How many five-digit multiples of 11 are there, if the five digits are 3, 4, 5, 6 and 7?
(a) 12
(b) 13
(c) 10
(d) None of these

VerifiedVerified
122.7k+ views
Hint: We start solving the problem by recalling the fact that in order to have a number divisible by 11, then the alternating sum of the digits present in it should be equal to 0 or multiple of 11. We then check the digits that need to be positioned in odd and even positions to get the alternating sum is a multiple of 11. Once we find the digits in odd and even positions, we then find the total number of arrangements that can be made with those digits in even and odd positions using the fact that the total number of ways of arranging n objects in n place is $ n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1 $ . We then multiply the arrangements obtained for odd and even positions to get the required answer.

Complete step by step answer:
According to the problem, we are asked to find the total number of five-digit multiples of 11 are there if the five digits are 3, 4, 5, 6, and 7.
We know that in order to have a number divisible by 11, then the alternating sum of the digits present in it should be equal to 0 or multiple of 11, i.e., if the number resembles $ abcde $, then $ a-b+c-d+e $ should be equal to 0 or multiple of 11.
Let us not find the sum of the given numbers which is $ 3+4+5+6+7=25 $ . We need to find the two digits which should be subtracted twice from the obtained sum for getting it as 11 (as 0 or 22 is not possible).
So, we get $ 3+4+5+6+7-2\left( a+b \right)=11 $ .
 $ \Rightarrow 25-2\left( a+b \right)=11 $ .
 $ \Rightarrow 2\left( a+b \right)=25-11 $ .
 $ \Rightarrow 2\left( a+b \right)=14 $ .
 $ \Rightarrow a+b=7 $ .
We can see that only 3 and 4 satisfies this.
So, we need to place 5, 6, and 7 in odd positions and 3 and 4 in even positions to get the 5 digits numbers that were divisible by 11.
Now, we need to find the number of arrangements that can be made in odd positions and even positions.
We know that the total number of ways of arranging n objects in n place is $ n!=n\times \left( n-1 \right)\times \left( n-2 \right)\times ......\times 2\times 1 $ .
So, we get the total number of the arrangement of digits in odd position is $ 3!=3\times 2\times 1=6 $ and the total number of the arrangement of digits in odd position is $ 2!=2\times 1=2 $ .
So, the total 5 digits number that can be formed is $ 6\times 2=12 $.
 $\therefore$ The correct option for the given problem is (a).

Note:
Whenever we get this type of problem, we first try to recall the condition for the numbers to be divisible by the given number. We can also use the fact that the result obtained by subtracting the last digit from the given number should be divisible by 11. Similarly, we can expect problems to find the 5 digits number formed by 3, 4, 5, 6, 7 that were divisible by 7.