AMC 10 · 2014 · #8
Grade 8 number-theoryWhich of the following numbers is a perfect square?
Pick an answer.
AMC 10 2014 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: Each answer choice has the form (two consecutive factorials multiplied together) divided by $2$. Decide which single choice is a perfect square — an integer equal to some whole number times itself.
Givens: Choice (A) is $\dfrac{14!\,15!}{2}$; Choice (B) is $\dfrac{15!\,16!}{2}$; Choice (C) is $\dfrac{16!\,17!}{2}$; Choice (D) is $\dfrac{17!\,18!}{2}$; Choice (E) is $\dfrac{18!\,19!}{2}$
Unknowns: Which one of the five choices is a perfect square
Understand
Restated: Each answer choice has the form (two consecutive factorials multiplied together) divided by $2$. Decide which single choice is a perfect square — an integer equal to some whole number times itself.
Givens: Choice (A) is $\dfrac{14!\,15!}{2}$; Choice (B) is $\dfrac{15!\,16!}{2}$; Choice (C) is $\dfrac{16!\,17!}{2}$; Choice (D) is $\dfrac{17!\,18!}{2}$; Choice (E) is $\dfrac{18!\,19!}{2}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Testing five giant factorial expressions one by one is hopeless — $18!$ alone has fifteen digits. Tool #5 spots that all five choices are the same object $\dfrac{n!\,(n+1)!}{2}$ with $n$ running $14,15,16,17,18$. Tool #4 then names that index $n$ and rewrites the shared form so the perfect-square question collapses to a tiny condition on $n$. Because $(n+1)! = (n+1)\cdot n!$, the product $n!\,(n+1)!$ becomes $(n!)^2(n+1)$ — a perfect square times a leftover factor. The whole thing is a perfect square exactly when that leftover, $\dfrac{n+1}{2}$, is itself a perfect square. Tool #3 finishes it: plug each choice's $n$ into that one small test and eliminate until one survives — no factorial ever has to be computed.
Execute — Answer: D
6.EE.A.2 Step 1 Name the shared form
- All five choices look identical except for the starting number.
- Write the common shape with a variable $n$ for the smaller factorial, so the five choices are just $n = 14, 15, 16, 17, 18$.
💡 One letter stands in for five nearly identical expressions, so you solve them all at once.
6.EE.A.1 Step 2 Pull out a perfect square
- The next factorial is the previous one times one extra factor: $(n+1)! = (n+1)\cdot n!$.
- Substitute that in, so the two factorials become $n!\cdot n!\cdot (n+1) = (n!)^2(n+1)$.
- Dividing by $2$ leaves a perfect square $(n!)^2$ multiplied by $\dfrac{n+1}{2}$.
💡 Two copies of $n!$ multiplied together is $(n!)^2$ — an automatic perfect square you can set aside.
8.EE.A.2 Step 3 Reduce to one small condition
- A perfect square times another number is a perfect square only when that other number is also a perfect square.
- Since $(n!)^2$ is already a perfect square, the whole expression is a perfect square exactly when $\dfrac{n+1}{2}$ is a perfect square.
💡 The square part carries itself; only the leftover factor can spoil or complete the square.
8.EE.A.2 Step 4 Test each choice
- Compute $\dfrac{n+1}{2}$ for each choice's $n$ and check whether it is a perfect square.
- It must first be a whole number, then a square.
💡 Only an even $n+1$ that is twice a perfect square passes — and $18 = 2\cdot 9$ is the one that does.
8.EE.A.2 Step 5 Confirm the winner
- Only $n = 17$ gives a perfect square for $\dfrac{n+1}{2}$, namely $9 = 3^2$.
- That is choice (D).
- Rebuilding it shows the full expression is a clean square.
💡 Multiplying the square $(17!)^2$ by another square $3^2$ keeps it a perfect square, so (D) works.
6.EE.A.2 All five choices look identical except for the starting number. Write the common 6.EE.A.1 The next factorial is the previous one times one extra factor: $(n+1)! = (n+1)\c 8.EE.A.2 A perfect square times another number is a perfect square only when that other n 8.EE.A.2 Compute $\dfrac{n+1}{2}$ for each choice's $n$ and check whether it is a perfect 8.EE.A.2 Only $n = 17$ gives a perfect square for $\dfrac{n+1}{2}$, namely $9 = 3^2$. Tha Review
Reasonableness: The test says a choice works only when $\dfrac{n+1}{2}$ is a perfect square, which forces $n+1$ to be twice a perfect square. Among $15, 16, 17, 18, 19$, the even ones are $16 = 2\cdot 8$ and $18 = 2\cdot 9$; of those only $18$ hands back a perfect square ($9$), so exactly one choice qualifies — as an AMC answer should. Checking (D) directly, $\dfrac{17!\,18!}{2} = (3\cdot 17!)^2$, a genuine square. The near-misses confirm it: (B) gives $8$ (between $2^2$ and $3^2$), and (A), (C), (E) are not even integers after halving. The answer is (D), matching the key.
Alternative: Argue by prime factors. In $\dfrac{n!\,(n+1)!}{2}$ the factor $(n!)^2$ contributes only even exponents, so a number is a perfect square exactly when $\dfrac{n+1}{2}$ contributes even exponents too — i.e. is itself a square. Scanning the choices, $\dfrac{17+1}{2} = 9 = 3^2$ is the only perfect square, giving (D). Same conclusion without expanding any factorial.
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Introducing the variable $n$ to capture all five choices as the single expression $\dfrac{n!\,(n+1)!}{2}$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Rewriting $n!\cdot n!$ as $(n!)^2$ to expose the built-in perfect square inside each choice.)8.EE.A.2Use square root and cube root symbols to represent solutions (Reasoning about perfect squares: reducing to the condition that $\dfrac{n+1}{2}$ be a perfect square and recognizing $9 = 3^2$ for choice (D).)
⭐ A product of consecutive factorials hides a perfect square $(n!)^2$ inside it — so all that matters is whether the leftover piece $\dfrac{n+1}{2}$ is a perfect square too.
⭐ A product of consecutive factorials hides a perfect square $(n!)^2$ inside it — so all that matters is whether the leftover piece $\dfrac{n+1}{2}$ is a perfect square too.
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