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How many five digit multiples of 11 are there, if the five digits are 3, 4, 5, 6 and 7?

Repetition of digits is now allowed?









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Option(C) is correct

A number is divisible by $11$, if the difference between the sum of digits at even places and odd places is either $0$ or divisible by $11$.

Numbers $5, 3, 6, 4, 7$ is a multiple of $11$

$\because (5+6+7)-(3+4)=11$, which is a multiple of $11$

The number at odd places $5, 6, 7$ can be arranged in $3!$ ways and $3,4$ can be arranged in $2!$ ways.

Therefore, the total number of ways of such numbers $= 3!×2! =\textbf{12}$

Edit: Based on Shreyansh and other user's input, the question has been modified and now instead of 'in the same order' it ends with 'repetition of digits is now allowed'.

Edit 2: For a detailed alternative solution, check comment by Manuj.

(12) Comment(s)


There is another possibility. Since the repetitions are allowed number can be 35365.




how is that order of the digits taken in the solution? Is it just a guess? as there is no order mentioned in the question !


When permutation is calculated $3!$ or $2!$, then the order is taken care off.

For example, Numbers 3 and 4 can be at the even places. So the possible choices are,

ABCDE=> A3C4E or A4C3E. Which gives us $2!$ choices. Similarly for the numbers, $5, 6, 7$ we have $3!$ choices.

I guess this makes sense now.


yes we can not change the order of the numbers so the answer would be zero.


Thank you Shreyansh for the input, question has now been updated.


We know from divisibility rules that for any five digit number to be a multiple of 11, if the number is of the form: abcde, then should be equal to either 0 or a multiple of 11.

In the present question, this difference between (a+c+e) and (b+d) should be 0, 11 or -11.

In case,

$[(a+c+e)] - (b+d)]=0$, then

$(a+c+e)= (b+d)$

Then, if we take the sum of all digits of this five digit number abcde, we get


$(b+d)+(b+d) = 2(b+d)$

or $(a+b+c+d+e) = 7+6+5+4+3=25$ which is odd.

But $2(b + d)$ will always be even. This implies that the difference will never be zero.

In case,

$[(a+c+e)] - (b+d)]= 11$, then

$(a+b+c+d+e) - 11= 2 (b+d)$

so that $(b+d) = 7$

This is only possible, when $b$ and $d$ are 3 and 4 in some order; and $a, c, e$ are 7, 6 & 5 in some order.

There will be $2 \times 3! = 12$ numbers satisfying this.

In case

$[(a+c+e)] - (b+d)]= 11$, then

$(a+b+c+d+e) - 11= 2 (b+d)$, so that

$(b+d)=18$ which again is not possible

Therefore, there will only be 12 numbers satisfying the conditions of the problem.


there are almost 5!=120 ways to change the digits given in the question and its not possible to check which pattern will gives us 0 or 11 i.e when one will be divisible by 11. Unless we know that, how we will perform further steps??


Well there are ways, that is why we learn Permutation and Combination so that counting can be done systematically. And doable as well.


I guess the order given in the question does'nt matter becoz the difference between the sum of digits at even places and odd places is not divisible by 0 or 11, but the order given in the solution gives you the result.Smile


If the question says 3,4,5,6,& 7 in same order, then why is the order changed in solution. Am I missing something...?
Thanks in advance.

S K Atwal

i have the same query Joe. Do tell me if you get to know.


By order it did not mean the sequential arrangement, but repetition of digits I guess.