Question

If $2\sqrt[3]{3x}+3\sqrt[3]{3y}$ such that x and y are positive integers, then sum of all possible values of x and y is

• A. 1600
• B. 2500
• C. 3200
• D. 3500

Answer 1

The correct option is D 3500

Given: $2\sqrt[3]{3x}+3\sqrt[3]{3y}=30$
$\Rightarrow 2\sqrt[3]{3x}=3(10-\sqrt[3]{3y})$ .......(i)
For $\sqrt[3]{3y}$ to be a positive integer, the possible values of y are only 8, 64, 216 and 512.
The perfect cube of odd numbers are neglected since the expression on left side of (i) is even.
And for perfect cube of even numbers which are greater than 512, $\sqrt[3]{3y}$ will be greater than 10, therefore, it will result in negative values of x. Thus, they are rejected. Substituting y = 8 in (i), we get
$2\sqrt[3]{3x}=3(10-\sqrt[3]{38})=3(10-2)$
$\Rightarrow \sqrt[3]{3x}=\frac{24}{2}=12$
$\Rightarrow x=(12{)}^3=1728$
Substituting y = 64 in (i), we get
$2\sqrt[3]{3x}=3(10-\sqrt[3]{364})$
$\Rightarrow \sqrt[3]{3x}=\frac{3(10-4)}{2}=\frac{3\times 6}{2}=3\times 3=9$
$\Rightarrow x=(9{)}^3=729$
Substituting y = 216 in (i), we get
$2\sqrt[3]{3x}=3(10-\sqrt[3]{3216})=3(10-6)$
$\Rightarrow 2\sqrt[3]{3x}=3\times 4$
$\Rightarrow \sqrt[3]{3x}=6$
$\Rightarrow x=(6{)}^3=216$
Substituting y = 512 in (i), we get
$2\sqrt[3]{3x}=3(10-\sqrt[3]{3512})$
$\Rightarrow 2\sqrt[3]{3x}=3(10-8)$
$\Rightarrow \sqrt[3]{3x}=\frac{3\times 2}{2}=3$
$\Rightarrow x=(3{)}^3=9$
$\therefore$ Required sum = (8 + 64 + 216 + 512) + (1728 + 729 + 216 + 27)
= 3500
Hence, the correct answer is option (d).