Solution: Find the largest integer $ p $ such that $ p^2 < 1000 $ and $ p $ is a multiple of 7.
["Title: Find the Largest Integer $ p $ Such That $ p^2 < 1000 $ and $ p $ Is a Multiple of 7", "When tasked with finding the largest integer $ p $ satisfying two conditions — $ p^2 < 1000 $ and $ p $ divisible by 7 — we combine basic number theory with strategic estimation. This problem is a great example of how rounding for inequalities and filtering by divisibility leads to efficient solutions.", "---", "### Step 1: Determine the largest integer $ p $ where $ p^2 < 1000 $", "We begin by solving the inequality:", "$$\np^2 < 1000\n$$", "Taking the square root of both sides:", "$$\np < \sqrt{1000}\n$$", "We know:", "$$\n\sqrt{1024} = 32 \quad \ ext{and} \quad 31^2 = 961, \quad 32^2 = 1024\n$$", "So:", "$$\n\sqrt{1000} \approx 31.62\n$$", "Thus, the largest integer $ p $ satisfying $ p^2 < 1000 $ is:", "$$\np < 31.62 \quad \Rightarrow \quad p \leq 31\n$$", "---", "### Step 2: Find the largest multiple of 7 less than or equal to 31", "Now we seek the largest multiple of 7 that is less than or equal to 31.", "List multiples of 7:", "$$\n7, 14, 21, 28, 35, \dots\n$$", "Among these, the largest one less than or equal to 31 is:", "$$\n28 \quad (\ ext{since } 35 > 31)\n$$", "Check:", "$$\n28^2 = 784 < 1000 \quad \ ext{✓}\n$$\n$$\n35^2 = 1225 > 1000 \quad \ ext{✗}\n$$", "---", "### Step 3: Confirm it’s the largest valid $ p $", "Is there any multiple of 7 between 28 and 31? No — 28 is the largest multiple of 7 in the range $ p \leq 31 $. So $ p = 28 $ satisfies both conditions:", "- $ p^2 = 784 < 1000 $\n- $ p = 28 $ is divisible by 7", "Thus, the largest integer $ p $ such that $ p^2 < 1000 $ and $ p $ is a multiple of 7 is:", "$$\n\boxed{28}\n$$", "---", "### Bonus Interpretation: Why This Method Works", "This approach efficiently narrows the solution space by:\n- First solving the inequality to define a strict upper bound,\n- Then filtering by divisibility to find the maximal eligible value.", "This pattern applies broadly in optimization and Diophantine problems where constraints involve both quadratic bounds and modular conditions.", "---", "### Key Summary\n- $ p < \sqrt{1000} \Rightarrow p \leq 31 $\n- Largest multiple of 7 ≤ 31 is $ 28 $\n- $ 28^2 = 784 < 1000 $, satisfies all constraints\n- Final answer: $ \boxed{28} $", "Use this method next time you need the largest multiple of $ k $ below a square root — straightforward and scalable!"]
