Question

please help me............​

Answer 1

Answer :

The required number is 42

Given :

• The sum of two digit number and the number formed interchanging the digit is 66
• The two digits differ by 2

To Find :

• The original number .

Solution :

Let us consider the digits be x and y respectively

Therefore, the number is :

10x + y

According to first condition :

$\sf10x + y + 10y + x = 66 \\ \\ \implies \sf11x + 11y = 66 \\ \\ \sf \implies11(x + y) = 66 \\ \\ \sf \implies x + y = 6..........(1)$

And from second condition :

$\sf \implies x - y = 2 \: \: ..........(2)$

Adding the equations (1) and (2) we have :

$\sf \implies x + y + x - y = 6 + 2 \\ \\ \implies \sf2x = 8 \\ \\ \implies \bf{x = 4}$

Putting the value of x in (1) we have :

$\sf \implies4 + y = 6 \\ \\ \sf \implies y = 6 - 4 \\ \\ \bf \implies y = 2 \: \:$

Thus the number is :

$\sf10 \times 4 + 2 \\ \\ = 40 + 2 \\ \\ \bold{ = 42}$

Answer 2

...

$\tt{\huge{\purple{ ................... }}}$

QUESTION -

The sum of a two digit number and the number formed by interchanging the digits is 66 and the two digits differ by 2.

Find the number.

SOLUTION -

Let the above number be a , a + 2 as the digits differ by 2.

Sum => 10 a + a + 2 = 11a + 2

Reversed Number => a+2 , a

=> Sum => 10 a + 20 + a = 11 a + 20

Their sum is 66

=> 22 a + 22 = 66

=> 22 a = 44

=> a = 2

=> a + 2 = 4.

Therefore the required number becomes 42.

ANSWER :

The required number is 42