# Guess the missing digits [duplicate]

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.

186*39=7254

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,

$$a_5=9$$.

Still being edited...

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