Question) Let us define a gang of integers as integers that are formed by rearranging the digits in the decimal representation of a positive integer. Example #1: 1123, 1213, 1231, 2113, 2131, 2311, 3112, 3121, and 3211 form a gang. Example #2: 770, 707 and 077 form a gang. Does there exist a gang, that is formed out of a 10-digit positive integer, in which we can find more than 13000 10-digit positive integers that are multiples of 7? Explain your answer.
How do i find the number of 10 digit integers which are multiples of 7, and how would i use the pigeonhole principle to solve this?
The total number of 10-digit positive integers is $N=10^{10}$.
Among them, the number of those divisible by $7$ is $N_7=\lfloor N/7\rfloor(=1,428,571,428)$.
The total number of gangs formed by $10$-digit positive integers is $G={19\choose10}(=92,378)$.
By pigeonhole, we have at least one gang who contains more than $\lceil N_7/G\rceil=15465$ members divisible by $7$.