AMC 8 · 2016 · #7
Grade 8 number-theoryProblem
Which of the following numbers is not a perfect square?
Pick an answer.
AMC 8 2016 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: Five numbers are given: $1^{2016}$, $2^{2017}$, $3^{2018}$, $4^{2019}$, $5^{2020}$. Exactly one of them is not a perfect square. Which one?
Givens: Choice (A) $= 1^{2016}$; Choice (B) $= 2^{2017}$; Choice (C) $= 3^{2018}$; Choice (D) $= 4^{2019}$; Choice (E) $= 5^{2020}$; A perfect square is an integer of the form $k^2$ for some integer $k$
Unknowns: The single choice that is not a perfect square
Understand
Restated: Five numbers are given: $1^{2016}$, $2^{2017}$, $3^{2018}$, $4^{2019}$, $5^{2020}$. Exactly one of them is not a perfect square. Which one?
Givens: Choice (A) $= 1^{2016}$; Choice (B) $= 2^{2017}$; Choice (C) $= 3^{2018}$; Choice (D) $= 4^{2019}$; Choice (E) $= 5^{2020}$; A perfect square is an integer of the form $k^2$ for some integer $k$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #5 Look for a Pattern, #9 Solve an Easier Related Problem
Five answer choices and a yes/no test on each — Tool #3 (Eliminate) is the natural lead. The test itself comes from Tool #5: by checking small cases like $2^2, 2^3, 2^4, 2^5$, the pattern is clear — a power $a^n$ (with $a$ prime) is a perfect square exactly when $n$ is even, because $a^{2k} = (a^k)^2$. Tool #9 (Easier Problem) is what lets us trust that pattern: we verify it on tiny exponents before applying it to 2016–2020. The one trap is choice (D), where the base $4$ is not prime; we rewrite $4 = 2^2$ first so the exponent rule applies to a prime base.
Execute — Answer: B
6.EE.A.1 Step 1 - Build the test on small cases (Tool #9).
- For base $2$: $2^2 = 4 = 2^2$ (square), $2^3 = 8$ (not a square), $2^4 = 16 = 4^2$ (square), $2^5 = 32$ (not a square).
- So $2^n$ is a perfect square exactly when $n$ is even.
- The same logic works for any prime base $p$: $p^{2k} = (p^k)^2$ is a square, and $p^{\text{odd}}$ is not.
💡 Walking through $n = 2, 3, 4, 5$ on base $2$ makes the even-exponent rule visible without any abstract proof — exactly the small-cases move.
6.EE.A.1 Step 2 - Eliminate (A).
- $1^{2016} = 1 = 1^2$.
- Perfect square.
- Cross off (A).
💡 Every power of $1$ is just $1$, and $1$ is trivially $1^2$.
6.EE.A.1 Step 3 - Test (B).
- Base $2$ is prime; the exponent is $2017$, which is odd.
- By the rule from Step 1, $2^{2017}$ is not a perfect square.
- This is the candidate answer — but we still verify the remaining three to make sure none of them also fails.
💡 Odd exponent on a prime base means you can't split the prime factors into two equal groups.
6.EE.A.1 Step 4 - Eliminate (C).
- Base $3$ is prime; the exponent $2018$ is even, so $3^{2018} = (3^{1009})^2$.
- Perfect square.
- Cross off (C).
💡 Half the exponent gives the integer whose square it is.
8.EE.A.1 Step 5 - Eliminate (D) — the trap choice.
- Base $4$ is not prime, so the exponent $2019$ (odd) does not directly decide the answer.
- Rewrite $4 = 2^2$ and apply the power-of-a-power rule $(a^m)^n = a^{mn}$: $4^{2019} = (2^2)^{2019} = 2^{4038}$.
- Now the prime-base exponent is $4038$, which is even, so $4^{2019} = (2^{2019})^2$.
- Perfect square.
- Cross off (D).
💡 Whenever the base is itself a square, the answer is automatically a square no matter what the outer exponent is.
6.EE.A.1 Step 6 - Eliminate (E).
- Base $5$ is prime; the exponent $2020$ is even, so $5^{2020} = (5^{1010})^2$.
- Perfect square.
- Cross off (E).
- Only (B) survives.
💡 Four of the five are squares; the only one left standing is (B).
6.EE.A.1 Build the test on small cases (Tool #9). For base $2$: $2^2 = 4 = 2^2$ (square), 6.EE.A.1 Eliminate (A). $1^{2016} = 1 = 1^2$. Perfect square. Cross off (A). 6.EE.A.1 Test (B). Base $2$ is prime; the exponent is $2017$, which is odd. By the rule f 6.EE.A.1 Eliminate (C). Base $3$ is prime; the exponent $2018$ is even, so $3^{2018} = (3 8.EE.A.1 Eliminate (D) — the trap choice. Base $4$ is not prime, so the exponent $2019$ ( 6.EE.A.1 Eliminate (E). Base $5$ is prime; the exponent $2020$ is even, so $5^{2020} = (5 Review
Reasonableness: Cross-check by prime factorization. (A) $1$ has exponent $0$ on every prime (even). (C) $3^{2018}$: prime $3$ to even exponent $2018$. (D) $4^{2019} = 2^{4038}$: prime $2$ to even exponent $4038$. (E) $5^{2020}$: prime $5$ to even exponent $2020$. All four pass the "every prime exponent is even" test. Only (B) $2^{2017}$ has an odd exponent on the prime $2$, so it cannot be written as $k^2$. The answer (B) is consistent.
Alternative: Tool #5 (Look for a Pattern) alone is enough if you skip the per-choice elimination: note that the exponents are $2016, 2017, 2018, 2019, 2020$ — exactly the pattern even, odd, even, odd, even. Bases $1, 3, 5$ are odd; bases $2, 4$ are even. The only "prime base $\times$ odd exponent" combination is $2^{2017}$ (since $4 = 2^2$ converts to even exponent). So (B) is the outlier by pattern inspection.
CCSS standards used (min grade 8)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Reading each choice $a^n$ as repeated multiplication and recognizing when it can be regrouped as $(a^k)^2 = a^{2k}$ — the Grade 6 entry point to exponent notation.)8.EE.A.1Know and apply the properties of integer exponents to generate equivalent numerical expressions (Rewriting choice (D) using the power-of-a-power rule: $4^{2019} = (2^2)^{2019} = 2^{2 \cdot 2019} = 2^{4038}$, which converts an odd outer exponent into an even prime exponent.)
⭐ A prime raised to an odd power can never be a perfect square — and that single Grade 8 exponent rule is enough to spot the odd-one-out among (A)–(E).
⭐ A prime raised to an odd power can never be a perfect square — and that single Grade 8 exponent rule is enough to spot the odd-one-out among (A)–(E).
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