AMC 10 · 2010 · #10
Grade 4 arithmeticMarvin had a birthday on Tuesday, May 27 in the leap year 2008. In what year will his birthday next fall on a Saturday?
Pick an answer.
AMC 10 2010 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.
Toolkit + CCSS Solution
Understand
Restated: Marvin's birthday, May 27, was a Tuesday in the leap year $2008$. Find the next year in which May 27 lands on a Saturday.
Givens: May 27, $2008$ is a Tuesday; $2008$ is a leap year; Answer choices: (A) $2011$, (B) $2012$, (C) $2013$, (D) $2015$, (E) $2017$
Unknowns: The next year after $2008$ in which May 27 is a Saturday
Understand
Restated: Marvin's birthday, May 27, was a Tuesday in the leap year $2008$. Find the next year in which May 27 lands on a Saturday.
Givens: May 27, $2008$ is a Tuesday; $2008$ is a leap year; Answer choices: (A) $2011$, (B) $2012$, (C) $2013$, (D) $2015$, (E) $2017$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #5 (Look for a Pattern): each year the same date marches forward a fixed number of weekdays, so once you know the yearly step you can just walk the days Tue, Wed, Thu, ... until you hit Saturday. Tool #7 (Subproblems): first figure out the shift for a common year, then the shift for the leap-year case — two small facts that build the whole table. Tool #3 (Eliminate): the target Saturday can be skipped over in a leap-year jump, so checking each year one by one rules out the choices where May 27 is not yet a Saturday.
Execute — Answer: E
4.NBT.B.6 Step 1 Common year shifts weekday by 1
- A common year has $365$ days.
- Divide by $7$: $365 = 52 \times 7 + 1$, so a common year is $52$ whole weeks plus $1$ extra day.
- That means the same date next year lands one weekday later.
💡 Whole weeks bring you back to the same weekday; only the leftover day past those weeks moves the date.
4.OA.B.4 Step 2 Leap year shifts weekday by 2
- A leap year has $366$ days: $366 = 52 \times 7 + 2$, an extra $2$ days.
- This extra day (Feb 29) sits before May 27, so the jump of $+2$ applies when the year you are stepping into is a leap year.
- Leap years are multiples of $4$, so between 2008 and 2017 the leap years are $2012$ and $2016$.
💡 An extra leap day inside the year pushes the date forward one more weekday than usual.
4.OA.C.5 Step 3 Step forward year by year
- Start at May 27, 2008 = Tuesday.
- Add $+1$ each year, but $+2$ when stepping into a leap year ($2012$ and $2016$): 2009 Wed, 2010 Thu, 2011 Fri, 2012 Sun (jumped $+2$, skipping Sat), 2013 Mon, 2014 Tue, 2015 Wed, 2016 Fri (jumped $+2$, skipping Thu), 2017 Sat.
💡 Following one simple step-rule year after year turns the calendar into a countable chain of weekdays.
4.OA.C.5 Step 4 Read off the first Saturday
- Scanning the chain, May 27 is not a Saturday in 2011, 2012, 2013, or 2015.
- The Saturday slot was actually leaped over in the 2012 jump (Fri straight to Sun), so it does not appear until $2017$.
- The next Saturday birthday is in $2017$, choice (E).
💡 Because a leap-year jump can hop over a weekday, the first Saturday can land later than a naive count would guess.
4.NBT.B.6 A common year has $365$ days. Divide by $7$: $365 = 52 \times 7 + 1$, so a commo 4.OA.B.4 A leap year has $366$ days: $366 = 52 \times 7 + 2$, an extra $2$ days. This ext 4.OA.C.5 Start at May 27, 2008 = Tuesday. Add $+1$ each year, but $+2$ when stepping into 4.OA.C.5 Scanning the chain, May 27 is not a Saturday in 2011, 2012, 2013, or 2015. The S Review
Reasonableness: Total days from May 27, 2008 to May 27, 2017 is $9$ years spanning two leap days (2012, 2016): $9 \times 365 + 2 = 3287$ days. Dividing by $7$, $3287 = 469 \times 7 + 4$, a shift of $+4$ from Tuesday, which lands on Saturday — matching the year-by-year table. Every earlier choice fails: 2011 is Friday, 2012 is Sunday, 2013 is Monday, 2015 is Wednesday, so only (E) $2017$ works.
Alternative: Instead of naming weekdays, track the shift as a running remainder mod $7$ with Tuesday $=0$. Add $1$ per year and $2$ on leap-year steps; the totals grow $1,2,3,5,6,0,1,3,4$ for 2009..2017. Saturday is $4$ (four days past Tuesday), first reached at $2017$, confirming (E).
CCSS standards used (min grade 4)
4.NBT.B.6Find whole-number quotients and remainders (Dividing $365$ and $366$ by $7$ to find the $+1$ and $+2$ weekday shifts.)4.OA.B.4Recognize multiples and determine whether a number is a multiple of a given number (Identifying $2012$ and $2016$ as leap years because they are multiples of $4$.)4.OA.C.5Generate a number or shape pattern that follows a given rule (Stepping the weekday forward year by year and reading off when Saturday first appears.)
⭐ A date slides forward one weekday each year, or two after a leap year — so just walk the weekdays year by year until you reach the one you want.
⭐ A date slides forward one weekday each year, or two after a leap year — so just walk the weekdays year by year until you reach the one you want.
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