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 $6$ 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?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Isabella has $6$ coupons total; She redeems one coupon every $10$ days (so $6$ redemptions span days $0, 10, 20, 30, 40, 50$ from the first); Pete's is closed on Sundays; None of the $6$ circled redemption dates falls on a Sunday; Answer choices: (A) Monday, (B) Tuesday, (C) Wednesday, (D) Thursday, (E) Friday

Unknowns: The day of the week on which Isabella redeems her FIRST coupon

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

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.

Execute — Answer: C

#5 Look for a Pattern 4.NBT.B.6 Step 1

Find how much the weekday shifts every $10$ days.
Because the week itself is $7$ days long, full weeks don't change the weekday — only the LEFTOVER days do.
Divide $10$ by $7$ to find that leftover.

$10 \div 7 = 1 \text{ remainder } 3$, so each $10$-day jump moves the weekday forward by $3$ days.

💡 Grade 4 long division with remainders tells us the weekday slides forward by $3$ each time.

#2 Make a Systematic List 4.OA.C.5 Step 2

Use Tool #2 to list the weekday of every redemption in order, calling the first weekday $D$.
After each jump, add $3$ days; whenever the running total goes past a full week ($7$ days), subtract $7$ to come back into the same week.

1st: $D$ \;\; 2nd: $D+3$ \;\; 3rd: $D+6$ \;\; 4th: $D+9 \to D+2$ \;\; 5th: $D+12 \to D+5$ \;\; 6th: $D+15 \to D+1$

💡 Generating a number pattern by a fixed rule ($+3$, then wrap around at $7$) is the Grade 4 "shape and number pattern" standard.

#5 Look for a Pattern 3.OA.D.9 Step 3

Collect those $6$ offsets in order from smallest to largest.
They are $\{D+0,\; D+1,\; D+2,\; D+3,\; D+5,\; D+6\}$ — six of the seven possible weekday offsets $\{0,1,2,3,4,5,6\}$.
The ONE offset that is missing is $D+4$.

Missing offset $= D + 4$

💡 Spotting that $\{0,1,2,3,5,6\}$ is the full set $\{0,1,2,3,4,5,6\}$ minus $4$ is a Grade 3 arithmetic-pattern observation.

#5 Look for a Pattern 4.OA.C.5 Step 4

Apply the "no Sundays" rule.
The $6$ redemption days hit $6$ of the $7$ weekdays and skip exactly one, and the problem says the skipped weekday must be Sunday.
So $D + 4$ lands on Sunday, which means $D$ itself is $4$ days BEFORE Sunday — count backward from Sunday: Saturday, Friday, Thursday, Wednesday.
So $D = $ Wednesday.

Sunday $-$ $4$ days $=$ Wednesday $\;\Longrightarrow\; D = $ Wednesday

💡 Counting backward $4$ weekdays from Sunday is a Grade 4 pattern-following step — no algebra needed.

#3 Eliminate Possibilities 4.OA.C.5 Step 5

Use Tool #3 to eliminate the other choices.
(A) Monday $\to$ skipped day is Friday, not Sunday.
(B) Tuesday $\to$ skipped day is Saturday.
(D) Thursday $\to$ skipped day is Monday.
(E) Friday $\to$ skipped day is Tuesday.
Only (C) Wednesday makes the skipped day Sunday, matching the puzzle's rule.

$$\textbf{(C) Wednesday}$$

💡 Plugging each multiple-choice candidate back into the rule is the safest AMC "check the answer" habit.

[1] #5 4.NBT.B.6 Find how much the weekday shifts every $10$ days. Because the week itself is $7$

[2] #2 4.OA.C.5 Use Tool #2 to list the weekday of every redemption in order, calling the first

[3] #5 3.OA.D.9 Collect those $6$ offsets in order from smallest to largest. They are ${D+0,\;

[4] #5 4.OA.C.5 Apply the "no Sundays" rule. The $6$ redemption days hit $6$ of the $7$ weekdays

[5] #3 4.OA.C.5 Use Tool #3 to eliminate the other choices. (A) Monday $\to$ skipped day is Frid

Review

Reasonableness: Verify by hand starting from Wednesday: Wed $\to$ Sat $\to$ Tue $\to$ Fri $\to$ Mon $\to$ 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.

Alternative: Tool #6 (Guess and Check) on each choice: for each of the five starting weekdays, list the six $+3$-stepped weekdays and check whether Sunday appears. Only Wednesday produces a Sunday-free schedule, so Wednesday wins. This is even more direct than the pattern argument and is a great backup whenever the "missing-day" trick feels slippery.

CCSS standards used (min grade 4)

3.OA.D.9 Identify arithmetic patterns and explain using properties of operations (Noticing that the six computed weekday offsets $\{0,1,2,3,5,6\}$ are missing exactly one value ($4$) from the full set $\{0,1,2,3,4,5,6\}$.)
4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends (Dividing $10 \div 7$ to get a remainder of $3$, which is the weekday shift per redemption.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Building the list of six redemption weekdays by repeatedly adding $3$ and wrapping past $7$, and counting backward $4$ days from Sunday to find the starting weekday.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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