The problem asks for the least number that leaves a remainder of 3 when divided by 4, 6, 8, 10, and 12.

Finding the Least Common Multiple (LCM)

To find the least number satisfying the condition, we first need to find the Least Common Multiple (LCM) of the divisors: 4, 6, 8, 10, and 12.

1. Prime Factorization: Find the prime factors of each number.
   • $4 = 2^2$
   • $6 = 2 \times 3$
   • $8 = 2^3$
   • $10 = 2 \times 5$
   • $12 = 2^2 \times 3$
2. Calculate LCM: The LCM is found by taking the highest power of each prime factor present in any of the numbers.
   • The prime factors involved are 2, 3, and 5.
   • Highest power of 2 is $2^3$.
   • Highest power of 3 is $3^1$.
   • Highest power of 5 is $5^1$.

   Therefore, $\text{LCM}(4, 6, 8, 10, 12) = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.

Calculating the Final Number

The problem states that the number leaves a remainder of 3 in all cases. This means the required number is 3 more than the LCM of the divisors.

Least Number = LCM + Remainder

Least Number = $120 + 3 = 123$.

Verification

Let's check if 123 leaves a remainder of 3:

• $123 \div 4 = 30$ remainder $3$ ($123 = 4 \times 30 + 3$)
• $123 \div 6 = 20$ remainder $3$ ($123 = 6 \times 20 + 3$)
• $123 \div 8 = 15$ remainder $3$ ($123 = 8 \times 15 + 3$)
• $123 \div 10 = 12$ remainder $3$ ($123 = 10 \times 12 + 3$)
• $123 \div 12 = 10$ remainder $3$ ($123 = 12 \times 10 + 3$)

The condition holds true for the number 123.