Auggie spent all of his money in 5 stores. In each store he spent Rs 4 more…

Concept: In a reverse-spending problem, work backward from the known final state using the inverse of the per-step rule. If the amount left after a step equals half the entering amount minus a fixed sum, the reverse relation recovers the entering amount from the amount left; apply it once per step, moving from the last step back to the first.

Application:

1. Let a1, a2, a3, a4, a5 be the amount Auggie had on entering each of the five stores in turn, and let a6 = 0 be what remained after the fifth store (he spent everything).

2. The spending rule for each store gives: amount left after a store = (amount entering)/2 - 4, i.e. next = current/2 - 4. Reversing it recovers the entering amount from what remained: current = 2 × (next + 4).

3. Working backward from the fifth store to the first: a5 = 2 × (a6 + 4) = 2 × (0 + 4) = 8.

4. a4 = 2 × (a5 + 4) = 2 × (8 + 4) = 24.

5. a3 = 2 × (a4 + 4) = 2 × (24 + 4) = 56.

6. a2 = 2 × (a3 + 4) = 2 × (56 + 4) = 120.

7. a1 = 2 × (a2 + 4) = 2 × (120 + 4) = 248.

Cross-check (forward test):

1. Store 1: half of 248 is 124, plus the fixed Rs 4 is 128 spent, leaving 120.

2. Store 2: half of 120 is 60, plus 4 is 64 spent, leaving 56.

3. Store 3: half of 56 is 28, plus 4 is 32 spent, leaving 24.

4. Store 4: half of 24 is 12, plus 4 is 16 spent, leaving 8.

5. Store 5: half of 8 is 4, plus 4 is 8 spent, leaving 0 — exactly matching the condition that Auggie spent everything.