Question

The arithemetic mean and the geometric mean of two distinct $2$-digit numbers $x$ and $y$ are two integers one of which can be obtained by reserving the digits of the other(in base 10 representation) Then $x+y$ equals

• A. $82$
• B. $116$
• C. $130$
• D. $148$

Answer 1

The correct option is C $130$

Given, AM and GM of two number $x$ and $y$ are integer

$1<a<90<b<9$

$\frac{x+y}{2}=10a+b\Rightarrow x+y=2(10a+b)\left(1\right)\sqrt{xy}=10b+a\Rightarrow xy=(10b+a{)}^2(2)$

Squaring equation $\left(1\right)$ we get

$x^2+y^2+2xy=400a^2+80ab+4b^2\left(3\right)$

On multiplying $\left(2\right)$ by $4$ we get

$4xy=400b^2+4a^2+80ab\left(4\right)$
Subtracting equation $\left(4\right)$ from $\left(3\right)$ we get

$(x-y{)}^2=396(a^2-b^2\left)\right(x-y)=6\sqrt{11(a^2-b^2)}$

For $11(a^2-b^2)$ to be a perfect square $(a^2-b^2)$ should be equal to $11n$

$(a^2-b^2)=11,44,...(a+b)(a-b)=11\Rightarrow (x-y)=66$

and $(a+b)(a-b)=44\Rightarrow |(x-y)|=132$

but $a$ can be atmost $9$ and $b$ can be atmost $9$

$a+b\le 18(a+b)(a-b)\le 44a+b=11,a-b=4$ but there is no interger value of $a$

Hence $(a+b)(a-b)=11(x-y)=66$

$6^2-5^2=11a=6,b=5(a,b)(6,5)$
$x+y=2(10a+b)=2(10\times 6+5)=130$