2

$$a_1 + \ldots + a_5 = 10$$
where $2\leq a_k \leq 6$ for all $k=1,2,\ldots,5$.

Let $x_k := a_k - 2$, so
$0 \leq x_k \leq 4$, and has the same number of solutions as
$$ x_1 + \ldots + x_5 + 2\times 5 = 10$$
$$x_1 + \ldots + x_5 = 0.$$
However, this has only one solution, that is, $x_1 = x_2 = \ldots = x_5 = 0$.
The original equation seems to have more than one solution.
What went wrong here?

Natash1
  • 1,379
  • Nothing went wrong. $x_k=0$ so $a_k=2$ and this is the only solution because of the restriction of $a_k$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 03 '17 at 07:26
  • But for the original equation, $(2,2,6,0,0)$ is a solution as well as $(2,0,6,2,0)$ – Natash1 Nov 03 '17 at 07:26
  • @Natash1 No it's not, because the $a_k$ have to be between $2$ and $6$. – Jack M Nov 03 '17 at 07:27
  • I thought none of $a_k$ could be $0$ since $2 \leq a_k \leq 6$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Nov 03 '17 at 07:28
  • Thanks, Sorry, I'm not sure what I was thinking – Natash1 Nov 03 '17 at 07:32
  • @Natash1 Note that if $a_k \geq 2$ for each $k$, then $a_1 + ... + a_5 \geq 10$ must hold. However, since equality holds here, it follows that $a_i = 2 $ for each $i$. So only one assignment of the $a_k$ exists, and therefore what you have done is correct, with the problem statement that you have given. If you feel it is incorrect, you can recheck your problem statement. Also, this trick works in general i.e. even if the number of solutions was more than $1$, because you can place the solutions in bijective correspondence by adding (subtracting) $2$ to(from) each $a_k$. – Sarvesh Ravichandran Iyer Nov 03 '17 at 07:57

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