AMC 8 · 2019 · #14

Grade 4 number-theorylogic

Archetype Modular Cycle Frequency Count └ Sub-archetype Calendar Day Lookup

modular-arithmeticsequences-arithmeticpattern-recognition pattern-recognitionsystematic-enumeration ↑ Prerequisites: modular-arithmeticsequences-arithmetic

📏 Medium solution 💡 3 insights

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Problem

Isabella has 6 free-ice-cream coupons and decides to use one every 10 days until they are gone. Pete's is closed on Sundays, and when she circles all 6 redemption dates on the calendar, not a single one lands on a Sunday. On which day of the week does she redeem the first coupon?

Pick an answer.

(A)

$text{Monday}$

(B)

$text{Tuesday}$

(C)

$text{Wednesday}$

(D)

$text{Thursday}$

(E)

$text{Friday}$

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Days of the week form a 7-day cycle, so jumping 10 days really means jumping 10 - 7 = 3 days forward each time. Tool #5 (Look for a Pattern) lets us discover that hidden "+3 each jump" rule. Tool #2 (Systematic List) then lays out all 6 redemption weekdays in order so we can SEE which weekdays are covered and which one is skipped. Finally, since the problem is multiple-choice and the "no Sunday" rule eliminates four of the five candidates, Tool #3 (Eliminate Possibilities) confirms the answer quickly.

1STEP 1

A week is 7 days, so only the leftover past a full week moves the weekday. As 10 ÷ 7 leaves 3, each jump shifts the weekday forward 3 days.

10 ÷ 7 = 1 remainder 3, so each 10-day jump moves the weekday forward by 3 days.

2STEP 2

Label the first weekday D, then add 3 each jump, subtracting 7 whenever you pass a full week, giving offsets D, D+3, D+6, D+2, D+5, D+1.

1st: D 2nd: D+3 3rd: D+6 4th: D+9 → D+2 5th: D+12 → D+5 6th: D+15 → D+1

3STEP 3

Sorted, the six offsets are D, D+1, D+2, D+3, D+5, D+6 — six of the seven possible offsets {0..6}; the one missing is D+4.

Missing offset = D + 4

4STEP 4

The skipped weekday must be Sunday, so D+4 is Sunday; counting 4 weekdays back from Sunday (Sat, Fri, Thu, Wed) gives D = Wednesday.

Sunday - 4 days = Wednesday ⟹ D = Wednesday

5STEP 5

Testing each start, only Wednesday skips Sunday — Monday skips Friday, Tuesday Saturday, Thursday Monday, Friday Tuesday — confirming (C).

Answer

text{Wednesday}

Verify by hand starting from Wednesday: Wed → Sat → Tue → Fri → Mon → Thu. That's {Wed, Sat, Tue, Fri, Mon, Thu} — exactly 6 different weekdays, and the only weekday missing is Sunday. Perfect match with the puzzle's "no Sunday" rule, so Wednesday is correct.

💡Key takeaway

This AMC 8 problem only needs Grade 4 division-with-remainders and pattern-counting that you already know!

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