π§ BRAIN-TEASER 05
Boatmen, Staircase & Number Puzzles
π Step-by-Step Explanation
πͺ Staircase Footmarks Puzzle
Problem: Three girls climb down a staircase. Girl A climbs 2 steps at a time, Girl B climbs 3 steps, Girl C climbs 4 steps. They start together from the top, leave footmarks, and all reach the bottom in complete steps. On which steps would there be exactly one pair of footmarks? Which steps have no footmarks?
❌ Steps with no footmarks: 1, 5, 7, 11
• Girl A (2 steps/stride) lands on: 2, 4, 6, 8, 10, 12
• Girl B (3 steps/stride) lands on: 3, 6, 9, 12
• Girl C (4 steps/stride) lands on: 4, 8, 12
Counting footmarks per step:
Step 1: none | Step 2: A only | Step 3: B only | Step 4: A and C (two) | Step 5: none | Step 6: A and B (two)
Step 7: none | Step 8: A and C (two) | Step 9: B only | Step 10: A only | Step 11: none | Step 12: all three
Therefore, steps 2,3,9,10 have exactly one footmark. Steps 1,5,7,11 have none.
π️ Soldiers in Rows – Chinese Remainder Theorem
Problem: A group of soldiers: when arranged in rows of 3, there is 1 extra. In rows of 5, there are 2 extra. In rows of 7, there are 3 extra. Find the minimum number of soldiers.
We need N such that:
N ≡ 1 (mod 3) → N leaves remainder 1 when divided by 3
N ≡ 2 (mod 5) → N leaves remainder 2 when divided by 5
N ≡ 3 (mod 7) → N leaves remainder 3 when divided by 7
Let's test numbers that satisfy the first two conditions:
Numbers ≡ 1 mod 3: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52...
Among these, which leave remainder 2 when divided by 5? Check: 7(remainder2✓), 22(remainder2✓), 37(remainder2✓), 52(remainder2✓)
Now check remainder when divided by 7:
7 ÷ 7 = 1 remainder 0 ❌ | 22 ÷ 7 = 3 remainder 1 ❌ | 37 ÷ 7 = 5 remainder 2 ❌ | 52 ÷ 7 = 7 remainder 3 ✅
Therefore, the minimum number is 52 soldiers.
π‘ This is a classic application of the Chinese Remainder Theorem, used in cryptography and computer science!
π’ Four 9's to Make 100
Challenge: Use exactly four 9's and any mathematical operations (+, -, ×, ÷, √, !, etc.) to make 100.
• 99 + (9 ÷ 9) = 99 + 1 = 100
• 9 × 9 + 9 + 9 = 81 + 18 = 99 (not 100, close!)
• (9 + 9/9) × 9 + 9? Let's check: (9+1)×9+9 = 10×9+9 = 99
• 999/9.99? That uses more than four 9's.
The simplest and most elegant is 99 + 9/9 = 100.
π‘ This puzzle teaches creative thinking and that there can be multiple valid solutions to the same problem!
π Number of Digits in 2³⁰
Question: How many digits are in the product 2 × 2 × 2 × ... × 2 (30 times)? That is, 2³⁰.
2³⁰ = 2¹⁰ × 2¹⁰ × 2¹⁰ = 1024 × 1024 × 1024 = 1,073,741,824 → This number has 10 digits.
Method 2 – Using Logarithms (for larger exponents):
Number of digits = floor(30 × log₁₀2) + 1
log₁₀2 ≈ 0.30102999566
30 × 0.30102999566 = 9.0308998698
floor(9.0308998698) = 9
9 + 1 = 10 digits
π‘ Fun fact: 2³⁰ = 1,073,741,824 is exactly 1 Gibibyte (1 GiB) in computer memory! This is why computers have RAM sizes like 1GB, 2GB, 4GB, 8GB, 16GB, 32GB — they are powers of two.
π Why These Puzzles Matter π
• The boatmen puzzle teaches relative speed and the concept of combined distance.
• The staircase puzzle introduces LCM and set theory in a fun way.
• The soldiers puzzle is a real application of the Chinese Remainder Theorem used in cryptography!
• Four 9's builds creative mathematical thinking.
• 2³⁰ digits connects math to computer science.
Every puzzle you solve builds a stronger, more flexible mind. Keep going!
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