Case 13 · operation ordering · Medium · About 8–12 minutes
A fictional market ledger begins with three credits and ends with twenty-five. Five instruction cards record everything that happened, but the cards have been shuffled.
Adding before multiplying can change the result. Recover the order, using the ledger's small set of timing notes as well as its final balance.
The opening balance is 3. Apply each card exactly once to the whole current balance:
| Card | Instruction |
|---|---|
| A | Add 4 |
| B | Multiply by 2 |
| C | Subtract 3 |
| D | Multiply by 3 |
| E | Add 2 |
After all five cards, the balance is exactly 25. Every intermediate balance is a positive whole number.
The timing notes say:
- A happens before B.
- D happens before C.
- C happens before E.
“Before” does not necessarily mean immediately before. No other cards, fees, rounding, or changes occur. Multiplication applies to the full balance at that moment.
What is the order of the five cards, and what balance follows each one?
You can test orders forward, but working backward from 25 is often shorter. When reversing a multiply card, the previous balance must still be a whole number. The timing notes also restrict which card could be last.
Hint 1 - reveal when you want a nudge
B cannot be last because 25 is odd. D cannot be last because 25 is not divisible by 3. Which remaining cards are excluded by the timing notes?
Hint 2 - reveal when you want a nudge
E must be last, taking 23 to 25. Work backward again: which of A,B,C,D can produce 23 while obeying the notes?
Reveal the solution and explanation
The order is D, A, B, C, E. The balances are 9, 13, 26, 23, 25.
At the end, B is impossible because 25 is odd. D is impossible because 25 is not divisible by 3. A cannot be last because B must follow it. C cannot be last because E must follow it. Therefore E is last, and the balance before it was 23.
Among the remaining cards, B cannot produce odd 23, and D cannot produce 23 from a whole number. A still requires a later B. Therefore C came next-to-last, taking 26 down to 23.
To produce 26, the last of A, B, D cannot be A, which must precede B. It cannot be D, since 26 is not divisible by 3. Thus B doubled 13 to 26.
Of A and D, only A can produce 13 from a whole-number balance; it added 4 to 9. D tripled the opening 3 to 9. Each backward step is forced, so the complete order is unique.
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