The product of digits of a 2-digit number is 24. If we add 45 to the number, the new number obtained is a number formed by interchanging the digits.What is the original number?

Let the two digit number be 10x + y. According to the question, xy = 24 .... (i) and, 10x + y + 45 = 10y + x $Rightarrow 10 y+x-10 x-y=45$ $Rightarrow 9 y-9 x=45$ $Rightarrow 9(y-x)=45$ $Rightarrow y-x=frac{45}{9}=5 ldots$ (ii) $herefore(x+y)^{2}=(y-x)^{2}+4 x y$ $=5^{2}+4 imes 24$ $=25+96=121$ $Rightarrow x+y=sqrt{121}=11 ldots$ (iii) On adding equations (ii) and (iii), y – x + x + y = 5 + 11 $Rightarrow 2 y=16 Rightarrow y=8$ $herefore x y=24 Rightarrow 8 x=24$ $Rightarrow x=frac{24}{8}=3$ Required number = 10x + y = 10 × 3 + 8 = 38

Alternative:

Let two digit number = 10x + y

According to the question,

xy = 24 ......(i)

And 10x + y + 45 = 10y + x

9x – 9y = – 45

x – y = – 5 .....(ii)

From option (3)

x = 3, y = 8 Satisfies equation (i) and (ii)

:. Original number = 10 × 3 + 8 = 38