Let's solve the problem step by step.

Let the total number of chocolates A initially had be $x$.

1. A gave away 60% of his chocolates: $\text{Chocolates given away} = 0.6x$ This means A has $0.4x$ chocolates left.

2. A gave 80% of the chocolates from the 60% to B: $\text{Chocolates given to B} = 0.8 \times 0.6x = 0.48x$

3. The remaining chocolates from the 60% were given to C: $\text{Chocolates given to C} = 0.6x - 0.48x = 0.12x$

4. The difference between the chocolates left with A and the chocolates with B is 144: $\text{Difference} = \text{Chocolates with B} - \text{Chocolates left with A}$ So, $0.48x - 0.4x = 144$

5. Simplify the equation: $0.08x = 144$ $x = \frac{144}{0.08} = 1800$

So, the total number of chocolates A had is $\boxed{1800}$.

Answer: (d) 1800

Would you like more details on this solution or have any other questions?

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Here are five related questions:

1. If A had 2000 chocolates initially, how many would B receive?
2. What if A gave 90% of his chocolates to B instead of 80%, what would be the new difference?
3. How many chocolates did C receive in the original problem?
4. If the difference was 200 instead of 144, what would be the total number of chocolates A had?
5. If A had initially 2400 chocolates, how many would he have left after giving 60% away?

Tip: Always double-check your percentage calculations, especially when dealing with multiple steps.