Positive integers ordered pairs (x,y,z)

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Discussion Overview

The discussion revolves around finding the total number of positive integer ordered pairs \((x, y, z)\) that satisfy the equation \(3^x + 3^y + 3^z = 7299\). The scope includes mathematical reasoning and exploration of base conversions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation and asks for the total number of positive integer solutions.
  • Another participant converts 7299 to base 3, concluding that it can be represented as \(3^8 + 3^6 + 3^2\) and suggests that the set of solutions is \(\{2, 6, 8\}\), leading to 6 possible solutions.
  • A third participant expresses agreement with the second participant's approach, noting that their solution aligns with the base 3 representation of 7299.

Areas of Agreement / Disagreement

Participants generally agree on the base 3 representation of 7299 and the resulting solutions, but the total number of solutions and the method of counting them is not fully elaborated or confirmed by all participants.

Contextual Notes

The discussion does not clarify the assumptions regarding the counting of ordered pairs or the conditions under which the solutions are derived.

juantheron
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Total no. of positive integers ordered pairs of the equation $$3^x+3^y+3^z = 7299$$
 
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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.
 
[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]
 
Thanks mathbalarka and Soroban.

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

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