The number of 6 digit numbers that can be formed using the digits 0,1,2,5,7 and 9 which are divisible by 11 and no digit is repeated, is:
Solutions · 1
Official solution
Using all 6 digits 0,1,2,5,7,9, divisibility by 11 requires (a1+a3+a5)−(a2+a4+a6)≡0(mod11).
Total sum of all digits =0+1+2+5+7+9=24. We need two groups of 3 digits each summing equally (each to 12) or differing by 11 (giving 6.5 and 17.5, not integer, so must be equal split): groups summing to 12 each: {1,2,9} and {0,5,7}.
Arrangements: within odd positions (3 slots) and even positions (3 slots), with first digit nonzero.
\[3!\times3! + \text{(adjustment for leading zero)} \]
Following the source's computation: 3!×3!+3!×2×2=36+24=60