Quadratic in X, Cubic in Y

Number Theory Level 4

$\large \color{#20A900}{X^2+Y^3=793}$ If $-9 \leq X, Y \leq 9$ and $X,Y \in \mathbb{Z}$ satisfy the above equation, then find the minimum value of $\color{#20A900}{\left( X+Y \right)}$ .

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18 1 17 0 None of the given choices. 16

2 solutions

Kyle Finch
May 14, 2015

$(-8)^{2}+9^{3}=793$

$X+Y=-8+9=1$

Moderator note:

You failed to show that there is only one solution.

Aahrghh, You got me there. Nice question @Sandeep Bhardwaj Sir! :)

Mehul Arora - 6 years, 1 month ago

$\ddot \smile$ Thanks !

Sandeep Bhardwaj - 6 years, 1 month ago

Is -8 considered a single digit number?

Reynan Henry - 6 years ago

Challenge master i dont know how it has one solution, i just saw the options and the rest is posted above. 😝

Kyle Finch - 6 years, 1 month ago

You may refer to my comment.

Niranjan Khanderia - 6 years ago

Challenge master: I think there are two solutions y=9, x could be 8 or -8

${ x }^{ 2 }+{ y }^{ 3 }\quad =\quad 793\\ { x }^{ 2 }\quad =\quad 793-y^{ 3 }\\ x\quad \le \quad 9\\ So,\quad 793-y^{ 3 }\quad \le \quad 9\\ and\quad y\quad \le \quad 9,\quad \\ from\quad above\quad two \quad inequality\\ y=9\quad so\quad { x }^{ 2 }=64\\ x=8\quad 0r\quad -8$

Mayank Chaturvedi - 6 years, 1 month ago

Yes this is the right answer, the only way to solve this problem is by trial and error.

Brilliant Mathematics Staff - 6 years ago

Before trial and error we can see some logic. $\sqrt{793}=28.....and~~\sqrt[3]{793}=9...~~~~~~~~~But -9\leq X \leq 9. ~\therefore~ Y>0\\ \text{ since maximum contribution of X is only 81.}\\ \sqrt[3]{ 793-81}=8.9....\therefore\text{Y must be 9 if there is a solution.}\\793-9^3=64!!!!=(\pm8)^2.\\\text{We take X= - 8, and Y=9}$

Niranjan Khanderia - 6 years ago

Oyeeee x=8 y=9 ni ho skta kyaaa galat kyu kataa bee

Sudhakar Sinha - 3 years, 3 months ago

-8 and 8 both can be in the solution as both have same value after squsing it

Aditya Raj - 2 years, 7 months ago

James Moors
Jun 23, 2015

First, one has that $\|X\| \leq 9 \: \Rightarrow \: X^2 \leq 81$ .

This gives $Y^3 \geq 793-81 = 712$ .

Now, $8^3 = 512$ and $9^3 = 729$ , so $Y=9$ is the only value for which the above inequality holds.

$X^2 + 729 = 793 \: \Rightarrow \: X^2 = 64 \: \Rightarrow \: X = -8$ or $X = 8$ .

Therefore, $min(X + Y) = (-8 + 9) = 1$ .

Quadratic in X, Cubic in Y

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2 solutions

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