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

10x + y = 7(x+y)
10x-7x +y-7y=0
3x-6y=0______1

10y+x=10x+y-18
10y-y+x-10x=-18
9y-9x=-18
y-x= -2______2

then find x and y by substituting

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.