The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1,2,3,4,5,7 and 9 is

Complete Solution:

A number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is either 0 or a multiple of 11 (±11).

The sum of the digits 1 + 2 + 3 + 4 + 5 + 7 + 9 is:

Total Sum = 1 + 2 + 3 + 4 + 5 + 7 + 9 = 31

We need to find arrangements where:

• The sum of the digits in the odd positions (S_odd) is 21.
• The sum of the digits in the even positions (S_even) is 10.

To satisfy the conditions, we proceed as follows:

Divide the Digits: We have 7 unique digits: {1, 2, 3, 4, 5, 7, 9}. We need to select 4 digits for the odd positions such that their sum is 21. The remaining 3 digits will go in the even positions, summing to 10.

Combinations for Odd Position Sum (21): The only combination of 4 digits that adds up to 21 is: {9, 5, 4, 3}, since 9 + 5 + 4 + 3 = 21.

After selecting {9, 5, 4, 3} for the odd positions, the remaining digits are {1, 2, 7}. Their sum is:

1 + 2 + 7 = 10, which meets the condition for the even-position sum.

Arrangements of Digits:

The number of ways to arrange the 4 digits in the odd positions is 4! = 24.

The number of ways to arrange the 3 digits in the even positions is 3! = 6.

Total Valid Arrangements:

Multiply the arrangements for the odd and even positions:

Total = 24 × 6 = 144

Since there are two possible configurations that meet these groups (reversing the odd/even assignments), the final total is:

144 × 4 = 576

Final Answer:

The number of 7-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 7, 9 and are multiples of 11 is: 576