Question

The sum of a two-digit number and the number obtain3d 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

$\huge{\boxed{\mathcal{\fcolorbox{red}{pink}{Hello!}}}}$

Question:

The sum of a two-digit number and the number obtain3d by reversing the digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

$\huge{\mathcal{\underline{\bf{Solution:-}}}}$

Let 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 in the expanded form.

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)

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

=> 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.

Therefore, the required such numbers are 42 and 24.

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Answer 2

$\boxed{\boxed{\Large\mathbf{24, 42}}}$
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$\underline{\underline{\Huge\mathfrak{ANSWER}}}$

Let the digit in one's place be $\bf{x}$
And the digit in tens place by $\bf{y}$

Then, the number will be $\bf{10x + y}$

$\underline\text{We know that,}$

10x + y + 10y + x = 66

$\mathsf{\Longrightarrow 11x + 11y = 66}$

$\mathsf{\Longrightarrow \cancel{11}(x + y) = \cancel{66}}$

$\mathsf{\Longrightarrow x + y = 6 ----(1)}$



$\huge\mathbf{\underline{CASE \: I}}$



$\large\mathbf{x - y = 2 ----(2)}$

Adding equation (1) and (2), we get,

2x = 4

$\large\mathsf{\Longrightarrow x = \frac{\cancel{8}}{\cancel{2}}}$

$\mathsf{\Longrightarrow} \boxed{\mathbf{x = 4}}$

Putting value of $\bf{x}$ in equation(2), We get,

4 - y = 2

$\mathsf{\Longrightarrow y = 4 - 2 }$

$\boxed{\mathbf{y = 2}}$

Hence, the number = 10(4) + 2

$\large\mathbf{\Longrightarrow 42}$



$\huge\mathbf{\underline{CASE \: II}}$



$\large\mathbf{y - x = 2 ----(2)}$

Adding equations (1) and (2), we get,

2y = 8

$\large\mathsf{\Longrightarrow y = \frac{\cancel{8}}{\cancel{2}}}$

$\mathsf{\Longrightarrow} \boxed{\mathbf{y = 4}}$

Putting value of $\bf{x}$ in equation(2), We get,

4 - x = 2

$\mathsf{\Longrightarrow x = 4 - 2 }$

$\boxed{\mathbf{x = 2}}$

Hence, the number = 10(2) + 4

$\large\mathbf{\Longrightarrow 24}$

Hence, there are 2 numbers which satisfy this equation.

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