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
Solution 2
1. To be multiple of 
Total of 
is 
dividing into two groups of numbers, the difference of sum of two group 
and 
need to be 
or multiple of 
i.e. 
only 
is possible.
Number can only be with 
One group of numbers make 
different arrangement, two groups make 
the group makes 
arrangement. The two group of numbers are alternating by digits.
Total number of multiple of is 
2. To be multiple of 
We noticed in each number group, there are two odd two even.
So the final answer is above divided by 
~Mathzu.club ~Latex by mathkiddus
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)
~Mathycoder
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. 
