Question

The sum of two digit no. And the number formed by reversing the order of digit is 66. If the two digits differ by 2, find the number. How many such numbers are there?

Answer 1

${\huge{\sf {\green{\underline{Answer}}}}}$

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$\boxed{ 42 \: / \: 24 }$

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${\huge{\sf {\blue{\underline{Explanation :}}}}}$

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Let, the digit in one's place be x.

So, the digit on the ten's place will be (x+2)

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The number = 10( x + 2 ) + x

= 10x + 20 + x

= 11x + 20

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We know,

Any number = 10 ( Ten's place ) + One's place

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The number obtained by reversing the digits,

= 10x + ( x + 2 )

= 10x + x + 2

= 11x + 2

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Given,

Sum of the number and the digits reversed = 66

11x + 20 + 11x + 2 = 66

22x + 22 = 66

22x = 44

x = 2

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Number = 11x + 20

= 11 (2) + 20

= 22 + 20

= 42

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Numbers reversed = 24.

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There are two numbers which when reversed and added sum up to 66 and the digits have a difference of 2.

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24 and 42