## 6. Three-digit number

Determine the largest three-digit number that is 21 times greater than the product of its digits.

## Result:

315
## Solution.

For the three-digit number $\overline{a b c}$, the following must hold: $100 a+10 b+c=21 a b c$.

The product of the digits $a b c$ multiplied by 21 is a three-digit number, so it must be

$$
5 \leq a b c \leq 47
$$

The possible products of $a b c$ of three digits are only:

$$
\{5,6,7,8,9,10,12,14,15,16,18,20,21,24,25,27,28,30,32,35,36,40,42,45\}
$$

because numbers that have a two-digit prime factor can be eliminated.

The corresponding numbers 21abc are

| $105=21 \cdot 5$ | $126=21 \cdot 6$ | $147=21 \cdot 7$ | $168=21 \cdot 8$ |
| :--- | :--- | :--- | :--- |
| $189=21 \cdot 9$ | $210=21 \cdot 10$ | $252=21 \cdot 12$ | $294=21 \cdot 14$ |
| $315=21 \cdot 15$ | $336=21 \cdot 16$ | $378=21 \cdot 18$ | $420=21 \cdot 20$ |
| $441=21 \cdot 21$ | $504=21 \cdot 24$ | $525=21 \cdot 25$ | $567=21 \cdot 27$ |
| $588=21 \cdot 27$ | $630=21 \cdot 30$ | $672=21 \cdot 32$ | $735=21 \cdot 35$ |
| $756=21 \cdot 36$ | $840=21 \cdot 40$ | $882=21 \cdot 42$ | $945=21 \cdot 45$ |

Now we directly check in which cases the product of the digits of the three-digit number on the left side of the equation is equal to the second factor on the right side. Many cases can be eliminated because, for example, no digit can be 0.

The condition is only met in the third case, as $3 \cdot 1 \cdot 5=15$. Therefore, the desired number is 315.