How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?

How many seven-digit numbers divisible by 11 have the sum of their digits equal to 59?

ANS- 40
Let abcdefg be a seven digit number.
Then a + b + c + d + e + f + g = 59         …(i)

Now, for the number to be divisible by 11,

|(a + c + e + g) – (b + d + f)| = 0 or 11 or 22 …

If |(a + c + e + g) – (b + d + f)| = 0, then from (i),

2(a + c + e + g) = 59, which is not possible.

Similarly, |(a + c + e + g) – (b + d + f)| cannot have any even value.

If |(a + c + e + g) – (b + d + f)| = 33, then from (i),

(a + c + e + g) = 92/2 = 46, which is not possible as the maximum sum of 4 digits is 9 × 4 = 36. Clearly, (b + d + f) = 46 is also not possible.

Similarly, |(a + c + e + g) – (b + d + f)| cannot have any value greater than 33 either.

∴ |(a + c + e + g) – (b + d + f)| = 11      …(ii)

From (i) and (ii), (a + c + e + g) = 35 or (b + d + f) = 35.

The second case is not possible as the maximum sum of 3 digits can be 27.

∴ (a + c + e + g) = 35 and (b + d + f) = 24

Now consider

a + c + e + g = 35

The maximum sum of a, b, c and d is 36 when a = c = e = g = 9

But our sum is 35. So we assign (−1) among the four variables. This can happen in 4 ways. Thus there are only 4 solutions to this equation, namely, (9, 9, 9, 8), (9, 9, 8, 9), (9, 8, 9, 9) and (8, 9, 9, 9).

Now consider

b + d + f = 24

The maximum sum of b, d and f is 27, when b = d = f = 9

But our sum is 24. So we distribute (−3) among the 3 variables.

Thus (b’ – j) + (d’ – k) + (f’ – l) = 24, where j + k + l = 3

The equation j + k + l has (3 + 3 – 1)C(3 – 1) = 5C2 = 10 solutions.

Thus there are 4 × 10 = 40 numbers that satisfy our conditions.