model. A science-kit organiser must transport all 137 indivisible specimen cards. Each envelope holds at most eight cards. How many envelopes are needed, and how many cards are in the last envelope?
- Underline all 137 cards and at most eight per envelope. Draw an envelope as a container with eight spaces. Because every card must travel, a partly filled final envelope is allowed and a fraction of an envelope is not an available container.
- Build the division record: 8 × 17 = 136 and 137 − 136 = 1, so 137 = 8 × 17 + 1. Label 17 as full envelopes and 1 as a leftover card. The leftover has a different unit from the quotient.
- Open one more envelope for that card. Check both sides of the decision: 17 envelopes have capacity 136, which is insufficient; 18 have capacity 144, which is sufficient. There are seven unused card spaces across the final envelope.
- Change only the job: if the organiser may make complete eight-card sets and keep spare cards, the result is 17 complete sets and one spare card. Compare this with the fractional ribbon shares in the transfer. The arithmetic record stays 17 remainder 1; the requested unit and allowed actions determine the decision.
Worked example: 18 envelopes are needed, with one card in the last envelope. Seventeen full envelopes hold 136 cards, so the remaining card needs an eighteenth envelope. The answer is a count of envelopes, not 17 remainder 1 envelopes.