Positive integers ordered pairs (x,y,z)

[juantheron]

Total no. of positive integers ordered pairs of the equation $$3^x+3^y+3^z = 7299$$

 

[mathbalarka]

Spoiler

WLOG $z > y > x$. Then $7299 = 3^x + 3^y + 3^z = 3^x(1 + 3^{y-x} + 3^{z-x})$. Note that $7299 = 3^2 \cdot 811$, thus $x$ is either $1$ or $2$. However $x = 2$ is the only possible candidate as $1 + 3^{y-x} + 3^{z-x}$ is not divisible by $3$. Thus. $3^y + 3^z = 7299 - 3^2 = 7290 = 3^6 \cdot 10$. Hence, as $y < z$, $3^y(1+3^{z-y}) = 3^6 \cdot 10$ and $y = 6$ as $1 + 3^{z-y}$ is not divisible by $3$. Hence $3^z = 7290 - 729 = 6561 = 3^8$. Thus $z=8$.

$(x, y, z) = (2, 6, 8)$ is the only possible solution upto rearrangement. Hence there are $3! = 6$ of them.

 

[soroban]

[sp]Convert 7299 to base 3: .$$101,000,100_3$$
Therefore: .$$7299 \:=\: 3^8 + 3^6 + 3^2 \quad\Rightarrow\quad \{x,y,z\} \,=\,\{2,6,8\}\;\text{ . . . 6 solutions}$$ [/sp]

 

[juantheron]

Thanks mathbalarka and Soroban.

My solution is same as Soroban (Base $3$ Representation of $7299$)