Question

The sum of a two digit number and the number formed by reversing the order of digit is 66 if the two digits differ by 2, find the number using one variable. How many such numbers are there

Answer 1

Explanation.

• GIVEN

Sum of two digit number and the number

formed by reversing the order is 66

if two digit differ by 2

How many such numbers are there.

SOLUTIONS.

Let the two digit number be x and y

According to the question

two digit number are differ by 2

we can write as,

$x - y = \pm2 \: ....(1)$

According to the question,

The number is found to be = 10y + x

after reversing the digit the number become

10x + y

According to the question,

10x + y + 10y + x = 66

11x + 11y = 66

x + y = 6 ....(2)

Now we can found two simultaneous equation

x - y = 2 ....(3)

x + y = 6 .....(4)

From equation (3) and (4)

we get,

2x = 8

x = 4

put x = 4 in equation (4)

we get,

4 + y = 6

y = 2

Hence,

the required number =

10y + x = 10(2) + 4 = 24

Now, second simultaneous equation is

x - y = -2 ....(5)

x + y = 6 ....(6)

From equation (5) and (6)

we get,

2x = 4

x = 2

put x = 2 in equation (6)

we get,

2 + y = 6

y = 4

Hence,

required number = 10y + x = 10(4) + 2 = 42

Therefore,

There are two possible number = 24 and 42