BRAIN-TEASERS 01

FUN WITH MATHEMATICS
BRAIN-TEASERS 01
PUZZLES & RIDDLES

🌟 Welcome to Fun with Mathematics! 🌟

These brain-teasers are designed to sharpen logical thinking, problem-solving skills, and mathematical reasoning. Each puzzle challenges you to think differently—whether it's about combinations, limiting factors, or number theory. Take your time, try to solve each one, then check the answers. Share these with friends and see who solves them fastest! Mathematics is not just about numbers—it's about patterns, creativity, and fun. Enjoy the journey!

🍕 Pizza Toppings Problem: In a pizza restaurant, you can get a basic pizza with cheese and tomato. You can choose from four extra toppings: olives, ham, mushrooms, and salami. Ross wants a pizza with two different extra toppings. How many different combinations can Ross choose?

✅ Answer: 6 combinations

📖 Why this works (Combination Formula):
This is a classic "combinations" problem. We have 4 toppings and need to choose any 2. The order does not matter (olives+ham is same as ham+olives). The mathematical formula is C(4,2) = 4! / (2! × 2!) = (4×3×2×1) / (2×1 × 2×1) = 24 / 4 = 6. The six combinations are: (Olives, Ham), (Olives, Mushrooms), (Olives, Salami), (Ham, Mushrooms), (Ham, Salami), (Mushrooms, Salami). This concept is used in probability, menu design, and even lottery calculations!

📚 Bookshelf Sets Problem

To complete one set of bookshelves, a carpenter needs: 4 long wooden panels, 6 short wooden panels, 12 small clips, 2 large clips, and 14 screws.
Stock available: 26 long panels, 33 short panels, 200 small clips, 20 large clips, 510 screws.
How many complete sets can the carpenter make?

✅ Answer: 5 complete sets

📖 Step-by-step reasoning (Bottleneck Principle):
This is a "limiting factor" problem. We divide each stock by the requirement per set:
• Long panels: 26 ÷ 4 = 6 sets (limit 6)
• Short panels: 33 ÷ 6 = 5 sets ← This is the bottleneck!
• Small clips: 200 ÷ 12 = 16 sets
• Large clips: 20 ÷ 2 = 10 sets
• Screws: 510 ÷ 14 = 36 sets
The carpenter cannot make more than 5 sets because short panels run out first. This concept applies to manufacturing, cooking recipes, and project management.

🥭 Mangoes in the Basket – Classic LCM Riddle

When counted in twos → 1 extra. In threes → 2 extra. In fours → 3 extra. In fives → 4 extra. In sixes → 5 extra. But counted in sevens → no extra. What is the minimum number of mangoes?

✅ Answer: 119 mangoes

📖 Mathematical solution (LCM & Divisibility):
The condition "remainder = k-1 when divided by k" means (N + 1) is divisible by 2, 3, 4, 5, and 6. The LCM of 2, 3, 4, 5, 6 is 60. So N + 1 = 60t, meaning N = 60t - 1. Also, N must be divisible by 7 (no remainder when counted in sevens). Trying t = 1 → N = 59 (not divisible by 7). t = 2 → N = 119 (119 ÷ 7 = 17 exactly). So 119 is the smallest solution. This type of problem is called the "Chinese Remainder Theorem" and appears in number theory and cryptography.

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