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 22, it must have an even unit's digit. Therefore, we can break it up into cases and let the last digit be either 2, 4, 6, or 8. This problem is symmetric so we may assume that once we find the number of cases for one of these numbers, say 2, then we multiply by 4. Now, we just need to find how to positions of the rest of the numbers can there be such that the unit's digit is 2 and the number is divisible by 11. If we remember the divisibility rule for 11, then we can denote the odd numbered positions to be a1, a3, a5, and a7 and the even numbered positions to be a2, a4, and a6. Now, by the divisibility rule, we must have:
(a1 + a3 + a5 + a7) - (a2 + a4 + a6 + 2) which is congruent to 0(mod 11). Therefore, by simplifying, we must have:
a1 - a2 + a3 - a4 + a5 - a6 + a7 to be congruent to 2(mod 11). Now, let's consider a1 + a2 + ... + a7. This is just 1 + 2 + ... + 8 - 2 as we have already used up 2 as our unit's digit. This sum simplifies to 34 which is congruent to 1(mod 11). Therefore, we can say that (a1 + a2 + ... + a7) - 2(a2 + a4 + a6) is congruent to 2(mod 11) which means a2 + a4 + a6 is congruent to 5(mod 11). Notice the least sum this could even have is 1 + 3 + 4 = 8 and the greatest sum of a2 + a4 + a6 is 6 + 7 + 8 = 21. So the only possible number congruent to 5(mod 11) in this range is 16. So now, we just have to count up all the possible sums of 16 using the values 1, 3, 4, 5, 6, 7, and 8. We get: 
, and 
counting a total of 4 pairs. Notice now, the arrangement of the odd-positioned numbers doesn't matter so we can say it is 4! or 24 ways. The arrangement of each of these new 4 pairs we have just found is 3! = 6. Thus, for the unit's digit to be 2, we have 24*6*4 possible ways. Since we claimed that this was symmetric over the rest of the even digit unit's digits, we have to multiply this by 4 again. So our total number of ways is 24*6*4*4 = 2304. Thus, the positive difference between N and 2025 is 2304 - 2025 = 
.
~ilikemath247365
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. 
