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Question

The sum of a two digit number and the number obtained by interchanging the digits is 99. If the digits differ by 3. find the number.

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Solution

Let the digit at units place be x and the digit at tens place be y.The number is 10y+x.The number obtained by interchanging the digits is 10x+y.Sum of these two numbers is 99.10y+x+10x+y=9911x+11y =9911x+y=99x+y=9 ...(i) The digits differ by 3; so, there will be two case.Case (1): x>yAs per case (1), x-y =3 ...(ii) By adding (i) and (ii), we get : x+y=9 + x-y=3 2x =12 x=6Substituting x=6 in x+y=9, we get: 6+y=9y=3So, the two digit number = 10y+x =10×3+6 =36 Case (2):x<yAs per case (2), y-x=3-x+y=3 ...(iii) By adding (i) and (iii), we get: x+y=9 -x+y=3 2y=12 y=6Substituting y=6 in x+y=9, we get: x+6=9x=3So, the two digit number =10y+x =10×6+3 =63

Therefore, the two digit number is 36 or 63

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