If a 7 digits no. "a b c d e f g" is divisible by 11, then
a+c+e+g = b+d+f
But I don't know the reason and the method to acquire that combinations.
Hello, prasum!
Seven digits from the digits {1, 2, 3, 4, 5, 6, 7, 8, 9} are written in a random order.
Find the probability that this seven-digit number is divisible by 11.
i have got total cases as: 9*8*7*6*5*4*3 = 181,440 . Right!
BookEnquiry has the right idea . . .
Let be the 7-digit number.
If is a multiple of 11,
. . then is a multiple by 11.
I found no formula for counting the multiples-of-11.
I was forced to use Brute Force listing.
I found 54 sets of digits which produce multiples-of-11.
. . . .
In each of the 54 sets, can be arranged in ways
. . and can be arranged in ways.
Hence, there are: . numbers divisible by 11.
Therefore: .