AMC 8 · 2014 · #4

Easy mode Grade 4

Problem

A prime number is a whole number bigger than $1$ that can only be divided evenly by $1$ and itself. The primes start $2, 3, 5, 7, 11, 13, \dots$ .

Two prime numbers add up to $85$ .

What do you get when you multiply those two primes together?

$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$

Pick an answer.

(A)

(B)

(C)

115

(D)

133

(E)

166

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Toolkit + CCSS Solution

Understand

Restated: Two prime numbers add to $85$. Find the product of those two primes.

Givens: There are exactly two prime numbers; Their sum is $85$; Answer choices: (A) $85$, (B) $91$, (C) $115$, (D) $133$, (E) $166$

Unknowns: The two primes themselves, and then their product

Understand

Restated: Two prime numbers add to $85$. Find the product of those two primes.

Givens: There are exactly two prime numbers; Their sum is $85$; Answer choices: (A) $85$, (B) $91$, (C) $115$, (D) $133$, (E) $166$

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #2 Make a Systematic List

The sum $85$ is odd, and odd $+$ odd $=$ even while even $+$ even $=$ even. So one of the two primes must be even and the other odd — that is a powerful elimination (Tool #3). The only even prime is $2$, so one prime is forced to be $2$ and the other is $85 - 2 = 83$. Tool #2 (Systematic List) is the backup: we can also list small primes $2, 3, 5, 7, 11, \ldots$, pair each with $85 - p$, and check primality — both routes land on the same pair.

Execute — Answer: E

#3 Eliminate Possibilities 2.OA.C.3 Step 1

Use parity to eliminate.
Odd $+$ odd $=$ even, and even $+$ even $=$ even, but our sum $85$ is odd.
So one prime must be even and the other odd — every other pairing is ruled out.

$$\text{odd} + \text{odd} = \text{even}, \quad \text{even} + \text{even} = \text{even}, \quad \text{odd} + \text{even} = \text{odd} = 85$$

💡 Knowing odd $+$ even $=$ odd is a Grade 2 fact, and we use it to rule out every pair of two odd primes in one shot.

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

Find the only even prime.
By definition, a prime is greater than $1$ with no divisors other than $1$ and itself.
Every even number bigger than $2$ has $2$ as a divisor, so it can't be prime.
That leaves $2$ as the only even prime — so one of our primes must be $2$.

$$\text{even primes} = \{2\}$$

💡 Identifying $2$ as the lone even prime is a Grade 4 prime/composite fact and finishes the elimination.

#3 Eliminate Possibilities 4.OA.B.4 Step 3

Find the other prime by subtracting.
If one prime is $2$, the other is $85 - 2 = 83$.

$$85 - 2 = 83$$

💡 A single subtraction pins down the second number; the only remaining job is to check it really is prime.

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

Verify $83$ is prime by trial division.
Since $\sqrt{83} < 10$, we only need to check primes $2, 3, 5, 7$.
$83$ is odd (not $\div 2$), digit sum $8+3=11$ (not $\div 3$), doesn't end in $0$ or $5$ (not $\div 5$), and $83 = 7 \times 11 + 6$ (not $\div 7$).
So $83$ is prime.

$$83 \div 2, 3, 5, 7 \;\Rightarrow\; \text{no clean divide} \;\Rightarrow\; 83 \text{ is prime}$$

💡 Listing the primes up to $\sqrt{83}$ in order is a tiny Tool #2 list — only four numbers to test.

#3 Eliminate Possibilities 3.OA.C.7 Step 5

Multiply the two primes to get the requested product.

$$2 \times 83 = 166 \;\Rightarrow\; \textbf{(E)}$$

💡 $2 \times 83$ is a Grade 3 multiplication; doubling $83$ gives $166$ and matches choice (E).

[1] #3 2.OA.C.3 Use parity to eliminate. Odd $+$ odd $=$ even, and even $+$ even $=$ even, but o

[2] #3 4.OA.B.4 Find the only even prime. By definition, a prime is greater than $1$ with no div

[3] #3 4.OA.B.4 Find the other prime by subtracting. If one prime is $2$, the other is $85 - 2 =

[4] #2 4.OA.B.4 Verify $83$ is prime by trial division. Since $\sqrt{83} < 10$, we only need to

[5] #3 3.OA.C.7 Multiply the two primes to get the requested product.

Review

Reasonableness: Quick sanity check: $2 + 83 = 85$ ✓ (sum condition holds) and $2 \times 83 = 166$ ✓ (matches choice E). The other choices fail the sum test — for example, $85 = 5 \times 17$ has $5 + 17 = 22$, $91 = 7 \times 13$ has $7 + 13 = 20$, $115 = 5 \times 23$ has $5 + 23 = 28$, $133 = 7 \times 19$ has $7 + 19 = 26$. None of those equal $85$, but $166 = 2 \times 83$ does. So (E) is the only consistent answer.

Alternative: Tool #6 (Guess and Check) applied to the answer choices: factor each choice into two primes and see whose factors add to $85$. (A) $85 = 5 \times 17 \to 22$. (B) $91 = 7 \times 13 \to 20$. (C) $115 = 5 \times 23 \to 28$. (D) $133 = 7 \times 19 \to 26$. (E) $166 = 2 \times 83 \to 85$ ✓. Only (E) survives — same answer reached without any parity argument.

CCSS standards used (min grade 4)

2.OA.C.3 Determine whether a group of objects has an odd or even number of members (Using parity (odd $+$ odd $=$ even, odd $+$ even $=$ odd) to argue that one of the two primes summing to the odd number $85$ must be even.)
4.OA.B.4 Find factor pairs and recognize prime/composite numbers within 100 (Recognizing $2$ as the only even prime, ruling out all even numbers $>2$, and verifying $83$ is prime by trial division up to $\sqrt{83}$.)
3.OA.C.7 Fluently multiply within 100 (and apply to larger products) (Computing the final product $2 \times 83 = 166$ by doubling.)

⭐ Whenever two primes add to an odd number, one of them has to be $2$ — that single Grade 4 parity trick cracks the whole problem.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 4)

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