What is the smallest seven-digit number divisible by 11 that can be formed by the digits 9, 8, 7, 4, 3, 1, and 0 each used exactly once?

Answer 1

Divisibility test for 11:
tag all the even-placed digits and the odd-placed digits
take the sum of the even-placed digits, and the sum of the odd-placed digits
If the absolute value of the difference of these 2 sums is divisible by 11,
so is the original number.
e.g. 456982
sum of odd-placed = 4+6+8 = 18
sum of evens = 5+9+2 = 16
difference = 18-16 = 2, not divisible by 11, neither is 456982

for 7,8,9,4,3,1,0
the sum of all the digits = 32
so we have to split 32 into 2 sum so the difference is a multiple of 11

30-2 = 28 , no
29 - 3 = 26, no
28 - 4 = 24, no
27 - 5 = 22 <----- YES, a multiple of 11
so I need a sum of 5, which would be 4+1+0
these must be in the even positions, and you want smallest, so
X0X1X4X , leaving me with 3,9,8,7 to be placed from smallest to largest
in the odd positions.

your number is
3071849