Solve 285a + 315b + 590c = 21,806. Does it have integer solutions, and what is the real solution set?

Step-by-step solution

1. 1

   Compute the gcd of the coefficients. $\gcd(285, 315) = 15$ and $\gcd(15, 590) = 5$, so

   $\gcd(285, 315, 590) = 5$

   Every term on the left is a multiple of $5$, so for any integers $a, b, c$ the whole left-hand side is a multiple of $5$.

2. 2

   Apply the divisibility test to the right-hand side. A linear Diophantine equation $\sum k_i x_i = N$ has integer solutions if and only if $\gcd(k_1,\dots,k_n)$ divides $N$. Here

   $21{,}806 = 5 \times 4361 + 1$

   so $21{,}806 \equiv 1 \pmod 5$ and the gcd condition fails.

3. 3

   Conclude: no integer solution exists. Because the left side is $\equiv 0 \pmod 5$ and the right side is $\equiv 1 \pmod 5$, no choice of integers can satisfy the equation. Note that "dividing through by 5" is not available here — $\tfrac{21806}{5} = 4361.2$ is not an integer, so writing $57a + 63b + 118c = 4361$ silently changes the problem.

4. 4

   Describe the real solution set instead. Over $\mathbb{R}$ there is one equation in three unknowns, so the solutions form a plane in $\mathbb{R}^3$. Solving for $a$ gives the parametrisation

   $a = \frac{21806 - 315b - 590c}{285}, \qquad b, c \in \mathbb{R}$

   two free parameters, as expected for a plane.

5. 5

   Check a sample point. Taking $b = c = 0$ gives $a = \tfrac{21806}{285} = 76.5123$, and $285 \times 76.5123 = 21806.0$. Taking $b = 10$, $c = 5$: $a = \tfrac{21806 - 3150 - 2950}{285} = \tfrac{15706}{285} = 55.1088$, and $285(55.1088) + 3150 + 2950 = 21806.0$. Neither $a$ is an integer, consistent with the mod-5 obstruction.

6. 6

   When would integers exist? If the target were $21{,}805$ or $21{,}810$ — any multiple of $5$ — the gcd test would pass and integer solutions would exist; for instance $285(1) + 315(0) + 590(36) = 21{,}525$ shows how close the lattice comes. It is the specific value $21{,}806$ that rules them out.