A logic problem by Paola Ramírez

Generally numbers that are closer together yield a greater product than numbers that are farther apart (if they are around the same value). A 2-digit by 2-digit almost always gives a higher product than a 3-digit by 1-digit. You want the higher digits to be in the tens place, because it will take more value. The 2 bigger digits are $9$ and $8$ . You are left with 2 choices, $91*86$ and $96*81$ . $91$ and $86$ are closer together than $96$ and $81$ , so they will give a greater product. The final answer is $91*86=$ 7826

$***$ If you don't believe that "numbers that are closer together yield a greater product," then check for yourself. $91*86$ is bigger than $96*81$ .

The way I thought of this was just from my lessons in Geometry - the area of a rectangle is maximized when it is a square. So, the way I approached this problem was aiming to make the two 'sides' (the two numbers being multiplied) as tight together as possible. I found this to happen with the values 86 and 91, which multiply out to 7826!

This is simply a matter of trial and error. The equation can only be a one-digit multiplied by a three-digit or a two-digit by a two-digit. Many people would first try to multiply a three-digit by a one-digit but later they would have realised that the result is high, but not high enough to satisfy the question.

Next, we have to try two-digits. If you keep on trying, you will eventually come to

$\quad \quad 91\\ \underline { \times \quad 86 } \\ \quad 7826$

as your final answer.

Check 9 × 861, 91 × 86 and 96 × 81.

9 × 861 = 7749
91 × 86 = 7826
96 × 81 = 7776