Which Solution Would You Feature? 11/26/2012

This post originally appeared on the Brilliant blog on 11/26/2012.

Below, we present a problem form the 11/19/2012 Algebra and Number Theory set, along with 3 student submitted solutions (none of them have been edited). In the comments section, please vote for the correct solution that you think should be featured, and state your reason for choosing the particular solution. For example "Solution C. Because it made sense."

These solutions are presented as is, and have not been edited.

Note: I will not reveal who wrote which solution in this blog post. The chosen featured solution will (as always) be credited to the originator.

Problem: What is the smallest 3-digit positive number whose product of the individual digits is still a 3-digit number?

Solution A

$9$ is the largest digit so mostly unit place $100/9=11.111$ so product of smallest digit with any digit>11.111 1 multiplied will not give so $2$ is one digit fixed to hundreds place as we want small number so 2 multiplied by smallest number which gives greater than $11.111$ is $6$ so it is fixed as tenth place and then if unit digit is $8$ it gives product of $96$ but if $9$ gives $108>100$ so the number is $269$!!!

Solution B

Let's call our number $\overline{abc}$. We want $abc \geq 100$. Remember $0 \leq a,b,c \leq 9$, this is important for the inequalities. Start by finding the lowest possible value of a. We know $bc \leq 81$. If a=1, $abc \leq 81$, never yielding a 3-digit product. When $a=2, abc \leq 162$. Only the smallest solution is important, so we can ignore values for a other than 2 (obviously, a solution in the 200s is less than a solution in the 300s or higher). From $abc \geq 100$ and a=2, $bc \geq 50$. Now let's look for the lowest possible value for b. If $b \leq 5$, the previous condition for bc cannot be met. Therefore, 6 is the lowest possible value for b. Again, we can ignore higher values of this digit. We now find that c=9 to meet the initial conditions. $\overline{abc}=269$. Sure enough, the product $2*6*9=108$ is a 3-digit number.

Solution C

You go through all the one hundreds, but realize that if the first number is 1, and the other two are 9, it is still only 81. Therefore, it can't start with a 1. So you go through the two hundreds. If the first number is 2, the second number is 2, then the third number needs to be 25, which is not a single digit. So, you check the second number with 3, which means the last number has to be about 17, which is still too high. Then you try the second number with 4, and the third number needs to be 12.5, still too high. The fourth number being 5, you would need a 10 as the last number which is still not a single digit. Finally, if you try a 2 and a 6, you get 8.33333, and if you choose the next-highest integer, it will be 9 (a single-digit). 12x9=144 and is the lowest possibility of a 3-digit number.

Remarks by Calvin

Most of the comments feel that Solution B, presented by Shefali N.

Solution C avoids using algebra, and does a step by step approach of how he would solve the problem. I agree that this is likely the approach taken by most students. However, it doesn't present a clear way of generalizing to a larger problem. For example, if we want to determine the 10-digit version of this problem, it appears that we will need a lot of case checking. Having successfully solve a problem, students should review it to see what the main important steps is. In this case, it is the realization that the steps are $a\times b \times c \geq 100, b\times c \leq 81 \Rightarrow a \geq 2$, and $b\times c \geq 50, c \leq 9 \Rightarrow b\geq 6$. As such, these ideas should be stressed (or made apparent) in the proof. Through constant review, it will be easier to apply similar ideas to different problems in future.

Solution A is actually incorrect. I'm slightly surprised that no one pointed that out. What is wrong with his approach?