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

Answer 1

Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number can be written as 10x + y in the expanded form.

When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x.

According to the given condition.

(10x + y) + (10y + x) = 66

11(x + y) = 66

x + y = 6 ... (1)

You are also given that the digits differ by 2. Therefore,

either x – y = 2 ... (2)

or y – x = 2 ... (3)

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

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

Thus, there are two such numbers 42 and 24.

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Answer 2

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Let the digit in the unit’s place be x and the digit in the tens place be y.

Then, the number = 10y + x

The number obtained by reversing the order of the digits = 10x + y

According to given conditions,

(10y + x) + (10x + y) = 66

⇒ 11(x + y) = 66

⇒ (x + y) = 6

According to second situation, digits differ by 2

So, either x – y = 2 or y – x = 2

Thus , we have the following sets of simuntaneous equations

x + y = 6 …I

x – y = 2 …II

or,

x + y = 6 …III

x – y = 2 …IV

solving equation I and II, we get x = 2 and y = 4

solving equation III and IV , we get x = 4 and y = 2

When x = 4 and y = 2,

Two digit number = (10y + x) = 10(4) + 2 = 42

When x = 2 and y = 4,

Two digit number = (10y + x) = 10(2) + 4 = 24

Hence, the required number is either 24 or 42.