't consider $2$ cases but these are symmetric so the number of ways to fill the second row is $56\cdot 6^3$.

There is $1$ way to place the last numbers in with $6^3$ orderings. Thus there are $2^7\cdot 3^4\cdot 5\cdot 7 \cdot 56\cdot 6^3 \cdot 6^3 = 2^{16}\cdot 3^{10}\cdot 5^1\cdot 7^2$ ways to fill out our mini Sudoku puzzle. Thus the answer is $2\cdot 16+3\cdot 10 + 5\cdot 1+7\cdot 2 = \boxed{81}.$

~ zhenghua

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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=2025_AIME_I_Problems/Problem_10&oldid=242296"