Attempt of the following sub questions:
A three-digit number is equal to 17 times the sum of its digits. If 198 is added to the number, the extreme digits get interchanged. The sum of first and third digit is 1 less than middle digit. Find the number.

Let the first digit be ‘a’ and third digit be ‘b’

Now according to 3^rd condition a + b = middle digit – 1

⇒ middle digit = a + b + 1

A three-digit number can be represented as

(Digit at hundredth place × 100) + (digit at tenth place × 10) + (digit at unit’s place)

So according to first condition

[a × 100] + [(a + b + 1) × 10] + b = 17 × [a + (a + b + 1) + b]

⇒ 100a + 10a + 10b + 10 + b = 17 × [2a + 2b + 1]

⇒ 110a + 11b + 10 = 34a + 34b + 17

⇒ 110a + 11b – 34a – 34b = 17 – 10

⇒ 76a – 23b = 7 …(i)

Now according to second condition

{[a × 100] + [(a + b + 1) × 10] + b} + 198=[b × 100] + [(a + b + 1) × 10] + a

⇒ (100a + 10a + 10b + 10 + b) + 198 = 100b + 10b + 10a + 10 + a

⇒ 110a + 11b + 208 = 110b + 11a + 10

⇒ 110a – 11a + 11b – 110b = 10 – 208

⇒ 99a – 99b = -198

Divide by 99

⇒ a – b = -2 …(ii)

⇒ a = b - 2

Put a in equation (i)

⇒ 76 × (b - 2) – 23b = 7

⇒ 76b - 152 – 23b = 7

⇒ 53b = 152 + 7

⇒ 53b = 159

∴ b = 3

Put b = 3 in equation (ii)

⇒ a – 3 = -2

⇒ a = 3 – 2

∴ a = 1

Middle digit = a + b + 1

∴ middle digit = 1 + 3 + 1 = 5

Therefore, the three-digit number is 153