Excellent observation! Since $456$, $42z$, and $39x$ are divisible by 3, it follows that $41y$ must also be divisible by 3. Given that 41 is not divisible by 3, $y$ must be a multiple of 3 for $41y$ to be divisible by 3.

Thus, $y$ must be one of the following values: $3$, $6$, or $9$.

We'll test each case systematically:

Case 1: $y = 3$

Substitute $y = 3$ into the equation: $39x + 41(3) + 42z = 456$ $39x + 123 + 42z = 456$ $39x + 42z = 333$ Simplify by dividing the equation by 3: $13x + 14z = 111$ Now, we need to find natural numbers $x$ and $z$ that satisfy this equation.

Testing small values for $z$:

• $z = 1$: 13x + 14(1) = 111 \implies 13x = 97 \quad \text{(not a solution, as x must be an integer)}
• $z = 2$: $13x + 14(2) = 111 \implies 13x = 83 \quad \text{(not a solution)}$
• $z = 3$: $13x + 14(3) = 111 \implies 13x = 69 \implies x = 5.31 \quad \text{(not a solution)}$

None of these lead to integer values for $x$, so $y = 3$ does not yield a solution.

Case 2: $y = 6$

Substitute $y = 6$ into the equation: $39x + 41(6) + 42z = 456$ $39x + 246 + 42z = 456$ $39x + 42z = 210$ Simplify by dividing the equation by 3: $13x + 14z = 70$ Again, let's test small values for $z$:

• $z = 2$: $13x + 14(2) = 70 \implies 13x = 42 \implies x = \frac{42}{13} \quad \text{(not a solution)}$
• $z = 4$: $13x + 14(4) = 70 \implies 13x = 14 \implies x = \frac{14}{13} \quad \text{(not a solution)}$

Again, no solution for $y = 6$.

Case 3: $y = 9$

Substitute $y = 9$ into the equation: $39x + 41(9) + 42z = 456$ $39x + 369 + 42z = 456$ $39x + 42z = 87$ Simplify by dividing the equation by 3: $13x + 14z = 29$ Testing small values for $z$:

• $z = 1$: $13x + 14(1) = 29 \implies 13x = 15 \implies x = \frac{15}{13} \quad \text{(not a solution)}$

Thus, $y = 9$ also does not yield a valid solution.

Conclusion

Since none of the cases $y = 3$, $6$, or $9$ provided valid solutions, we can conclude that there is no solution to the equation $39x + 41y + 42z = 456$ where $x$, $y$, and $z$ are natural numbers.

Relative Questions:

1. Why does the divisibility condition help narrow down possible values for $y$?
2. How does simplifying the equation by dividing through by common factors help?
3. What are some other methods to solve Diophantine equations efficiently?
4. Could there be any special conditions under which the original equation might have solutions?
5. How can modular arithmetic be applied in more complex equations?

Tip:

When dealing with linear Diophantine equations, breaking down the problem using divisibility rules and simplifying the equation can significantly reduce the complexity and number of cases to check.