To solve this problem, we need to determine the number of whole number solutions $(x, y)$ for the linear Diophantine equation:

$3x + 5y = 100$

Given that one solution is $x = 15$ and $y = 11$, we are to find all such whole number solutions and then determine $N$, the total count of solutions.

Step 1: General Solution of the Diophantine Equation

For a linear Diophantine equation $ax + by = c$, if there exists one particular solution $(x_0, y_0)$, then the general solution is given by:

$x = x_0 + k \cdot \frac{b}{\gcd(a, b)}$ $y = y_0 - k \cdot \frac{a}{\gcd(a, b)}$

where $k$ is an integer, and $\gcd(a, b)$ is the greatest common divisor of $a$ and $b$.

In this case:

• $a = 3$, $b = 5$, and $c = 100$.
• $\gcd(3, 5) = 1$, so solutions exist since 1 divides 100.
• One particular solution is $(x_0, y_0) = (15, 11)$, as given.

Thus, the general solution for $x$ and $y$ becomes:

$x = 15 + 5k$ $y = 11 - 3k$

Step 2: Determine Valid Range for $k$

For $x$ and $y$ to be whole numbers, they must be non-negative. We need to find the range of $k$ values such that both $x \geq 0$ and $y \geq 0$.

1. Condition for $x \geq 0$: $15 + 5k \geq 0$ $k \geq -3$

2. Condition for $y \geq 0$: $11 - 3k \geq 0$ $k \leq \frac{11}{3}$ $k \leq 3.67$

Since $k$ must be an integer, the possible integer values for $k$ range from $-3$ to $3$.

Step 3: Count the Solutions

Now we substitute each integer value of $k$ from $-3$ to $3$ into the general solution to find the corresponding $(x, y)$ pairs:

• For $k = -3$: $x = 15 + 5(-3) = 0, \quad y = 11 - 3(-3) = 20 \quad \Rightarrow (x, y) = (0, 20)$

• For $k = -2$: $x = 15 + 5(-2) = 5, \quad y = 11 - 3(-2) = 17 \quad \Rightarrow (x, y) = (5, 17)$

• For $k = -1$: $x = 15 + 5(-1) = 10, \quad y = 11 - 3(-1) = 14 \quad \Rightarrow (x, y) = (10, 14)$

• For $k = 0$: $x = 15 + 5(0) = 15, \quad y = 11 - 3(0) = 11 \quad \Rightarrow (x, y) = (15, 11)$

• For $k = 1$: $x = 15 + 5(1) = 20, \quad y = 11 - 3(1) = 8 \quad \Rightarrow (x, y) = (20, 8)$

• For $k = 2$: $x = 15 + 5(2) = 25, \quad y = 11 - 3(2) = 5 \quad \Rightarrow (x, y) = (25, 5)$

• For $k = 3$: $x = 15 + 5(3) = 30, \quad y = 11 - 3(3) = 2 \quad \Rightarrow (x, y) = (30, 2)$

Step 4: Conclusion

We have found 7 distinct solutions:

$(x, y) = (0, 20), (5, 17), (10, 14), (15, 11), (20, 8), (25, 5), (30, 2)$

Thus, $N = 7$.

Answer: $N = 7$

Would you like additional details or have any questions?

Related Questions to Explore

1. How can you determine if a Diophantine equation has solutions in whole numbers?
2. What is the general solution for linear Diophantine equations with different coefficients?
3. How can we check the existence of solutions for $ax + by = c$ without solving for $x$ and $y$?
4. What are the applications of Diophantine equations in real-world problems?
5. How can we adapt this method to find solutions when $x$ and $y$ are restricted to positive integers only?

Tip

In linear Diophantine equations, if you know one solution, you can use it to find all possible solutions using integer multiples, as shown in the general solution formula.