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.
Replacing each letter with a zero gives the 29-digit number:
70204054995045801090603017862 which is 582 mod 3168.
Each letter represents a digit times 10^power, where for A,B,...,H,I the powers are:
[27, 25, 23, 17, 13, 11, 9, 7, 5]
All odd powers of 10, where the power is 5 or more, are 1792 mod 3168.
So each letter contributes (d*1792) mod 3168 to the final answer.
The sum of all the d's 1+...+9 is 45.
45*1792 = 80640
Add the mod value of the original number with zeros in place of each letter.
582+1440 = 2022
So EVERY permutation of ABCDEFGHI yields a remainder of 2022.
9! = 362880
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Posted by Larry
on 2023-07-06 06:50:31 |
Analytic solution
