2025 AIME I Problems/Problem 5
Problem
There are 
eight-digit positive integers that use each of the digits 
exactly once. Let 
be the number of these integers that are divisible by 
. Find the difference between and 
.
Solution 1
Notice that if the 8-digit number is divisible by , it must have an even units digit. Therefore, we can break it up into cases and let the last digit be either 
or 
. Due to symmetry, upon finding the total count of one of these last digit cases (we look at last digit 
here), we may multiply the resulting value by 
.
Now, we just need to find the number of positions of the remaining numbers such that the units digit is and the number is divisible by 
. Denote the odd numbered positions to be 
and the even numbered positions to be 
(recall 
). By the divisibility rule of , we must have:
![\[(a_1 + a_3 + a_5 + a_7) - (a_2 + a_4 + a_6 + 2)\]](http://latex.artofproblemsolving.com/5/f/b/5fb4eedecf4014902707fb5bcb67fe1af24af83b.png)
which is congruent to 
. Therefore, after simplifying, we must have:
![\[a_1 - a_2 + a_3 - a_4 + a_5 - a_6 + a_7\equiv2\hspace{2mm}(\text{mod}\hspace{1mm}11)\]](http://latex.artofproblemsolving.com/5/e/b/5eb8beaf2b4b98cead7f9f52d2573ec4f11160f3.png)
Now consider 
. Therefore,
![\[(a_1 + a_2 + \ldots+ a_7) - 2(a_2 + a_4 + a_6)\equiv2\hspace{2mm}(\text{mod}\hspace{1mm}11)\]](http://latex.artofproblemsolving.com/4/6/2/4629f18645daccc0af27477bf5a059501d0d4471.png)
which means that
![\[a_2 + a_4 + a_6\equiv5\hspace{2mm}(\text{mod}\hspace{1mm}11)\]](http://latex.artofproblemsolving.com/0/d/f/0df1c20207d2067e76cbf68f2c23ba727c53216d.png)
Notice that the minimum of 
is 
and the maximum is 
. The only possible number congruent to 
in this range is 
. All that remains is to count all the possible sums of using the values 
. There are a total of four possibilities:
![\[(1, 7, 8), (3, 5, 8), (3, 6, 7), (4, 5, 7)\]](http://latex.artofproblemsolving.com/9/d/8/9d8093a83a564f93f689cebba169b9a78f1b1d09.png)
The arrangement of the odd-positioned numbers (
) does not matter, so there are 
arrangements of these numbers. Recall that the triplets above occupy 
; the number of arrangements is 
. Thus, we have 
possible numbers such that the units digit is . Since we claimed symmetry over the rest of the units digits, we must multiply by , resulting in 
eight-digit positive integers. Thus, the positive difference between and is 
.
~ilikemath247365
~LaTeX by eevee9406
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=P6siafb6rsI (also the person in the Youtube video wrote the final answer wrong, it was supposed to be 279 and he accidentally wrote it as 729)
See also
| 2025 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
1. To be multiple of 11: Total of 1,2,3,4,5,6,7,8 is 36, dividing into two groups of 4 numbers, the difference of sum of two group x and y need to be 0 or multiple of 11, i.e. x+y=36, x-y=0,11,22… only x=y=18 is possible. Number 8 can only be with (8,1,4,5),(8,1,2,7),(8,1,3,6),(8,2,3,5). One group of 4 numbers make 4! different arrangement, two groups make 4!*4!, the 2 group makes 2! arrangement. The two group of numbers are alternating by digits. Total number of multiple of 11 is 4*2!*4!*4! 2. To be multiple of 2: We noticed in each number group, there are two odd two even. So the final answer is above divided by 2, 4*2!*4!*4!/2=2304. 2304-2025=279.
~Mathzu.club
