Question

The sum of two natural numbers is 4923. If we write the number 7 to the right side of the first number and erase the last digit of the
second number, the obtained numbers are equal. Find the initial numbers.
2​

Answer 1

Given : The sum of two natural numbers is 4923.

If we write the number 7 to the right side of the first number and erase the last digit of the second number, the obtained numbers are equal.

To Find : the initial numbers.

Solution:

Let say initial number are

X  and  Y

X + Y  = 4923

write the number 7 to the right side of the first number  

X7   = 10X  + 7

erase the last digit of the second number,

= [ Y/10]

10X  + 7  =    [ Y/10]

so Tens place of Y  must be  7

as Y is a maximum 4 digit number hence

[ Y/10] would be max 3 digit number

so X can be max 2 digit numbers

X can not be 1 digit number as then Y would be 3 digit number and it would not be possible to get sum  4923

X  = AB     where A and  B represent digits

Y  = AB7C  

           AB7C

      +        AB

     __________

          4923

 7 + A  = 2    => A must be 4 of 5  depending upon carry over from last operation

Similarly B can be  8 or 9  

from last digit A muse be 4 as its not possible to have carry forward from last operation as B is only digit

Using A = 4

           4B7C

      +        4B

     __________

           4923

C + B must be 13

and B + 1  = 9  => B = 8

=> C = 5

           4875

      +        48

     __________

           4923

initial numbers. were    48 and   4875  

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sum of two natural numbers is 4923.

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