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:
which is congruent to
. Therefore, after simplifying, we must have:
Now consider
. Therefore,
which means that
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:
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
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 | ||
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