Question

The sum of the digits of a three‐digit number is 14. The unit’s

digit is half the ten’s digit. When the digits are reversed the

resulting number is 198 more than the original number.

What is the original number?

Answer 1

With the information for the number "ABC" we can write the following equations or assumptions:

(a) For the sum of digits: $A+B+C=14$

(b) The unit's digit is half the ten’s digit: $C=B/2 \iff B=2C$

(c) When the digits are reversed the resulting number is 198 more than the original number:

$"ABC"=100 \times A+10 \times B+1 \times C \\ "CBA"=100 \times C+10 \times B+1 \times A \\ "CBA"="ABC"+198 \\ \text{we proceed to substitute and find} \\100C+\cancel{10B}+A=100A+\cancel{10B}+C+198 \\ 99C=99A+198 \\ \implies C=A+2 \iff A=C-2$

If we use the conclusions for (b) and (c) and we substitute this in (a) we can find a relation in terms of C:

$A+B+C=(C-2)+(2C)+C=4C-2=14 \\ \implies 4C=16 \\ \implies C=4 \\ \implies B=2(4)=8 \\ \implies A=4-2=2 \\ \implies "ABC"=284$

In conclusion, the original number ABC is 284.