Divisors of a 11-digit number

Given the number 109999999ab and the number must be divisible by 99, which can also be expressed as at the same time divisible by 11 and 9.

For the number to be divisible by 9, the sum of the digits must be divisible by 9. So, 64 + a + b = 9c. But since a and b must be between 0 and 9 inclusive, a + b = 8 or a + b = 17.

For the number to be divisible by 11, the difference of its alternating digits from ones to the highest possible digit and from the second to the highest digit to the tens digit must either be a multiple of 11 or 0. In symbols: 10 + b - a = 11 (This is from the given.)

Solving the systems: a+b = 17 and b-a = 1 for the first equation for divisibility for 9 doesn't give integral solutions. We have a = 8 and b = 9. Hence, the product is 72.

we will divide 109999999ab by 99 and get remainder as 10 ab , and now will try some combination such that it will be equal to 10ab 99x10 = 990 it is wrng 99x11 = 1089 it is correct 99x12 = 1188 it is wrong again so we will put 1089 = 10ab therefore ab=89 8x9 =72 ans

A number which is divisible by 99,should be also divisible by 9 and 11 since 9 and 11 are the factors of 99.

Divisibility test for 9 First we test whether the number is divisible for 9 i.e. sum of all the digits should be divisible by 9 i.e. 1+0+9+9+9+9+9+9+9+a+b should be divisible by 9. As there are seven digits as 9,we will consider only 1+a+b i.e. 1+a+b should be divisible by 9. This implies that a+b should be either 8 or at-most 17. Thus we shall consider all the single digit numbers which add up to give 8 or 17.

Divisibility for 11 For the number to be divisible by 11, the difference of the sum of digits at odd places and the sum of its digits at even places should be either 0 or divisible by 11 i.e. [(37+b)-(27+a)] should be either 0 or divisible by 11. Therefore (10+b-a) should be 0 or divisible by 11. We consider 11 as the possible number to be equal to (10+b-a) since other numbers which are divisible by 11 will not satisfy these conditions 0<=a,b<10 and a,b E Z.

By considering all the above conditions, we a=9 and b=8 which makes ab=72.

the correct is 8*9=72

The answer is 72

Make it simple, it must be able to be div by some factor of 99. (11 & 9)

since the 11 digit is in between 10999999900 and 10999999999, we divide both number with 99, and get 111111110.1 and 111111111.1, thus the value is 111111111 x 99 = 10999999989, 8x9 = 72

If the number is divisible by 99, it is also divisible by 11 and 9. The divisibility rule for 9 is that it's digits must add up to a multiple of nine. The sum of the 7 nine's in $\overline{1099999999ab}$ are obviously a multiple of nine. Then, $1+a+b=9$ or $18$ , because the sum of 2 digits is at most 18. So, $a+b=$ either $8$ or $17$ .

Now, notice that $11000000000$ is divisible by 11. By subtracting 11 repeatedly from this number until the last 2 digits add up to either 8 or 17, we can determine a and b. It turns out that $11000000000-11=10999999989$ , and $8+9=17$ . So, $a*b=8*9=\boxed{72}$