AMC 8 · 2014 · #23

Grade 6 number-theorylogic

Archetype Small-Domain Constrained Search

prime-numberslogical-deductionbound-inequality-then-enumerate bound-inequality-then-enumeratesystematic-enumerationcasework ↑ Prerequisites: prime-numberslogical-deduction

📏 Medium solution 💡 4 insights

Problem

Three members of the Euclid Middle School girls' softball team had the following conversation.

Ashley: I just realized that our uniform numbers are all $2$ -digit primes.

Bethany : And the sum of your two uniform numbers is the date of my birthday earlier this month.

Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.

Ashley: And the sum of your two uniform numbers is today's date.

What number does Caitlin wear?

$\textbf{(A) }11\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad \textbf{(E) }23$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Ashley, Bethany, and Caitlin each wear a $2$-digit prime number on their softball uniforms. The sum of any two of their numbers is a calendar date in the current month: Ashley + Caitlin = Bethany's birthday (earlier this month), Ashley + Bethany = Caitlin's birthday (later this month), and Bethany + Caitlin = today's date (between the two birthdays). Find Caitlin's uniform number.

Givens: All three uniform numbers are $2$-digit primes (so each is at least $11$); Ashley + Caitlin $=$ Bethany's birthday (earlier this month); Bethany + Caitlin $=$ today's date (this month); Ashley + Bethany $=$ Caitlin's birthday (later this month); Answer choices: (A) $11$, (B) $13$, (C) $17$, (D) $19$, (E) $23$

Unknowns: Caitlin's uniform number

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #2 Make a Systematic List, #7 Identify Subproblems

The candidate set is tiny: $2$-digit primes whose pairwise sums fit on a calendar (at most $31$). Tool #2 (Systematic List) lets us write that short list explicitly, and Tool #3 (Eliminate) knocks out every prime $\ge 19$ because pairing it with any other $2$-digit prime gives a sum past $31$. Tool #7 (Identify Subproblems) splits the question into two cleaner pieces: (a) which three primes are used, and (b) which girl wears which number. Part (b) is decided by chaining the date inequalities, never algebra.

Execute — Answer: A

#2 Make a Systematic List 4.OA.B.4 Step 1

List the $2$-digit primes in order: $11, 13, 17, 19, 23, 29, 31, \ldots$ We only need the smallest few because any pairwise sum must be $\le 31$ (the largest possible date).

$2$-digit primes: $11, 13, 17, 19, 23, 29, 31, \ldots$

💡 Checking which numbers from $10$ to $31$ are prime is exactly the Grade 4 "prime or composite" skill.

#3 Eliminate Possibilities 4.OA.A.3 Step 2

Eliminate any prime $\ge 19$.
If one uniform number is $19$, the smallest possible partner is $11$, giving $19 + 11 = 30$ — still OK — but a third prime has to pair with $19$ too, and even the next smallest partner $13$ gives $19 + 13 = 32 > 31$.
So $19, 23, 29, 31, \ldots$ are all ruled out.
Only $11, 13, 17$ survive, and we need exactly three distinct primes, so the set must be $\{11, 13, 17\}$.

$13 + 19 = 32 > 31$ $\Rightarrow$ $19$ (and larger) cannot appear $\Rightarrow$ set $= \{11, 13, 17\}$

💡 Knocking out each answer choice by checking the sum against $31$ is the Grade 4 multi-step arithmetic check.

#7 Identify Subproblems 4.OA.A.3 Step 3

Check that the set $\{11, 13, 17\}$ really works: $11 + 13 = 24$, $11 + 17 = 28$, $13 + 17 = 30$.
All three sums are valid dates ($\le 31$) and all three are different, so we can match them to the three distinct dates in the story.

Pairwise sums: $11+13 = 24,\; 11+17 = 28,\; 13+17 = 30$

💡 Splitting "find the primes" from "find who wears which" is the Tool #7 subproblems move; the arithmetic check belongs to the first subproblem.

#7 Identify Subproblems 6.EE.B.8 Step 4

Now decide which girl wears which number.
Let $A$, $B$, $C$ stand for Ashley's, Bethany's, Caitlin's numbers.
The story orders the dates as Bethany's birthday $<$ today $<$ Caitlin's birthday, i.e.
$A + C < B + C < A + B$.
Subtract $C$ from the first inequality: $A < B$.
Subtract $B$ from the second: $C < A$.
Combined: $C < A < B$.

$$A + C < B + C < A + B \;\Longrightarrow\; C < A < B$$

💡 Reading the date order as a chain of inequalities and subtracting the common term is Grade 6 inequality reasoning.

#3 Eliminate Possibilities 6.EE.B.8 Step 5

Assign the primes by the order $C < A < B$: Caitlin gets the smallest, Ashley the middle, Bethany the largest.
So $C = 11$, $A = 13$, $B = 17$.
Caitlin's uniform number is $11$ $\Rightarrow$ choice $\textbf{(A)}$.

$$C = 11,\; A = 13,\; B = 17 \;\Rightarrow\; \text{Caitlin} = 11 \;\Rightarrow\; \textbf{(A)}$$

💡 Matching the sorted primes to the sorted variables eliminates the four other answer choices in one stroke.

[1] #2 4.OA.B.4 List the $2$-digit primes in order: $11, 13, 17, 19, 23, 29, 31, \ldots$ We only

[2] #3 4.OA.A.3 Eliminate any prime $\ge 19$. If one uniform number is $19$, the smallest possib

[3] #7 4.OA.A.3 Check that the set $\{11, 13, 17\}$ really works: $11 + 13 = 24$, $11 + 17 = 28$

[4] #7 6.EE.B.8 Now decide which girl wears which number. Let $A$, $B$, $C$ stand for Ashley's,

[5] #3 6.EE.B.8 Assign the primes by the order $C < A < B$: Caitlin gets the smallest, Ashley th

Review

Reasonableness: Plug back in: Bethany's birthday $= A + C = 13 + 11 = 24$, today $= B + C = 17 + 11 = 28$, Caitlin's birthday $= A + B = 13 + 17 = 30$. The dates $24 < 28 < 30$ all fall in the same month, and the story's order (Bethany earlier, today now, Caitlin later) is satisfied. Every sum is $\le 31$, so every date is real. The answer $11$ is the smallest answer choice, which matches the deduction that Caitlin wears the smallest of the three primes.

Alternative: Tool #6 (Guess and Check) on the answer choices: only the three smallest $2$-digit primes can possibly co-exist (Step 2 argument), so among the choices (A) $11$, (B) $13$, (C) $17$, (D) $19$, (E) $23$, the three primes must come from $\{11, 13, 17\}$, and the chain $C < A < B$ forces Caitlin to be the smallest. The only choice that is the minimum of $\{11, 13, 17\}$ is $11$, so $(A)$. Choices $(D)$ and $(E)$ are eliminated immediately because $19$ and $23$ are not even in the surviving set.

CCSS standards used (min grade 6)

4.OA.B.4 Find factor pairs and determine whether a whole number is prime or composite (Listing the $2$-digit primes ($11, 13, 17, 19, 23, \ldots$) that are the candidate uniform numbers.)
4.OA.A.3 Solve multistep word problems using the four operations (Computing pairwise sums like $11+13=24$, $11+17=28$, $13+17=30$ and comparing each to the calendar cap of $31$ to eliminate impossible prime sets.)
6.EE.B.8 Write and reason about inequalities representing a constraint or condition (Translating "Bethany's birthday $<$ today $<$ Caitlin's birthday" into $A + C < B + C < A + B$ and cancelling common terms to get $C < A < B$.)

⭐ This AMC 8 problem only needs the Grade 6 idea that you can subtract the same number from both sides of an inequality — the prime-number list and date cap do all the rest of the work!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

More like this