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Question

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

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Solution

Let the digit at one's place be x.

Then, the digit at ten's place be (x2). [ Digits differ by 2]

The original two digit number =10(x2)+x=10x20+x=11x20

When the digits are reversed, then digit at ones place =(x2)

And, the digit at tens place =x

So, the reversed number =10x+(x2)=11x2

It is given that the sum of the original two digit number and the reversed number is 66.

(11x20)+(11x2)=66

22x22=66

22x=6622

22x=88

x=8822=4

So, digit at ones place =x=4

and, the digit at tens place =x2=42=2

Hence, the original two digit number is 24.

There will be two such numbers i.e, one will be 24 and the other one will be 42.


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