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Home » Play Like a Pro with The Riddler—Solving His Puzzles Is Addictive! - inBeat

Play Like a Pro with The Riddler—Solving His Puzzles Is Addictive! - inBeat

📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 Waitâperhaps the problem says âdivisible by 7, 11, and 13â but allows smaller multiples? Noâ1001 is the smallest.

📌 Unless the number is not required to be divisible by the product, but by **some**âbut no.

📌 Alternatively, perhaps the problem meant: divisible by **one** of them, but says âandââhighlighting shared constraint.

📌 Given the difficulty, reconsider: maybe the number is divisible by **each** 7, 11, 13âso must be multiple of 1001.

📌 Thus, **no** three-digit number qualifies.

📌 But this is impossibleâso likely a misinterpretation.

📌 Waitâperhaps âdivisible by 7, 11, and 13â is meant to be âdivisible by 7 and 11 and 13â, i.e., divisible by lcm=1001âbut again, too big.

📌 But thereâs a well-known problem: smallest number divisible by 7, 11, 13 is 1001.

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The rationalized form is $ \boxed{\sqrt{15} + \sqrt{10}} $.
Question: A science journalist reports that a lab uses two reagents, A...
Solution: Let $ a $ = units of A, $ b $ = units of B.
$$ a + b = 15 $$
$$ 12a + 8b = 156 $$
From the first equation, $ b = 15 - a $. Substitute into the second:
$$ 12a + 8(15 - a) = 156 $$
$$ 12a + 120 - 8a = 156 $$
$$ 4a = 36 $$
$$ a = 9 $$

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