Problem 14 · 2019 AMC 8
Medium
Number Theory
mod-10divisibility
Isabella has 6 coupons that can be redeemed for free ice cream cones at Pete's Sweet Treats. In order to make the coupons last, she decides that she will redeem one every 10 days until she has used them all. She knows that Pete's is closed on Sundays, but as she circles the 6 dates on her calendar, she realizes that no circled date falls on a Sunday. On what day of the week does Isabella redeem her first coupon?
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Answer: C — Wednesday.
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Hint 1 of 2
A 10-day jump is the same as a 3-day jump on the weekly cycle (10 = 7 + 3, and a full week of 7 lands on the same weekday). So each coupon is 3 weekdays after the last.
Still stuck? Show hint 2 →
Hint 2 of 2
Stepping +3 six times lands on six different weekdays — they cover six of the seven, skipping exactly one. That skipped day has to be Sunday, which tells you where to start.
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Approach: 10 days = +3 weekdays; find the one weekday left untouched
- Adding 10 days moves the weekday forward by 10 − 7 = 3 each time (a full week doesn't change the weekday). So from the start day the six redemptions land at +0, +3, +6, +9, +12, +15 weekdays = +0, +3, +6, +2, +5, +1 after reducing by 7s.
- Those six offsets are all different and cover every weekday except +4. Since none of the six is a Sunday, Sunday is precisely the missing +4 day.
- So the start is 4 weekdays before Sunday: counting back Sat, Fri, Thu, Wed.
- Why this transfers: repeated equal steps around a 7-day cycle reduce to stepping by the remainder mod 7; because 3 shares no factor with 7, six steps hit six distinct days — so the single avoided day fixes everything.
Another way — just test the start days:
- Try starting Wednesday and step +3 weekdays each time: Wed, Sat, Tue, Fri, Mon, Thu — six dates, none on Sunday. It works.
- Any earlier choice (Mon, Tue) lands on a Sunday within the six steps, so the answer is Wednesday.
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