Problem
Fewest river crossings under rules
33 people stand on one bank of a river. There is one boat and it holds at most 5. Someone must ride back to bring the boat over. Find the fewest crossings to get everyone across.
Reasoning · Grade 6-1 Deductive Logic (2)
Your answer
Watch out Five per trip is tempting, but the rower comes back — only 4 settle per round trip.
How to solve
Strategy Make a Systematic List — 33 people is too many to picture at once, so I first run a small version -- five or nine people -- and watch what one round trip actually accomplishes. That reveals the rule that drives everything: five go over but one has to row back, so a round trip resettles only 4 people. A table with a row per round trip then carries the count up to 33 and shows a schedule that finishes in a definite number of crossings, which proves that many crossings are enough. But 'enough' is not 'fewest'. To show no schedule can do better I stop tracking who is where and instead sort the crossings into just two kinds -- across and back -- and put a limit on what each kind can do to the far-bank count. That second, differently organised view is what turns an answer that works into the answer that is smallest.
1STEP 1
Try a small group first to see what a round trip really does
A round trip leaves 4 people across.
5 - 1 = 4 people resettled per round trip
Shrinking 33 to 9 makes the whole trip something you can act out with counters in under a minute, and the rule you discover there -- one rower always has to come back -- does not change when the group gets bigger.
2.OA.A.1Solve An Easier Related Problem2STEP 2
State the pattern: 4 per round trip, and the last trip is one-way
Only the last boatload does not return.
round trip: +5 - 1 = +4 (2 crossings), final trip: +5 (1 crossing)
Spotting that the same +5 then -1 repeats over and over is the pattern-finding a second grader already does with skip counting, and it replaces a long story with one repeating step.
2.OA.A.1Look For A Pattern3STEP 3
Find how many round trips are needed before the final boatload
Before the last five, 7 round trips are needed.
4 × r + 5 = 33 → 4 × r = 28 → r = 7, 4 × 7 + 5 = 33
This is one multiplication fact and one subtraction: take the last full boatload of 5 off the total first, then see how many groups of 4 fill the remaining 28.
4.OA.A.3Look For A Pattern4STEP 4
Build the schedule and count the crossings -- 15 is achievable
Building the schedule gives 15 crossings.
7 × 2 + 1 = 15 crossings
Writing one row per round trip keeps the two counts from drifting apart, and a fourth grader can verify the near-bank column drops by exactly 4 each row until the final 5 are left.
4.OA.A.3Make A Systematic List5STEP 5
Show that fewer than 15 crossings is impossible
Fewer cannot move everyone.
5 × (b+1) - 1 × b = 4 × b + 5 ≥ 33 → b ≥ 7 → 2 × b + 1 ≥ 15
Instead of asking who sits where, ask only 'how much can one crossing change the far-bank count' -- the best case is +5 going over and the least damage is -1 coming back, and multiplying those best cases by how many crossings of each kind there are gives a ceiling nothing can beat.
4.OA.A.3Organize Information In More WaysFewer than fifteen crossings is impossible, because each full round trip moves the group forward by only a fixed amount.
Why?
Every round trip nets the same number of people across, so the progress climbs by a fixed step and no trip can do better.
🧱A constant step reaches any termWhen a list climbs by the same amount every time, any term is the starting value plus that step repeated, so you never have to write out the ones between.
Why?
Counting how many whole round trips fit before the final one-way trip, and what is left over, is exactly a division with a remainder.
🧱Division with remainderAny number splits into a whole count of equal groups plus a leftover, and the leftover is always smaller than the group size.
6STEP 6
Put the two halves together
So the answer is 15.
15 ≤ answer and answer ≤ 15 → answer = 15
A 'fewest' question always has two halves -- show it can be done, and show it cannot be done in less -- and answering only the first half is the usual way to get this kind of problem wrong.
4.OA.A.3Make A Systematic ListAnswer
15 crossings
7 × 2 + 1 = 15
The answer is a count of one-way trips, so it must be a whole number, and it is odd -- which it has to be, since the boat starts on the near bank and finishes on the far bank. A quick sanity bracket: if the boat could magically cross with nobody rowing back, 33 people at 5 per boatload would need 7 boatloads (5 x 7 = 35, so 7 crossings), and 15 is a bit more than double that, which is exactly what you expect when almost every trip out has to be paid for with a trip home. Counting the people also balances: 7 round trips move 4 people each, 7 x 4 = 28, plus the final boatload of 5 gives 33, with no one crossing who does not need to and no one left behind.
Takeaway
Someone has to row the boat back, so each round trip really moves only 4 people -- and to answer a 'fewest' question you must both show a way that works and explain why no shorter way could.
- Try a small group first to see what a round trip really does
- State the pattern: 4 per round trip, and the last trip is one-way
- Find how many round trips are needed before the final boatload
- Build the schedule and count the crossings -- 15 is achievable
- Show that fewer than 15 crossings is impossible
- Put the two halves together