2010 AMC 10A Problem 24

The exponent of $5$ in $90!$ is $\lfloor 90/5 \rfloor + \lfloor 90/25 \rfloor = 18 + 3 = 21$, and the exponent of $2$ is $45 + 22 + 11 + 5 + 2 + 1 = 86$. So $90!$ ends in $21$ zeros and $n \equiv \dfrac{90!}{10^{21}} \pmod{100}$. This quotient still contains $2^{65}$, so $n \equiv 0 \pmod 4$; every choice is a multiple of $4$, so the work is mod $25$.

A useful fact. For any $k$, $(5k+1)(5k+2)(5k+3)(5k+4) \equiv 1 \cdot 2 \cdot 3 \cdot 4 = 24 \equiv -1 \pmod{25}$ (the cross terms are multiples of $5k \cdot 50$, hence of $25$).

Remove the 5s. Write $90! = 5^{18} \cdot 18! \cdot P$, where $P$ is the product of the $72$ non-multiples of $5$ up to $90$. Those form $18$ blocks of four, so $P \equiv (-1)^{18} = 1$. Similarly $18! = 5^3 \cdot 3! \cdot Q$ with $Q = (1\cdot2\cdot3\cdot4)(6\cdots9)(11\cdots14)(16\cdot17\cdot18) \equiv (-1)^3 \cdot 16 \cdot 17 \cdot 18$. Since $16 \cdot 17 = 272 \equiv -3$ and $-3 \cdot 18 = -54 \equiv -4$, we get $Q \equiv 4$. Hence
$$\frac{90!}{5^{21}} = 6 \cdot Q \cdot P \equiv 24 \equiv -1 \pmod{25}.$$

Remove the 2s. $2^{10} = 1024 \equiv -1$, so $2^{21} = (2^{10})^2 \cdot 2 \equiv 2 \pmod{25}$. Then $n = \dfrac{90!/5^{21}}{2^{21}}$ satisfies $2n \equiv -1 \equiv 24$, so $n \equiv 12 \pmod{25}$.

Among the choices, $32 \equiv 7$, $48 \equiv 23$, $52 \equiv 2$, $68 \equiv 18 \pmod{25}$; only $12$ works.

The answer is $\boxed{\textbf{(A)}\ 12}$.

"Last two nonzero digits" means the quotient by $10^{\text{(number of trailing zeros)}}$ taken mod $100$, and $100 = 4 \cdot 25$ splits the job. The factors of $2$ are plentiful, so mod $4$ is trivial; mod $25$ the factorial is tamed by stripping $5$s in layers ($90! \to 18! \to 3!$) and by the block identity that turns runs of four consecutive units into $-1$. Dividing by $2^{21}$ at the end is a modular inverse, not a real division.

• Reducing 90! mod 100 directly (it is 0) or forgetting that removing 21 factors of 5 also requires removing exactly 21 factors of 2.
• Applying the block identity to $18!$ without separately handling its own multiples of $5$ ($5, 10, 15$), which contribute $5^3 \cdot 3!$.