# How many $10-$digit numbers are divided by $11.111$ and all the digits are different?

The Problem:

How many $$10-$$digit numbers are divided by $$11.111$$ and all the digits are different?

A) $$3250$$

B) $$3456$$

C) $$3624$$

D) $$3842$$

E) $$4020$$

The Problematic point is, "all digits must be different".

I could find only all $$10-$$ digit numbers.

$$999990000+11.111k≤9.999.999.999$$

$$1≤k≤810.009$$

The problem is, I have no method how can I calculate "all digits are different numbers."

• Downvote without any comment?
– user548054
Jan 6, 2019 at 8:51
• I don't quite understand the downvote and close vote. It may be because most countries do not use a period to separate the thousands like you've written so it looks like you are asking for 10 digit numbers that divide by a number slightly larger than 11 rather than 11111. Jan 6, 2019 at 9:04
• @JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English
– user548054
Jan 6, 2019 at 9:09
• A clarification on mathematical English grammar: "is/are divided by" is incorrect. Either "can be divided by" or "is divisible by" would work; the latter adjective form is most standard. Also, in English, it's standard to use a period for the decimal point separator and a comma to break out blocks of digits (when we do so). Jan 6, 2019 at 9:34
If all digits are different, they must be all ten digits. In particular, the digit sum is $$45$$ and hence our number is a multiple of $$9$$. Thus we are in fact looking for certain multiples of $$99999$$. This reduces your $$k$$ range down to about $$90000$$ possibilities - still unfeasible do work out by hand.
If the ten digit number is $$abcdefghij$$, then after subtracting $$99999\cdot abcde$$ we still have a multiple of $$99999$$, namely $$fghij+abcde$$. As this sum is certainly $$>0$$ and $$<99999+99999$$, we conclude that $$fghij+abcde=99999.$$ In particular, $$j+e=9$$ without carry. Then also $$i+d=9$$ without carry, and so on. Thus the digit pairs $$\{a,f\},\{b,g\},\{c,h\},\{d,i\},\{e,j\}$$ must be the pairs $$\{0,9\},\{1,8\},\{2,7\},\{3,6\},\{4,5\}$$ in some order. There are $$5!$$ such permutations and then for each pair there are $$2$$ ways to match. This gives us $$2^5\cdot 5!$$ numbers of the desired form. However, among these are $$2^4\cdot 4!$$ where we attempt to set $$a=0$$ (and $$f=9$$), i.e., that are not really ten-digit numbers. Hence the final answer is $$2^5\cdot 5!-2^4\cdot 4! = 3456.$$