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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 (x−2). [∵ Digits differ by 2]The original two digit number =10(x−2)+x=10x−20+x=11x−20When the digits are reversed, then digit at ones place =(x−2)And, the digit at tens place =xSo, the reversed number =10x+(x−2)=11x−2It is given that the sum of the original two digit number and the reversed number is 66.∴(11x−20)+(11x−2)=66⇒22x−22=66⇒22x=66−22⇒22x=88⇒x=8822=4So, digit at ones place =x=4and, the digit at tens place =x−2=4−2=2Hence, 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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