the sum of a two digit number is 16. if digit at tens place is increased by 8 and digit at units place is decreased by 8 the digits of number found to be reversed. find the original number.
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QUESTION
the sum of a two digit number is 16. if digit at tens place is increased by 8 and digit at units place is decreased by 8 the digits of number found to be reversed. find the original number.
ANSWER
Had the constraint “ The sum of 2 digits is 16” been not there, the number would have been 19 and after adding 8 to 10th Place digit and after subtracting 8 from the unit place digit, the new number would have been 91, which is the number after reversing the digits in 19. But that’s not the case. Further, if we just think logically, we realize that such a number is not possible in decimal system or any number system below 10.
The next (popular) Number System after 10 is Hexadecimal System i. e. a Number System with Base 16.
Let there be a Number XY in Hexadecimal System. From the given, we have:
X + Y = 16 (Let’s consider 16 as a Decimal Number). Let this be equation No. 1.
16*X + Y ————— Original Number. Let this be number ‘I’.
16*Y + X ————— New Number after Reversing Digits of the Original Number. Let this be number ‘II’.
16*(X+8) + Y - 8 ———- Number formed after adding 8 to the 10th Place Digit and subtracting 8 from the Unit Digit. Let this be number ‘III’.
As per the problem statement the Numbers ‘II’ and ‘III’ are equal and therefore, we have
16*Y + X = 16*X + 128 + Y - 8, which gives
15*Y - 15*X = 120, which further gives
Y - X = 8 ————- Let this be equation No. 2.
Using equations No. 1 and 2, it is very easy to find that X and Y are 4 and 12 respectively.
answer by Arman..