Problem
Locate a number in a repeating cycle
Six students sit in seats A to F. Play runs clockwise, so after F it is A again. A says 1, the next says 2, and so on. Find the seat that says 100.
Reasoning · Grade 4-1 Number Arrangement Tables
Your answer
Watch out Reading the remainder as a seat breaks at 0 — that case is the last seat.
How to solve
Strategy Look for a Pattern — Saying 100 numbers out loud around the hexagon would work but is far too slow. Because there are only six seats, the seating chart repeats itself every 6 numbers, so I list the first couple of laps to see exactly how the repeat works, describe the pattern by the remainder after dividing by 6, and then apply it once to 100.
1STEP 1
Number the seats around the hexagon
Number the seats 1 to 6.
A=1, B=2, C=3, D=4, E=5, F=6
Turning the six circles in the drawing into the numbers 1 to 6 keeps the clockwise order but makes it something a solver can compute with.
4.OA.C.5Draw A Diagram2STEP 2
List the first two laps
Two laps show it repeats every six.
1 2 3 4 5 6_A B C D E F 7 8 9 10 11 12_A B C D E F 13 → A
Six seats means the turn comes back to where it started after exactly six calls, which a solver can see just by walking around the drawing twice.
4.OA.C.5Make A Systematic List3STEP 3
Turn the cycle into a remainder rule
So the remainder on dividing by 6 names the seat.
10 ÷ 6 = 1 R 4 → 4th seat = D ✓
The complete laps are wasted effort — only the leftover part of the last lap decides where the counting stops, and testing the rule on 10 shows it is set up correctly.
4.NBT.B.6Solve An Easier Related ProblemThe complete laps around the hexagon are wasted effort, so only the leftover part of the last lap decides where the counting stops.
Why?
Six seats mean the turn comes back to where it started after exactly six calls, so a whole lap changes nothing.
🧱A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.
Why?
Splitting 100 calls into whole laps and a leftover is possible in exactly one way, and the leftover is what the question needs.
🧱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.
4STEP 4
Divide 100 by 6
100 divided by 6 leaves 4.
100 ÷ 6 = 16 R 4, 6 × 16 + 4 = 96 + 4 = 100
Dividing with a remainder is exactly the question 'how many whole laps fit, and how far into the next one do we get?'
4.NBT.B.6Look For A Pattern5STEP 5
Read off the winning seat
The fourth seat wins, so D.
97 → A, 98 → B, 99 → C, 100 → D
Counting out the last four calls by hand is a short, concrete finish that confirms what the remainder rule predicted.
4.OA.A.3Make A Systematic ListAnswer
D
100 ÷ 6 = 16 R 4
The answer is one of the six printed seat labels, as it must be. The rule can be spot-checked at both ends: the first number, 1, gives 1 divided by 6 = 0 remainder 1, which points to A — correct, since A starts. And 96 gives remainder 0, pointing to F, which fits because 96 = 6 x 16 lands exactly at the end of a lap. Seat D collects the numbers 4, 10, 16, 22, ... , and 100 fits that list because 100 - 4 = 96 is a multiple of 6, so D is consistent.
Takeaway
With 6 seats the count repeats every 6, so just divide 100 by 6 and let the leftover 4 walk you around: A, B, C, D — seat D wins.
- Number the seats around the hexagon
- List the first two laps
- Turn the cycle into a remainder rule
- Divide 100 by 6
- Read off the winning seat