Question

The number of elements in the set S = {(x, y, z) : x, y, z ∈ Z, x + 2y + 3z = 42, x, y, z ≥ 0} equals ____

Answer 1

Correct Answer: 169

We need to find the number of non-negative integer solutions to the equation \(x + 2y + 3z = 42\), where \(x, y, z \in \mathbb{Z}_{\ge 0}\).

Concept Used:

We can count the number of non-negative integer solutions by fixing the value of \(z\) and then counting the number of solutions for \(x\) and \(y\) for each fixed \(z\).

Step-by-Step Solution:

Step 1: Rewrite the equation in terms of \(z\).

\[ x + 2y = 42 - 3z \]

Since \(x, y \ge 0\), we require \(42 - 3z \ge 0 \Rightarrow z \le 14\). So \(z\) can range from 0 to 14.

Step 2: For a fixed \(z\), find the number of non-negative integer solutions \((x, y)\) to \(x + 2y = 42 - 3z\).

Let \(N = 42 - 3z\). Then \(x = N - 2y\), and we require \(N - 2y \ge 0 \Rightarrow y \le \frac{N}{2}\).

So \(y\) can range from 0 to \(\left\lfloor \frac{N}{2} \right\rfloor\).

Thus, for fixed \(z\), the number of solutions is:

\[ \left\lfloor \frac{N}{2} \right\rfloor + 1 = \left\lfloor \frac{42 - 3z}{2} \right\rfloor + 1 \]

Step 3: Compute the number of solutions for each \(z\) from 0 to 14.

We need to consider the parity of \(42 - 3z\):

• If \(z\) is even, \(3z\) is even, so \(42 - 3z\) is even.
• If \(z\) is odd, \(3z\) is odd, so \(42 - 3z\) is odd.

Let's compute \(\left\lfloor \frac{42 - 3z}{2} \right\rfloor + 1\) for \(z = 0, 1, 2, \dots, 14\):

For \(z = 0\): \(N = 42\), even, \(\left\lfloor \frac{42}{2} \right\rfloor + 1 = 21 + 1 = 22\)

For \(z = 1\): \(N = 39\), odd, \(\left\lfloor \frac{39}{2} \right\rfloor + 1 = 19 + 1 = 20\)

For \(z = 2\): \(N = 36\), even, \(\left\lfloor \frac{36}{2} \right\rfloor + 1 = 18 + 1 = 19\)

For \(z = 3\): \(N = 33\), odd, \(\left\lfloor \frac{33}{2} \right\rfloor + 1 = 16 + 1 = 17\)

For \(z = 4\): \(N = 30\), even, \(\left\lfloor \frac{30}{2} \right\rfloor + 1 = 15 + 1 = 16\)

For \(z = 5\): \(N = 27\), odd, \(\left\lfloor \frac{27}{2} \right\rfloor + 1 = 13 + 1 = 14\)

For \(z = 6\): \(N = 24\), even, \(\left\lfloor \frac{24}{2} \right\rfloor + 1 = 12 + 1 = 13\)

For \(z = 7\): \(N = 21\), odd, \(\left\lfloor \frac{21}{2} \right\rfloor + 1 = 10 + 1 = 11\)

For \(z = 8\): \(N = 18\), even, \(\left\lfloor \frac{18}{2} \right\rfloor + 1 = 9 + 1 = 10\)

For \(z = 9\): \(N = 15\), odd, \(\left\lfloor \frac{15}{2} \right\rfloor + 1 = 7 + 1 = 8\)

For \(z = 10\): \(N = 12\), even, \(\left\lfloor \frac{12}{2} \right\rfloor + 1 = 6 + 1 = 7\)

For \(z = 11\): \(N = 9\), odd, \(\left\lfloor \frac{9}{2} \right\rfloor + 1 = 4 + 1 = 5\)

For \(z = 12\): \(N = 6\), even, \(\left\lfloor \frac{6}{2} \right\rfloor + 1 = 3 + 1 = 4\)

For \(z = 13\): \(N = 3\), odd, \(\left\lfloor \frac{3}{2} \right\rfloor + 1 = 1 + 1 = 2\)

For \(z = 14\): \(N = 0\), even, \(\left\lfloor \frac{0}{2} \right\rfloor + 1 = 0 + 1 = 1\)

Step 4: Sum all these values.

\[ \text{Total} = 22 + 20 + 19 + 17 + 16 + 14 + 13 + 11 + 10 + 8 + 7 + 5 + 4 + 2 + 1 \]

Group them for easier addition:

First group: \(22 + 20 = 42\)

Second group: \(19 + 17 = 36\)

Third group: \(16 + 14 = 30\)

Fourth group: \(13 + 11 = 24\)

Fifth group: \(10 + 8 = 18\)

Sixth group: \(7 + 5 = 12\)

Seventh group: \(4 + 2 + 1 = 7\)

Now sum these group totals:

\[ 42 + 36 = 78,\quad 78 + 30 = 108,\quad 108 + 24 = 132,\quad 132 + 18 = 150,\quad 150 + 12 = 162,\quad 162 + 7 = 169 \]

Hence, the number of elements in the set \(S\) is 169.