Xiao Ming has some candies. When he divides them among 3 friends, 2 candies are left. But when he divides them among 4 friends, it divides evenly. What is the minimum number of candies he could have?

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Xiao Ming has some candies and wants to share them with his friends. When he divides the candies among 3 friends, there are 2 candies left over. But when he divides them among 4 friends, the candies divide evenly with no remainder. We need to find the minimum number of candies he could have that satisfies both conditions. Now let's translate this word problem into mathematical language. Let n represent the number of candies. The first condition means n divided by 3 leaves remainder 2, which we write as n congruent to 2 modulo 3. The second condition means n divided by 4 leaves remainder 0, written as n congruent to 0 modulo 4. We can visualize this on number lines where red dots show numbers that leave remainder 2 when divided by 3, and blue dots show numbers divisible by 4. The green dot shows where both conditions are satisfied. Now let's develop a systematic approach to solve this problem. Since we know n must be congruent to 0 modulo 4, n must be a multiple of 4. So we can list the multiples of 4 and check which ones leave remainder 2 when divided by 3. Let's check: 4 divided by 3 gives remainder 1, so it doesn't work. 8 divided by 3 gives remainder 2, so this works! 12 divided by 3 gives remainder 0, so it doesn't work. 16 divided by 3 gives remainder 1, doesn't work. 20 divided by 3 gives remainder 2, so this also works. We can see that 8 is our first solution. Let's verify that 8 is indeed the correct answer by checking both conditions step by step. For condition 1: 8 divided by 3 equals 2 with remainder 2. This satisfies our first requirement. For condition 2: 8 divided by 4 equals 2 with remainder 0, which means it divides evenly. This satisfies our second requirement. We can visualize this: 8 candies divided among 3 friends gives 2 candies to each friend with 2 left over. And 8 candies divided among 4 friends gives exactly 2 candies to each friend with no remainder. Therefore, the minimum number of candies Xiao Ming could have is 8. Now let's verify our solution and explore the general pattern. If 8 is a solution, then all numbers of the form 8 plus 12k, where k is any non-negative integer, are also solutions. Why 12? Because 12 is the least common multiple of 3 and 4. Let's verify: when k equals 0, n equals 8, which we know works. When k equals 1, n equals 20: 20 divided by 3 gives 6 remainder 2, and 20 divided by 4 gives 5 remainder 0, so it works. When k equals 2, n equals 32, which also satisfies both conditions. This pattern continues indefinitely, but 8 remains the minimum positive solution to our candy problem.