Question

A two digit positive number is such that the product of the digits is 24 if 18 is subtracted from the number the digits are reversed find the numbers

Answer 1

$\huge{\underline{\purple{\rm{Solution:}}}}$

$\large{\underline{\red{\rm{Given:}}}}$

• A two digit positive number is such that the product of the digits is 24 if 18 is subtracted from the number the digits are reversed

$\large{\underline{\red{\rm{Let:}}}}$

• $\sf{The\: ones\: digit=X}$
• $\sf{The\: tens\: digit=Y}$
• $\sf{The\: Number=10Y+X}$

$\small{\underline{\red{\rm{As\:per\:the\: above\: information:}}}}$

$\sf{The\: product\:of\:its\: digits=24}$

$\sf{\implies XY=24}$

$\sf\green{\implies XY=24-----(i)}$

$\sf{\implies X=\dfrac{24}{Y}}$

$\sf{If\:18\:is\:subtracted\: from\:the\: number\:the\: digits\:are\: reversed}$

$\sf{\implies (1OY+X)-18=10X+Y}$

$\sf{\implies 10X-X+Y-1OY=-18}$

$\sf{\implies 9X-9Y=-18}$

$\sf{\implies Dividing\:by\:9\:on\:both\: side}$

$\sf\green{\implies X-Y=-2---(ii)}$

$\sf{\implies now,\: putting\:the\:value\:of\:x}$

$\sf{\implies \dfrac{24}{Y}-Y=-2}$

$\sf{\implies \dfrac{24-Y^2}{Y}=-2}$

$\sf{\implies Y^2-2Y-24=0}$

$\sf{\implies Y^2-6Y+4Y-24=0}$

$\sf{\implies Y(Y-6)+4(Y-6)=0}$

$\sf{\therefore Y=6\:and\:-4}$

• Negative value can't be kept here so the value of Y=6 putting it in EQ 1

$\sf{\implies X*6=24}$

$\sf{\implies 6X=24}$

$\sf{\implies X=4}$

Hence,

$\rm\green{\implies The\: Number=10Y+X=10*6+4=64}$

Answer 2

Since the product of the digits is $\sf{24,}$ if one digit is taken as $\sf{x,}$ then the other digit will be $\sf{\dfrac{24}{x}.}$

Hence let our two digit number be $\sf{10x+\dfrac{24}{x}.}$ Or,

$\longrightarrow\sf{10x+\dfrac{24}{x}=\dfrac{10x^2+24}{x}}$

So the two digit number obtained on reversing the digits will be,

$\longrightarrow\sf{10\left(\dfrac{24}{x}\right)+x=\dfrac{240}{x}+x}$

$\longrightarrow\sf{10\left(\dfrac{24}{x}\right)+x=\dfrac{240+x^2}{x}}$

This number is obtained on subtracting 18 from our original number. Thus,

$\longrightarrow\sf{\dfrac{10x^2+24}{x}-18=\dfrac{240+x^2}{x}}$

$\longrightarrow\sf{\dfrac{10x^2+24}{x}-\dfrac{240+x^2}{x}=18}$

$\longrightarrow\sf{\dfrac{10x^2+24-240-x^2}{x}=18}$

$\longrightarrow\sf{\dfrac{9x^2-216}{x}=18}$

$\longrightarrow\sf{9x^2-18x-216=0}$

$\longrightarrow\sf{x^2-2x-24=0}$

$\longrightarrow\sf{x^2-6x+4x-24=0}$

$\longrightarrow\sf{x(x-6)+4(x-6)=0}$

$\longrightarrow\sf{(x-6)(x+4)=0}$

Since $\sf{x}$ is a positive integer,

$\longrightarrow\sf{x=6}$

And,

$\longrightarrow\sf{\dfrac{24}{x}=4}$

Hence our two digit number is,

$\longrightarrow\sf{\underline{\underline{10x+\dfrac{24}{x}=64}}}$

And the two digit number obtained on reversing the digits is,

$\longrightarrow\sf{\underline{\underline{10\left(\dfrac{24}{x}\right)+x=46}}}$