Question

The sum of a two-digit number and the number obtained by reversing the
digits is 66. If the digits of the number differ by 2, find the number. How
many such numbers are there?​

Answer 1

Answer:

The number is 42(or)24. Only these two numbers are possible for this condition.

Step-by-step explanation:

Let the two digit number be 10x+y

Then the reversed number is 10y+x

Given their sum is equal to 66.

=>11x+11y=66

=>x+y=6-----(1)

Also given the numbers differ by 2,

=>x-y=2------(2)

Solving (1)&(2),

2x=8

=>x=4.

Substituting x=4 in (2),

y=2.

Therefore the number is 42.

Verification: 42+24=66.

Here 24 is obtained by reversing 42.

Hence it is verified.

Also there are only two such numbers for this condition.

1)42+24=66.

2)24+42=66.

Answer 2

$\huge\sf{Answer}$ 

• The ten’s digit in the first number be x.

• Unit's digit in the first number be y. 

Therefore, the first number can be written as 10x + y

When the digits are reversed, x and y becomes the unit's digit and ten's digit respectively.

Now, the number will be in the expanded notation is 10y+ x

According to the given condition,

⟹ (10x + y) + (10y + x) = 66

⟹ 11(x + y) = 66

⟹ x + y = 66/11

⟹ x + y = 6 ___ (1)

Hence,

⟹ x – y = 2 ___(2)

⟹ y – x = 2 ___(3)

If x – y = 2, then solving (1) and (2) by elimination, x = 4 and y = 2. In this case, the number is 42

If y – x = 2, then solving (1) and (3) by elimination, x = 2 and y = 4. In this case, the number is 24

Therefore,

the required such numbers are 42 and 24

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