AoPS Academy
be a convex quadrilateral inscribed in a circle and satisfying 
. Points 
and 
are chosen on sides 
and 
such that 
and 
. Prove that 
.
eight-digit positive integers that use each of the digits 1, 2, 3, 4, 5, 6, 7, 8 exactly once. Let N be the number of these integers that are divisible by 
. Find the difference between 
and 2025.
and get 
, which gives us a final answer of 
However, 
same issue wtih 
Thus both alternating set of digits sums to 
ways.
, so it is impossible for 3 digits to be in either one of ![$[1, 4], [5, 8]$](http://latex.artofproblemsolving.com/5/d/9/5d9f6f78361cd851c801b6bcd5a328c8ec93bc07.png)
by symmetry. Thus it is split 2-2. We do casework on the mean 
of the smaller 2, since the mean of the larger 2 must then symmetricly be 
.
we have 1 case. 
gives 1 case. 
gives 
cases for each pair so 
total. 
both give 1 case. Note that in each case, 2 numbers are even and 2 are odd. So for each of the 
cases there are 
ways to order.
, we simply want 
and 
is a multiple of 2
, if 
, but 
is not achievable since 
and you reverse the elements in each pair. be a multiples of 2, there are in total 
ways, the answer is then 
such that 
, 
, 
, 
, 
, and the sums of 
and 
differ by a multiple of 
.
such pairs. We claim that 
. even digits we can put at the end. Then, by the divisibility rule for , there are 
ways to make and take up alternating digits.
, so our claim is proven.
.
, we want the sum of to be 
or 
. Since 
, the sum of must be .
, so our answer is .
, so the individual sums of each alternating sequence is either 
or 
. Only the latter is possible, now we carefully list partitions that sum to with the given digits:![\[ 9621 \quad 9531 \quad 9432 \quad 8721 \quad 8631 \quad 7631 \quad 7542 \quad 6543 \]](http://latex.artofproblemsolving.com/4/d/6/4d650ceac5f8e5c6f7d0da0257be50d3909536e3.png)
There are partitions. Now the units digit is even so the alternating sum with the units digit has 
possibilities, and we can arbitrarily permute the other alternating sum ways. The requested answer is![\[ \left| 8 \cdot \dfrac{4!}{2} \cdot 4! - 2025 \right| = \left|2304 - 2025 \right| = \boxed{279}. \]](http://latex.artofproblemsolving.com/4/4/1/44137544dcfd094ca4e0a929b3712b5fd1eeeaff.png)
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 = .
be the number, and 
, and 
.
, if and only if 
. Since 
and 
, we easily see that only 
works., are![\[
(1, 2, 7, 8), (1, 3, 6, 8), (1, 4, 5, 8), (1, 4, 6, 7), (2, 3, 5, 8), (2, 3, 6, 7), (2, 4, 5, 7), (3, 4, 5, 6)
\]](http://latex.artofproblemsolving.com/d/8/0/d80edc38c068442f6fb9700f7adada8dc3c3f222.png)
and their permutations. Each even, and odd numbers. Since 
must be even, for each one above, there are 
combinations for 
, and 
for 
. The number of such 
are 
. In particular, the answer is 
.
be the sum of even positions and 
be sum of odd positions. Note 
but 
and we may see that only 
works. We find quadruplets of four digits which sum to 
and similarly four odds fails so by parity we must have two evens and two odds to sum to ![\[8 \cdot 12 \cdot 24 = 2304 = 2025 + \boxed{279}\]](http://latex.artofproblemsolving.com/d/9/9/d99176ce8e4fbbe8983f70e03162c0f100fe2c40.png)
extraction
and 
.
