Question

The digit in the tens place of a 2-digit number is twice the units digit. If the digits are reversed, the new
number is 18 less than the original number. Find the original number.
dimit number is f more than a times the sum of its digits If?7 is added to the number, its digits are​

Answer 1

Answer:-

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

The number = 10x + y

Given:

The ten's digit is twice the units digit.

→ x = 2y -- equation (1)

And,

If the digits are reversed the new number is 18 less than the original number.

Number formed by reversing the digits = 10y + x.

According to the above condition,

→ (10x + y) - 18 = 10y + x

→ 10x + y - 10y - x = 18

→ 9x - 9y = 18

→ 9(x - y) = 18

Substitute the value of x from equation (1)

→ 2y - y = 18/9

→ y = 2

Putting the value of y in equation (1) we get,

→ x = 2*2

→ x = 4

The number = 10*4 + 2 = 40 + 2 = 42.

Therefore, the required number is 42.

Answer 2

Step-by-step explanation:

let digit in units place = x

digit in tens place =2x

two digit number = 10(2x) + x= 21x

after interchanging the digits,

= 10x + 2x = 12x

new number = original no. - 18

12x = 21x- 18

12x - 21x= - 18

-9x = -18

x = -18/-9

x=2

original number = 21x

= 21× 2= 42