The product of two positive integers is 504. If the first integer is 6 greater than 3 times the second...

1. TRANSLATE the problem information

• Given information:
  • Product of two positive integers = 504
  • First integer = 6 more than 3 times the second integer
  • Need to find the smaller integer
• What this tells us: We need two equations with two unknowns

2. TRANSLATE into mathematical equations

Let \(\mathrm{x}\) = first integer, \(\mathrm{y}\) = second integer

• From the conditions:
  • \(\mathrm{xy = 504}\)
  • \(\mathrm{x = 3y + 6}\)

3. INFER the solution strategy

• Since we have \(\mathrm{x}\) expressed in terms of \(\mathrm{y}\), substitute this into the first equation
• This will give us one equation with one unknown (\(\mathrm{y}\))

4. SIMPLIFY by substitution and algebraic manipulation

Substitute \(\mathrm{x = 3y + 6}\) into \(\mathrm{xy = 504}\):

\(\mathrm{(3y + 6)y = 504}\)

\(\mathrm{3y^2 + 6y = 504}\)

\(\mathrm{3y^2 + 6y - 504 = 0}\)

Divide everything by 3:

\(\mathrm{y^2 + 2y - 168 = 0}\)

5. SIMPLIFY using the quadratic formula

Using \(\mathrm{y = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}}\) where \(\mathrm{a = 1}\), \(\mathrm{b = 2}\), \(\mathrm{c = -168}\):

\(\mathrm{y = \frac{-2 ± \sqrt{4 + 672}}{2}}\)

\(\mathrm{y = \frac{-2 ± \sqrt{676}}{2}}\)

Calculate \(\mathrm{\sqrt{676} = 26}\) (use calculator):

\(\mathrm{y = \frac{-2 + 26}{2} = 12}\)

or

\(\mathrm{y = \frac{-2 - 26}{2} = -14}\)

6. APPLY CONSTRAINTS to select valid solutions

• Since we need positive integers: \(\mathrm{y = 12}\)
• Therefore: \(\mathrm{x = 3(12) + 6 = 42}\)

7. INFER the final answer

The two integers are 42 and 12, so the smaller integer is 12.

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE reasoning: Students struggle to correctly set up the system of equations from the word problem, particularly misinterpreting "6 greater than 3 times the second integer."

Some students write \(\mathrm{x = 6 + 3y}\) instead of \(\mathrm{x = 3y + 6}\) (though these are equivalent), or worse, they might write \(\mathrm{x = 6y + 3}\) or \(\mathrm{x + 6 = 3y}\). These setup errors lead to completely different quadratic equations and wrong solutions, causing them to select incorrect answer choices or get stuck completely.

Second Most Common Error:

Poor APPLY CONSTRAINTS execution: Students correctly solve the quadratic but fail to reject the negative solution \(\mathrm{y = -14}\), either forgetting the "positive integers" constraint or not understanding why negative values don't make sense in context.

This leads them to work with \(\mathrm{y = -14}\), getting \(\mathrm{x = 3(-14) + 6 = -36}\), then incorrectly selecting the "smaller" value as -36, which doesn't match any answer choice, leading to confusion and guessing.

The Bottom Line:

This problem tests whether students can bridge the gap between word problems and algebraic manipulation. The key challenge is translating complex verbal relationships into precise mathematical equations, then systematically solving while respecting real-world constraints.