Question

the sum of two digit number and the number obtained by reversing are order of the digit is 66 . if the digits of the number differ by 2. find the number.​

Answer 1

AnsWer :

42.

To Find :

Find the number.

Solution :

Let the tenth place digits be x

and units place digits be y.

A/Q,

★ Case 1.

• The sum of two digit number and the number obtained by reversing are order of the digit is 66 .

$\tt \dagger \: \: \: \: \:( 10x + y )+ (10y + x) = 66. \: \: \: \: \: - (1)$

★ Case 2.

• If the digits of the number differ by 2.

$\tt \dagger \: \: \: \: \: x - y = 2.$

Taking Equation ( 1 )

$\tt : \implies( 10x + y )+ (10y + x) = 66.$

$\tt : \implies 11x + 11y= 66.$

$\tt : \implies 11(x +y)= 66.$

$\tt : \implies x + y= 6. \: \: \: \: \: -( 3)$

Now, By Adding both Equation 2 and 3, We get.

$\tt : \implies 2x = 8.$

$\tt : \implies x = 4.$

Now, Putting the value of x in Equation 3, We get.

$\tt : \implies x + y = 6$

$\tt : \implies 4 + y = 6$

$\tt : \implies y = 2.$

Now, Required no. be 10x + y = 40 + 2 => 42.

Therefore, the required answer is 42.