**Problem Analysis**

Let's assume that Raju had 'x' chocolates.

When Raju made groups of 5 chocolates, he found out that 1 chocolate was left. This means that the number of chocolates 'x' is 1 more than a multiple of 5.

Similarly, when Raju made groups of 6 chocolates, he found out that 1 chocolate was left. This means that the number of chocolates 'x' is 1 more than a multiple of 6.

And when Raju made groups of 8 chocolates, he found out that 1 chocolate was left. This means that the number of chocolates 'x' is 1 more than a multiple of 8.

**Finding the Smallest Number**

To find the smallest number of chocolates Raju could have, we need to find the smallest number that satisfies all three conditions mentioned above.

We can start by finding the least common multiple (LCM) of the numbers 5, 6, and 8. LCM is the smallest multiple that is divisible by all the given numbers.

Let's find the LCM of 5, 6, and 8:

- Prime factors of 5: 5
- Prime factors of 6: 2, 3
- Prime factors of 8: 2, 2, 2

LCM = 5 * 2 * 2 * 2 * 3 = 120

So, the smallest number 'x' that satisfies all three conditions is 120.

**Explanation**

If Raju had 120 chocolates, he could make groups of 5, 6, and 8 chocolates.

- When he makes groups of 5 chocolates, there would be 24 groups of 5 chocolates (24 * 5 = 120), and 1 chocolate would be left.
- When he makes groups of 6 chocolates, there would be 20 groups of 6 chocolates (20 * 6 = 120), and 1 chocolate would be left.
- When he makes groups of 8 chocolates, there would be 15 groups of 8 chocolates (15 * 8 = 120), and 1 chocolate would be left.

So, the smallest number of chocolates Raju could have is 120.

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