Problem 22 · 1990 AJHSME
Stretch
Number Theory
mod-arithmeticdivisors
Several students are seated at a large circular table. They pass around a bag of 100 pieces of candy. Each person takes one piece and passes the bag to the next person. If Chris takes the first and the last piece of candy, then the number of students at the table could be
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Answer: B — 11.
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Hint 1 of 2
The bag comes back to Chris once every full lap around the table. So Chris gets pieces 1, then 1+(one lap), then 1+(two laps), … The real question is: how far apart are his pieces?
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Hint 2 of 2
Chris grabs the 1st piece and the 100th piece — that's a gap of 99 pieces, which must be a whole number of laps. So the number of students has to divide evenly into 99. Test the choices for who splits 99 with nothing left over.
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Approach: the gap between Chris's first and last piece must be a whole number of laps
- With n students, the bag returns to Chris every n pieces (one lap). So Chris takes pieces 1, 1+n, 1+2n, … — each of his pieces is one full lap after the last.
- He gets the 1st *and* the 100th piece. The distance from piece 1 to piece 100 is 100 − 1 = 99 pieces, and that must be an exact whole number of laps. So n must divide 99 with nothing left over.
- Check the choices: 10, 19, 20, 25 all leave a remainder, but 99 ÷ 11 = 9 exactly. So there could be 11 students (Chris gets every 11th piece: 1, 12, 23, …, 89, 100).
- *Why this transfers:* 'same person at the start and at position k' around a circle means the number of people must divide the gap (here k−1) evenly — it's really a 'which numbers go in evenly' (divisor) question in disguise.
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