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Can you guess the missing digits in the following multiplication?

??? x 3? = ????

Digits from 1 to 9 appear exactly once each.

The goal is to solve it with as little calculation as possible.

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186*39=7254

How to get the answer:

Let's label the numbers from left to right as $a_1,...,a_8$. Note that $a_1$ is at most $3$ since the result of the product must be $3$ digits. Since $3$ is already used, $a_1=1$ or $2$. Let's first run with the assumption that $a_1=2$.

Hence,

If $a_1=2$, since $256*34>8700$, $a_5=8$ or $a_5=9$. If $a_5=8$, then $a_3=4$ as $254*36>9000$ (It's obvious that $a_3, a_4\neq5$.) And, we know that $a_6>6$ from the same calculation. Via $3$ multiplications, we can see that $a_2\neq5\to a_2>5$. But, $34*267>9000$.

So,

$a_5=9$.

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  • $\begingroup$ That was quick, curious to see how you proceeded. My solution still requires ~60 multiplications. $\endgroup$
    – Jacques Gaudin
    Jan 6, 2020 at 21:22
  • $\begingroup$ You may want to hide the method with a spoiler tag $\endgroup$
    – Jacques Gaudin
    Jan 6, 2020 at 21:27
  • $\begingroup$ I'm going to finish this later: have a class to attend. It's a bunch of casework as above. $\endgroup$
    – Rushabh Mehta
    Jan 6, 2020 at 21:41

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