AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 8 Easy

Which of the following numbers is a perfect square?

  • A.

    \frac{23!24!}{3}

  • B.

    \frac{24!25!}{3}

  • C.

    \frac{25!26!}{3}

  • D.

    \frac{26!27!}{3}

  • E.

    \frac{27!28!}{3}

Answer:D

Note that for all positive n, we have \dfrac{n!\left( n+1\right)!}{3}=\dfrac{\left( n!\right)^{2}\cdot\left( n+1\right)}{3}=\left( n!\right)^{2}\cdot\dfrac{n+1}{3}.

We must find a  value of n such that \left( n!\right)^{2}\cdot\dfrac{n+1}{3} is a perfect  square. Since(n!)^2 is a perfect square, we must also have \dfrac{n+1}{3} be a perfect square. In order for \dfrac{n+1}{3} to be a perfect square, n+1 must be twice a perfect square. From the answer choices, n+1=27 works, thus, n=26 and our desired answer is (\rm D)\dfrac{26!27!}{3}.

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