Find all 8-digit numbers 273A49B5 which are divisible by 11 as well as 25.

For the divisibility by 25,

A number is divisible by 25, if and only if, it ends in 00, 25, 50 and 75.

The number 273A49B5 ends in B5.

⇒ B can be either 2 or 7.

For the divisibility by 11,

A number is divisible by 11, if and only if, (Sum of digits at odd places) – (Sum of digits at even places) = Multiple of 11

So, (5 + 9 + A + 7) – (B + 4 + 3 + 2) = Multiple of 11

⇒ (21 + A) – (9 + B) = Multiple of 11

⇒ 21 + A – 9 – B = Multiple of 11

⇒ A – B + 12 = Multiple of 11 …(i)

Now, if B = 2,

We get from equation (i),

A – 2 + 12 = Multiple of 11

⇒ A + 10 = Multiple of 11

Put A = 1 (0 ≤ A ≤ 9) arbitrarily, we get

1 + 10 = Multiple of 11

⇒ 11 = Multiple of 11

So, one pair of solution is (A, B) = (1, 2).

If B = 7,

From equation (i), we get

A – 7 + 12 = Multiple of 11

⇒ A + 5 = Multiple of 11

Put A = 6 (0 ≤ A ≤ 9) arbitrarily, we get

6 + 5 = Multiple of 11

⇒ 11 = Mulitple of 11

And, another pair of solution is (A, B) = (6, 7).

So, the number from solution (1, 2) comes out to be 27314925.

And, the number from solution (6, 7) comes out to be 27364975.

Thus, the numbers are 27314925 and 27364975.