Question

A two digit number is 7 times the sum of its two digits. The number that is
formed by reversing its digits is 18 less than the original number. What is
the number?

Answer 1

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Answer 2

Solution :-

Let the digit at the ten's place be x ; and the digit at the unit's place be y.

Then,

The number will be = 10x + y.

A/Q, (1st condition)

$10x + y = 7(x + y) \\ \implies10x + y = 7x + 7y \\ \implies3x = 6y \\ \implies \: x = 2y \: \: \: \: \: ....(i)$

The number formed by reversing the digits is 10y + x.

A/Q, (2nd condition)

$10y + x = (10x + y) - 18 \\ \implies \: 10y - y = 10x - x - 18 \\ \implies \: 9y = 9x - 18 \\ \implies \: y = x - 2 \\ \implies y = 2y - 2 \: \: \: \: \: ...{using(i)} \\ \implies \: y = 2.$

From eq, (i)

x = 2 × 2 = 4.

Hence the required number is 42.