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JEE Main 2023
Permutations & Combinations
Permutations and Combinations
Medium

Question

The number of elements in the set S={(x,y,z):x,y,zZ,x+2y+3z=42,x,y,z0}\mathrm{S}=\{(x, y, z): x, y, z \in \mathbf{Z}, x+2 y+3 z=42, x, y, z \geqslant 0\} equals __________.

Answer: 2

Solution

Key Concepts and Formulas

  • Diophantine Equation: An equation where only integer solutions are of interest.
  • Non-negative Integer Solutions: Solutions to an equation where the variables can only take non-negative integer values (0, 1, 2, ...).
  • Iterative Approach: When dealing with multiple variables, iterate through possible values of one variable to simplify the problem.

Step-by-Step Solution

Step 1: Isolate x and establish bounds for z.

We are given the equation x+2y+3z=42x + 2y + 3z = 42, where x,y,z0x, y, z \ge 0 are integers. We want to find the number of possible triples (x,y,z)(x, y, z) that satisfy this equation. First, isolate xx: x=422y3zx = 42 - 2y - 3z Since x0x \ge 0, we have 422y3z042 - 2y - 3z \ge 0, which implies 2y+3z422y + 3z \le 42. Since y0y \ge 0, we must have 3z423z \le 42, which means z14z \le 14. Also, since z0z \ge 0, we know 0z140 \le z \le 14.

Step 2: Iterate through possible values of z.

We will now iterate through possible integer values of zz from 0 to 14. For each value of zz, we will determine the possible values of yy. Since 2y+3z422y + 3z \le 42, we have 2y423z2y \le 42 - 3z, so y423z2y \le \frac{42 - 3z}{2}. Since y0y \ge 0, we have 0y423z20 \le y \le \frac{42 - 3z}{2}.

  • For a fixed zz, the number of possible integer values for yy is 423z2+1\left\lfloor \frac{42 - 3z}{2} \right\rfloor + 1. (We add 1 to include the case where y=0y=0).
  • Since x=422y3zx = 42 - 2y - 3z, for each pair of yy and zz we will have exactly one value of xx satisfying the equation x+2y+3z=42x+2y+3z=42. Thus, the number of solutions is the sum of the number of possible values for yy for each value of zz.

Step 3: Calculate the number of solutions for each z and sum them up.

We need to calculate the sum: z=014(423z2+1)\sum_{z=0}^{14} \left( \left\lfloor \frac{42 - 3z}{2} \right\rfloor + 1 \right)

We can expand this summation as follows: (422+1)+(4232+1)+(4262+1)++(423(14)2+1)\left( \left\lfloor \frac{42}{2} \right\rfloor + 1 \right) + \left( \left\lfloor \frac{42-3}{2} \right\rfloor + 1 \right) + \left( \left\lfloor \frac{42-6}{2} \right\rfloor + 1 \right) + \dots + \left( \left\lfloor \frac{42-3(14)}{2} \right\rfloor + 1 \right) =z=014(423z2+1)=z=014423z2+z=0141=z=014423z2+15 = \sum_{z=0}^{14} \left( \left\lfloor \frac{42 - 3z}{2} \right\rfloor + 1 \right) = \sum_{z=0}^{14} \left\lfloor \frac{42 - 3z}{2} \right\rfloor + \sum_{z=0}^{14} 1 = \sum_{z=0}^{14} \left\lfloor \frac{42 - 3z}{2} \right\rfloor + 15

Now we can expand the first summation term by term: 422+392+362+332+302+272+242+212+182+152+122+92+62+32+02\left\lfloor \frac{42}{2} \right\rfloor + \left\lfloor \frac{39}{2} \right\rfloor + \left\lfloor \frac{36}{2} \right\rfloor + \left\lfloor \frac{33}{2} \right\rfloor + \left\lfloor \frac{30}{2} \right\rfloor + \left\lfloor \frac{27}{2} \right\rfloor + \left\lfloor \frac{24}{2} \right\rfloor + \left\lfloor \frac{21}{2} \right\rfloor + \left\lfloor \frac{18}{2} \right\rfloor + \left\lfloor \frac{15}{2} \right\rfloor + \left\lfloor \frac{12}{2} \right\rfloor + \left\lfloor \frac{9}{2} \right\rfloor + \left\lfloor \frac{6}{2} \right\rfloor + \left\lfloor \frac{3}{2} \right\rfloor + \left\lfloor \frac{0}{2} \right\rfloor =21+19+18+16+15+13+12+10+9+7+6+4+3+1+0=154= 21 + 19 + 18 + 16 + 15 + 13 + 12 + 10 + 9 + 7 + 6 + 4 + 3 + 1 + 0 = 154 Thus, the total number of solutions is 154+15=169154 + 15 = 169.

Common Mistakes & Tips

  • Forgetting the +1: Remember to add 1 when counting the number of possible values for yy, to account for the case where y=0y=0.
  • Integer Division: Be careful with the floor function. x\lfloor x \rfloor gives the largest integer less than or equal to xx.
  • Systematic Approach: Iterating through possible values of one variable and solving for the others is a reliable strategy for these types of problems.

Summary

We found the number of non-negative integer solutions to x+2y+3z=42x + 2y + 3z = 42 by isolating xx, establishing bounds for zz, and then iterating through possible values of zz. For each value of zz, we calculated the number of possible values for yy and summed these up to find the total number of solutions. This resulted in a total of 169 solutions.

Final Answer

The final answer is \boxed{169}.

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