AMC 10 · 2004 · #6
Grade 8 number-theoryWhich of the following numbers is a perfect square?
Pick an answer.
AMC 10 2004 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 candidate numbers are each a product of two factorials drawn from $98!,\,99!,\,100!,\,101!$. Decide which one of the five is a perfect square.
Givens: The five choices are (A) $98!\cdot99!$, (B) $98!\cdot100!$, (C) $99!\cdot100!$, (D) $99!\cdot101!$, (E) $100!\cdot101!$.; $n!$ means the product of all whole numbers from $1$ up to $n$.; A perfect square is a whole number equal to some whole number times itself, for example $100=10^2$.
Unknowns: Which single choice is a perfect square.
Understand
Restated: Five candidate numbers are each a product of two factorials drawn from $98!,\,99!,\,100!,\,101!$. Decide which one of the five is a perfect square.
Givens: The five choices are (A) $98!\cdot99!$, (B) $98!\cdot100!$, (C) $99!\cdot100!$, (D) $99!\cdot101!$, (E) $100!\cdot101!$.; $n!$ means the product of all whole numbers from $1$ up to $n$.; A perfect square is a whole number equal to some whole number times itself, for example $100=10^2$.
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The numbers here are too big to ever write down, the signature trigger for Tool #9 (Solve an Easier Related Problem): trade the giant factorials for a small, manageable structure and answer the question there. Tool #4 (Introduce a Variable) supplies that structure — call the smaller index $n$ and use $(n+1)!=(n+1)\cdot n!$ to fold each product into a perfect square times one tiny leftover. Tool #3 (Eliminate Possibilities) then sweeps the five choices, keeping the one whose leftover is a perfect square and discarding the rest.
Execute — Answer: C
6.EE.A.3 Step 1 Shrink the giants to a leftover
- Never compute $98!$ or $101!$ — they are hopeless by hand.
- Instead use the one fact that links neighboring factorials: the bigger one is just the smaller one with extra factors tacked on, $(n+1)!=(n+1)\cdot n!$.
- So a product of two factorials whose indices are $n$ and $n+1$ collapses to $n!\cdot(n+1)!=n!\cdot(n+1)\cdot n!=(n!)^2\cdot(n+1)$.
- The monstrous part $(n!)^2$ is packaged as a perfect square, and all that is left to judge is the small number $n+1$.
💡 Two neighboring factorials always pair off into a perfect square times one small extra factor.
4.OA.B.4 Step 2 A square times what is still a square
- A whole number is a perfect square exactly when every prime in its factorization appears an even number of times.
- The block $(n!)^2$ already contributes an even count of every prime, so multiplying by a leftover $k$ never disturbs those even counts — it only adds $k$'s own primes.
- Therefore $(n!)^2\cdot k$ is a perfect square if and only if the leftover $k$ is itself a perfect square.
- The whole problem reduces to checking one small number per choice.
💡 A perfect square factor keeps every prime's count even, so squareness is decided entirely by the leftover.
8.EE.A.2 Step 3 Test the neighboring pairs
- Choices (A), (C), (E) pair consecutive indices, so each is $(n!)^2\cdot(n+1)$ and the leftover is simply $n+1$.
- For (A) $98!\cdot99!$ the leftover is $99$; for (C) $99!\cdot100!$ it is $100$; for (E) $100!\cdot101!$ it is $101$.
- Among these only $100=10^2$ is a perfect square — $99$ sits between $9^2=81$ and $10^2=100$, and $101$ is a prime just past $100$.
- So (C) survives and (A), (E) are eliminated.
💡 For back-to-back factorials the leftover is just the larger index, so ask only whether that index is a perfect square.
8.EE.A.2 Step 4 Clear the gapped pairs and conclude
- Choices (B) and (D) skip an index, so their leftover is a product of two numbers.
- For (B) $98!\cdot100!=(98!)^2\cdot(99\cdot100)=(98!)^2\cdot9900$, and $9900=2^2\cdot3^2\cdot5^2\cdot11=30^2\cdot11$ carries the lone prime $11$, so it is not a square.
- For (D) $99!\cdot101!=(99!)^2\cdot(100\cdot101)=(99!)^2\cdot10100$, and $10100=10^2\cdot101$ carries the lone prime $101$, so it is not a square either.
- Every choice but one fails the leftover test, so the perfect square is (C).
💡 A leftover with any prime appearing an odd number of times can never be a perfect square.
6.EE.A.3 Never compute $98!$ or $101!$ — they are hopeless by hand. Instead use the one f 4.OA.B.4 A whole number is a perfect square exactly when every prime in its factorization 8.EE.A.2 Choices (A), (C), (E) pair consecutive indices, so each is $(n!)^2\cdot(n+1)$ an 8.EE.A.2 Choices (B) and (D) skip an index, so their leftover is a product of two numbers Review
Reasonableness: The five leftovers are $99$, $9900$, $100$, $10100$, and $101$, and exactly one of them, $100=10^2$, is a perfect square — matching the promise of a single correct choice. The traps line up with real mistakes: the prime leftovers $99$ (which is $9\cdot11$), $101$, and the lone-prime products $9900=30^2\cdot11$ and $10100=10^2\cdot101$ each hide one prime to an odd power, so none can be a square. Landing on (C), whose leftover is exactly $10^2$, is a strong confirmation.
Alternative: Work purely with prime-exponent parity: a number is a perfect square exactly when every prime in its factorization occurs an even number of times. Since $n!\cdot(n+1)!=(n!)^2\cdot(n+1)$ for the consecutive choices, you need only ask whether the bridging number is a perfect square — $99$ (no), $100$ (yes), $101$ (no) — which pinpoints (C) without even expanding the gapped pairs (B) and (D).
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting $n!\cdot(n+1)!$ as the equivalent form $(n!)^2\cdot(n+1)$ using $(n+1)!=(n+1)\cdot n!$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring each leftover into primes ($99=9\cdot11$, $9900=30^2\cdot11$, $10100=10^2\cdot101$) to judge whether every prime appears an even number of times.)8.EE.A.2Use square root and cube root symbols to represent solutions (Recognizing perfect squares among the leftovers, notably $100=10^2$, and ruling out $99$, $101$, $9900$, $10100$.)
⭐ A product of two factorials is a perfect square exactly when the small leftover that bridges them is itself a perfect square, so pair them up and just check that leftover.
⭐ A product of two factorials is a perfect square exactly when the small leftover that bridges them is itself a perfect square, so pair them up and just check that leftover.
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