Consider
x = 3168 and
y = 7A2B4C54995D458E1F9G6H3I17862, where
ABCDEFGHI represents some permutation of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9.
Determine the total number of such
permutations such that the remainder of y upon division by x is exactly 2022.
First thing is I'll take 7A2B4C54995D458E1F9G6H3I17862 - 2022 = 7A2B4C54995D458E1F9G6H3I15840.
We will call this N. Then we want N to be a multiple of 3168.
3168 = 32*9*11.
Divisibility by 32=2^5 is simply divisibility of the last 5 digits of N:
15840/32 = 495. So N is divisible by 32, no matter what the rest of the digits are.
Divisibility by 9 is the divisibility of the sum of the digits of N:
(7+A+2+B+4+C+5+4+9+9+5+D+4+5+8+E+1+F+9+G+6+H+3+I+1+5+8+4+0) / 9 = 11 + (A+B+C+D+E+F+G+H+I) / 9
But since the unknown digits are given as a permutation of the digits 1-9 we know that sum is 45. Then the remainder is zero; therefore N is divisible by 9 for any such permutation.
Divisibility by 11 is the divisibility of the alternating sum and difference of the digits of N:
(7-A+2-B+4-C+5-4+9-9+5-D+4-5+8-E+1-F+9-G+6-H+3-I+1-5+8-4+0) / 11 = (45-(A+B+C+D+E+F+G+H+I))/11
Again, the sum of all the unknown digits equals 45, so then the remainder is 0; therefore N is divisible by 11 for any such permutation.
Since all divisibilities are satisfied we must have N is divisible by 3168.
Therefore all 9!=362880 permutations of digits 1,2,3,4,5,6,7,8,9 satisfy the problem statement.
P&P Solution
