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.



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

  • 2
    $\begingroup$ Downvote without any comment? $\endgroup$
    – user548054
    Jan 6, 2019 at 8:51
  • 1
    $\begingroup$ 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. $\endgroup$
    – JessicaK
    Jan 6, 2019 at 9:04
  • $\begingroup$ @JessicaK if you understand the mean of question, can you edit ? If there is a mistake in translation into English $\endgroup$
    – user548054
    Jan 6, 2019 at 9:09
  • $\begingroup$ 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). $\endgroup$
    – jmerry
    Jan 6, 2019 at 9:34
  • $\begingroup$ Please see this article on MathSE protocol. $\endgroup$ May 11 at 0:31

1 Answer 1


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 $$2^5\cdot 5!-2^4\cdot 4! = 3456. $$

  • $\begingroup$ (+) Thank you very much. I'm trying to understand your answer. $\endgroup$
    – user548054
    Jan 6, 2019 at 9:33

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