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 _____________.

Solution

Maths sir · 4 steps

Use the 11 rule: the 4 odd-place digits and the 3 even-place digits must differ in sum by a multiple of 11.

1. Divisibility by 11. A 7-digit number has 4 odd places and 3 even places. Let Sₒ be the sum of the digits in the odd places and Sₑ the sum in the even places. The number is a multiple of 11 exactly when Sₒ - Sₑ is a multiple of 11.
2. Fix the sums. The digits add up to 1+2+3+4+5+7+9 = 31, so Sₒ + Sₑ = 31. Since 31 is odd, Sₒ - Sₑ is odd too, so it can only be ± 11. A difference of ± 33 would need one sum to exceed 31. That leaves (Sₒ, Sₑ) = (21, 10) or (10, 21).

   Case	Digit groups that work	Arrangements
   Sₑ = 10, Sₒ = 21	3 even-place digits: {1,2,7}, {1,4,5}, {2,3,5}	3 × 3! × 4! = 432
   Sₒ = 10, Sₑ = 21	4 odd-place digits: only {1,2,3,4}, leaving {5,7,9}	1 × 4! × 3! = 144

3. Why these counts. Once the split of digits is chosen, the 4 odd-place digits can be arranged in 4! ways and the 3 even-place digits in 3! ways. Zero is not among the digits, so every arrangement is a genuine 7-digit number.
4. So. Total = 432 + 144 = 576.

The common slip: 432. Counting only the case where the even-place digits sum to 10 gives 432. The case with odd-place digits {1,2,3,4} adds another 144.

The idea to keep. For multiples of 11 with given digits, first split the digits by the sum rule, then arrange within odd and even places. Check both signs of the difference.

Answer from the NTA final answer key. Solution by PrepWiser.