2025 AIME I Problems/Problem 10

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

Solution 2

We will call the sudoku blocks 1-3 from left to right, and rows 1-3 from up to down. Lets start from the entire first block - we have 9! options of choosing it as you just arrange the numbers.

Starting from the second block, we need to do some casework. We need to discuss the different possibilities that the first row of the second block contains.

Case 1: The first row of the second block has the exact 3 numbers from the second row of first block.

We have 3P3 choices as you just arrange the numbers for the first row.

Then, considering the second row of the second block, you must take the exact 3 numbers from the third row of first block, and we could arrange then in different order. This will gives another 3P3 choices.

The third row of the second block must contain the 3 numbers of the first row in first block. We will then have another 3P3 of arranging the numbers.

Summing up case 1, we have 3P3 * 3P3 * 3P3 = 6^3 = 2^3 * 3^3.

Case 2: The first row of second block contains 2 numbers from the second row of first block and 1 number from the third row of first block.

To satisfy my case 2 statement, we will need 3C1 * 3C2 * 3P3.

To look at the second row of the second block, we must first choose the remaining 2 numbers that haven't got picked from the third row of first block, or else the sudoku won't work. This means that we can only pick 1 remaining number from the numbers in first row of first block - because the 6th number is the number in the second row and sudoku doesn't allow repeated numbers in a row. Therefore, the second row has the choice of 2C2 * 3C1 * 3P3.

The third row of the second block gets the remaining 3 numbers that is left from the second block, which is 3P3.

Summing up case 2, we have 3C1 * 3C2 * 3P3 * 2C2 * 3C1 * 3P3 * 3P3 = 3^3 * 2^3 * 3^3 = 2^3 * 3^6.

Notice that case 3 (3 numbers repeating from the third row of first block) and case 4 (2 numbers from block 1 row 3 and 1 number from block 1 row 2) are symmetric to case 1 and 2, respectively. Therefore the second block results with (1 + 3^3 + 1 + 3^3) * 2^3 * 3^3 = 56 * 2^3 * 3^3 = 7 * 2^6 * 3^3

Third block: easy

Notice that the 6 numbers of each row has been filled already, the choices of the rest 3 numbers has been done. You only need to arrange them in different orders, which gives 3P3 * 3P3 * 3P3 = 2^3 * 3^3.

In total, our answer is: 9! * 7 * 2^6 * 3^3 * 2^3 * 3^3 = 2 * 3 * 2^2 * 5 * 2 * 3 * 7 * 2^3 * 3^2 * 7 * 2^9 * 3^6 = 2^16 * 3^10 * 5 * 7^2 Our required answer is therefore 2*16 + 3*10 + 5*1 + 7*2 = 32 + 30 + 5 + 14 = 081

(Can anyone fix this poor latex please and thank you) ~Mitsuihisashi14

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2025 AIME I Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See also