1. **Problem statement:** Find the least number which when divided by 24 and 32 leaves a remainder of 6 in each case. 2. **Understanding the problem:** If a number $N$ leaves a remainder 6 when divided by 24 and 32, then $N - 6$ is divisible by both 24 and 32. 3. **Mathematical formulation:** Let $M = N - 6$. Then $M$ is divisible by 24 and 32, so $M$ is a common multiple of 24 and 32. 4. **Find the least common multiple (LCM) of 24 and 32:** - Prime factorization: - $24 = 2^3 \times 3$ - $32 = 2^5$ - LCM takes the highest powers of primes: - $\text{LCM} = 2^5 \times 3 = 32 \times 3 = 96$ 5. **Calculate the least number $N$:** $$N = M + 6 = 96 + 6 = 102$$ 6. **Verification:** - $102 \div 24 = 4$ remainder $6$ - $102 \div 32 = 3$ remainder $6$ **Final answer:** The least number is $\boxed{102}$.