A 25-digit number is such that its extreme digits are same and all other 23 digits are same but different from extreme digits. If that number is divisible by 11, then end digit can’t be:

Given:

A 25-digit number such that:

Extreme digits are the same.

All other 23 digits are identical but different from the extreme digits.

The number is divisible by 11.

Formula used:

A number is divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is divisible by 11.

Calculation:

Let the extreme digit be 'a' and the identical middle digits be 'b'.

Odd positions: 1st, 3rd, 5th, ..., 23rd, 25th.

Even positions: 2nd, 4th, ..., 24th.

Sum of digits at odd positions:

⇒ (1st digit) + 11 × (middle digits) + (25th digit) = a + 11 × b + a = 2a + 11b

Sum of digits at even positions:

⇒ 12 × (middle digits) = 12 × b

Difference between odd and even positions:

⇒ (2a + 11b) - (12b)

⇒ 2a - b

For divisibility by 11:

⇒ 2a - b must be divisible by 11.

Checking the options:

Option 1: a = 9

⇒ 2 × 9 = 18

⇒ 2a - b = 18 - b

⇒ If b = 7, 18 - 7 = 11 (divisible).

Option 2: a = 8

⇒ 2 × 8 = 16

⇒ 2a - b = 16 - b

⇒ If b = 5, 16 - 5 = 11 (divisible).

Option 3: a = 5

⇒ 2 × 5 = 10

⇒ 2a - b = 10 - b

⇒ 10 - b cannot be divisible by 11 as it does not meet the condition.

Option 4: a = 4

⇒ 2 × 4 = 8

⇒ 2a - b = 8 - b

⇒ If b = -3, 8 - (-3) = 11 (divisible).

∴ The correct answer is option 3.