Question

a two digit number is 7 times the sum of its digits. the number obtained by reversing the digit is less then the original number by 18. find the original number.

Answer 1

Hey user

Here is your answer :-

Let the units place digit be " n " and tens place digit be " m " .

So the number is 10m + n

By the first condition given in your question we get equation,

10m + n = 7 ( m + n )

》10m + n = 7m + 7n

》3m - 6n = 0

》m - 2n = 0. . . . ( 1 )

By the second condition given in your question we get equation,

10n + m + 18 = 10m + n

》9m - 9n = 18

》m - n = 2. . . . ( 2 )

Subtracting equation ( 1 ) from equation ( 2 ) we get

n = 2

Putting n = 2 in equation ( 2 ) we get ,

m - 2 = 2

》m = 2 + 2

》m = 4

So the number is 10m + n = 10 ( 4 ) + 2

= 40 + 2

= 42

So the number is 42 .

Thank you . '

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.