Discussion
A has some coins. A gives 1/4 of the total coins and 3 coins more to B. B gives 1/4 of the total coins he receives and 5 coins more to C. Further, C gives 1/3 of the coins he receives and 1 coin more to D. The number of coins D has now is 20. How many coins did A have in the beginning?
Answer: C
A had 820 coins.
Work backwards from D, who received everything he now has from C.
\[ \begin{aligned} \tfrac{c}{3} + 1 &= 20 \Rightarrow c = 57 \\ \tfrac{b}{4} + 5 &= 57 \Rightarrow b = 208 \\ \tfrac{a}{4} + 3 &= 208 \Rightarrow a = 820 \end{aligned} \]
Here c, b and a are the coins C received, B received and A had at the start. Check forward: A gives \(205 + 3 = 208\) to B, B gives \(52 + 5 = 57\) to C, and C gives \(19 + 1 = 20\) to D. 683 and 799 are not divisible by 4, and 1100 would give B 278 coins, a quarter of which is not a whole number.
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