$16!$ without calculator

$16!$ is divisible by $125$ since it's divisible by $5\times10\times15$, and by $8$, since it's divisible by $2\times4$.

Therefore, $16!$ must be a multiple of $1000$, and the only acceptable choice is a).


Since $16!$ is divisible by $9$, the sum of its digits must also be divisible by $9$. $$20\ 922\ 789\ 888\ 000 \to 63$$ $$18\ 122\ 471\ 235\ 500 \to 41$$ $$17\ 223\ 258\ 843\ 600 \to 51$$ Only answer a) is divisible by $9$.

We could also have looked for divisibility by $11$ with alternate sum of the digits. $$20\ 922\ 789\ 888\ 000 \to 11$$ $$18\ 122\ 471\ 235\ 500 \to -5$$ $$17\ 223\ 258\ 843\ 600 \to -7$$ Only answer a) is divisible by $11$.


\begin{align}16!&=1\times2\times3\times4\times5\times6\times7\times8\times9\times10\times11\times12\times13\times14\times15\times16 \\&=(2\times5\times10)\times3\times4\times6\times7\times8\times9\times11\times12\times13\times14\times15\times16 \\&=100\times(3\times4\times6\times7\times8\times9\times11\times12\times13\times14\times15\times16)\end{align} So the final answer will have $2$ zeros at the end. If you want the first digit after the zeros, you can multiply the remaining numbers modulo $10$.

\begin{align}&3\times4\times6\times7\times8\times9\times11\times12\times13\times14\times15\times16 \pmod {10} \\&= 3\times4\times6\times7\times8\times9\times1\times2\times3\times4\times5\times6 \pmod {10} \\&= 2\times6\times7\times8\times9\times2\times3\times4\times5\times6 \pmod {10} \\&= 2\times7\times8\times9\times2\times3\times4\times5\times6 \pmod {10} \\&= 4\times8\times9\times2\times3\times4\times5\times6 \pmod {10} \\&= 2\times9\times2\times3\times4\times5\times6 \pmod {10} \\&= 8\times2\times3\times4\times5\times6 \pmod {10} \\&= 6 \times3\times4\times5\times6 \pmod {10} \\&= 8\times4\times5\times6 \pmod {10} \\&= 2\times5\times6 \pmod {10} \\&= 60 \pmod {10} \\&= 0 \pmod {10} \end{align}

So the last 3 digits of $16!$ are $000$ and the answer must be a).

Tags:

Discrete Mathematics

Factorial

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