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 \_\_
Answer: 576
Solutions · 1
Official solution
For a 7-digit number using all of 1,2,3,4,5,7,9 (sum =31) to be divisible by 11, the alternating sum (a1+a3+a5+a7)−(a2+a4+a6) must be a multiple of 11. With 4 digits in odd positions and 3 in even positions, and total sum 31, we need the odd-position sum Sodd and even-position sum Seven=31−Sodd satisfying 2Sodd−31≡0(mod11).
This yields specific digit-subset splits for the 4 odd-position digits; the number of valid digit assignments times arrangements (4!×3!) per valid split gives:
\[\text{Total} = 576\]