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Question
A two-digit number is such that the ten's digit exceeds thrice the unit's digit by 3 and the number obtained by interchanging the digits is 2 more than twice the sum of the digits. Find the number.
Solution
Let x be the digit at ten's place and y be the digit at unit's place.
Then, the number is 10x + y.
Number obtained by reversing the digits = 10y + x
According to given information, we have
x = 3y + 3 ....(i)
And,
10y + x = 2(x + y) + 2
⇒ 10y + x = 2x + 2y + 2
⇒ 8y - x = 2
⇒ 8y - (3y + 3) = 2 ....[From (i)]
⇒ 8y - 3y - 3 = 2
⇒ 5y = 5
⇒ y = 1
⇒ x
= 3(1) + 3
= 3 + 3
= 6
∴ Required number
= 10x + y
= 10 x 6 + 1
= 60 + 1
= 61.
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