A Quest to Solve an Egg Riddle: Exploring LCM and Modular Arithmetic

Imagine a riddle about an egg basket: when the number of eggs is divided by 2, 3, 4, 5, or 6, there is always one egg left over. Moreover, the basket gets emptied when the eggs are taken out in groups of 7. The challenge is to determine the number of eggs in the basket. This problem involves modular arithmetic and the least common multiple (LCM) to find a solution that satisfies all given conditions.

The Problem and Initial Observations

The riddle can be described mathematically as finding a number N that satisfies these conditions:

N ≡ 1 (mod 2) N ≡ 1 (mod 3) N ≡ 1 (mod 4) N ≡ 1 (mod 5) N ≡ 1 (mod 6) N ≡ 0 (mod 7)

The key insight is to recognize that for the first five conditions, N - 1 must be divisible by 2, 3, 4, 5, and 6. We can use the least common multiple (LCM) to find the smallest such number.

Calculating the Least Common Multiple (LCM)

First, we need to find the LCM of 2, 3, 4, 5, and 6. To do this, we look at the prime factorizations of each number:

2 2 3 3 4 22 5 5 6 2 × 3

The LCM is obtained by taking the highest power of each prime factor:

Highest power of 2: 22 Highest power of 3: 31 Highest power of 5: 51

Therefore, the LCM is:

LCM(2, 3, 4, 5, 6)  22 × 3 × 5  4 × 3 × 5  60

This means N - 1 is a multiple of 60, so we can write:

N 60k 1 for some integer k.

Applying the Seventh Condition

The seventh condition states that N ≡ 0 (mod 7). Substituting N 60k 1 into this condition, we get:

60k 1 ≡ 0 (mod 7)

This simplifies to:

60k ≡ -1 (mod 7)

Next, we calculate 60 mod 7:

60 ÷ 7 8 remainder 4

So, 60 ≡ 4 (mod 7)

Substituting this into the equation, we get:

4k ≡ -1 (mod 7)

To solve for k, we can rewrite -1 as 6 (since -1 7 6):

4k ≡ 6 (mod 7)

Finding the Multiplicative Inverse

We need to find the multiplicative inverse of 4 mod 7. Testing values, we find:

4 × 1 4 ≡ 4 (mod 7) 4 × 2 8 ≡ 1 (mod 7)

The multiplicative inverse of 4 mod 7 is 2. Multiplying both sides of the equation 4k ≡ 6 (mod 7) by 2, we get:

2 × 4k ≡ 2 × 6 (mod 7)

8k ≡ 12 (mod 7)

Since 8 ≡ 1 (mod 7) and 12 ≡ 5 (mod 7), this simplifies to:

k ≡ 5 (mod 7)

This means we can write:

k 7m 5 for some integer m.

Final Calculation and Verification

Substituting back into the equation for N, we get:

N 60(7m 5) 1

N 420m 300 1

N 420m 301

To find the smallest positive N, we set m 0 and thus:

N 301

We verify that 301 satisfies all the initial conditions:

ConditionCalculationRemainder N ≡ 1 (mod 2)301 ÷ 2 150 remainder 11 N ≡ 1 (mod 3)301 ÷ 3 100 remainder 11 N ≡ 1 (mod 4)301 ÷ 4 75 remainder 11 N ≡ 1 (mod 5)301 ÷ 5 60 remainder 11 N ≡ 1 (mod 6)301 ÷ 6 50 remainder 11 N ≡ 0 (mod 7)301 ÷ 7 43 remainder 00

All conditions are satisfied, and thus the number of eggs in the basket is 301.