Similar Problem

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[Originally posted in the Discussions]
Hi, Prof. Loh,I was having trouble with a similar problem. I wonder if you could help me figure it out.
A rectangular box measures \(a \times b \times ca×b×c,\) where \( a, b,a,b,\) and \(cc\) are integers and \(1 \leq a \leq b \leq c1≤a≤b≤c.\) The volume and the surface area of the box are numerically equal. How many ordered triples \( (a,b,c)(a,b,c) \) are possible? Although this has an explanation, I still don't get it. If you could explain the solution, that would be great!

That's a great question! I'll show you how I learned to do this type of problem with three variables. It's definitely harder than the other questions in this lesson, but it uses some of the main ideas.
Since the surface area and volume are equal, we know that 2(ab+ac+bc) = abc. If we divide each side by abc and 2, we get 1/a + 1/b + 1/c = 1/2, a very simple equation.
This step might be counterintuitive: Aren't fractions bad in a number theory problem? They are! In this case though, I want to divide by that abc to make things much simpler and cleaner (in fact, what's important is that each of a, b, and c now appear only once!). This allows us to do an important trick called bounding.
The idea here is that there are many possibilities for each of a, b, and c. So, I'd like to reduce those possibilities to only a small number, so this becomes manageable. To do this, we want to somehow get some inequality like a < 7. Hm, where would we get that?
Actually, we're already given an inequality. It's:
\( 1 \leq a \leq b \leq c\)Right now, we're looking at the equation 1/a + 1/b + 1/c = 1/2. So if \( 1 ≤ a ≤ b ≤ c\), how do you think 1/a, 1/b, and 1/c are related to each other? You can probably figure out that it's basically the reverse inequality:
\( 1 \geq \frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}\)
Now here is the key step: I know that 1/a+1/b+1/c and 1/2 are equal.
\( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}\)
Now how do 1/a+1/a+1/a and 1/2 compare?
\( \frac{1}{a} + \frac{1}{a} + \frac{1}{a} \bigcirc \frac{1}{2} \)
Well, they could be equal, but probably not all the time. If a, b, and c are different, then basically what I did to the fractions was change the 1/b to 1/a, and change the 1/c to 1/a. From the previous inequality, I know that if I go from 1/b to 1/a, I should get bigger (or maybe stay the same). And if I go from 1/c to 1/a, I should also get bigger. That means that overall, if I go from 1/a+1/b+1/c to 1/a+1/a+1/a, I should get bigger! (Or maybe stay the same, if they are all equal). So that means:
\( \frac{1}{a} + \frac{1}{a} + \frac{1}{a} \geq \frac{1}{2} \)
To reiterate, it's greater than or equal to 1/2, because the changes I made to the left side must increase the value of the left side (or stay the same). Another way to see this is to write down 1/a + 1/a + 1/a ≥ 1/a + 1/b + 1/c, but I know that 1/a + 1/b + 1/c = 1/2.
Alright, now what? Well, if I simplify that left side of the inequality, I get 3/a ≥ 1/2. But that means that 6 ≥ a. Hm, it seems like there are only now 6 possibilities for a! Now this problem seems doable, just try every case, right?
Actually, it's easier than it looks! Right now, we think the possible values for a are 1, 2, 3, 4, 5, 6. Let's look back at the equation 1/a + 1/b + 1/c = 1/2. Think for a moment: Are there any values for a that just won't make sense?
That's right, a = 1 and a = 2 would be pretty absurd! They make the left side way too big. That means we only need to look at a = 3, 4, 5, 6. What does the equation turn into for each of these values? Well, let's look back at the original equation, 2ab + 2bc + 2ca = abc, and try out, maybe, a = 3. Then 6b + 2bc + 6c = 3bc, and so 6b + 6c = bc. How would we solve this?
Hey, this sounds like something that we know to solve, from the video, right? Wasn't there some kind of factoring trick? Yes! Let's move everything to one side to get bc  6b  6c = 0. Now let's add 36 to both sides to get bc  6b  6c + 36 = 36. But why would we ever do that? So the left side factors! Now it's just (b6)(c6) = 36, and we can look at the factors of 36. This might seem like a lot of work because 36 has a lot of factors. Fortunately, it's not too bad, partly because we know that a, b, and c have to be positive integers. In fact, we can say that the only possible values for b6 and c6 are:
(b6, c6) = (1, 36) or (2, 18) or (3, 12) or (4, 9) or (6,6)
Keep in mind that we know that b ≤ c, so we only need to go "halfway" through the different ways to factor 36. Now we just add 6 to all the numbers, and we get five solutions for (a, b, c):
(a, b, c) = (3, 7, 42) or (3, 8, 24) or (3, 9, 18) or (3, 10, 15) or (3, 12, 12)
This case was quick! I won't do all the rest of the three cases, but I actually want to look at the case a = 5, because it is a little trickier.
If a = 5, then 10b + 2bc + 10c = 5bc, so 3bc  10b  10c = 0. Huh, how do we use that factoring trick here? Last time it was so easy, because there wasn't a number in front of the bc. Now what?
Don't fret, let's just get rid of that number! Divide by 3:
\( bc  \frac{10}{3}b  \frac{10}{3}c = 0 \)
"Wait, aren't fractions bad in a number theory problem?" Yeah, you're completely right! But right now we're just trying to do some algebra tricks to factor the expression. We'll fix the fractions later.
Anyways, we can use that factoring trick now! I want my left side to factor to (b  10/3)(c  10/3), but it's not quite there yet. What's the missing term? Yup, it's 100/9:
\( bc  \frac{10}{3}b  \frac{10}{3}c + \frac{100}{9} = \frac{100}{9} \)
\( (b  \frac{10}{3})(c  \frac{10}{3}) = \frac{100}{9} \)
This looks great! Well, except for those fractions. Man, how should we get rid of those fractions? Well, why not just multiply by 9?
\( 9(b  \frac{10}{3})(c  \frac{10}{3}) = 100 \)
\( 3(b  \frac{10}{3})\cdot3(c  \frac{10}{3}) = 100 \)
\( (3b  10)(3c  10) = 100 \)
As you can see, we can use that 9 and split it up into two 3's, and each of those 3's can get rid of one of those 10/3 fractions. Now it looks great!
Hm, what are the possible values of b and c in this case? For example, we can't just do (3b  10) = 10 and (3c  10) = 10, because then b and c won't be integers. I'll leave this for you to figure out!
I'm also going to let you try the cases a = 4 and a = 6 on your own. It's good practice for using the cool factoring trick in this lesson. And, when counting up all the solutions, make sure that 1 ≤ a ≤ b ≤ c is true, otherwise you're going to get some repeat solutions. You should get an answer of 10.
I hope this explanation made sense. If you don't understand some part of it, or you have any other questions, feel free to ask. Happy learning!
The Daily Challenge Team

Thank you so much, Support team. I asked this question. the solution is so detailed.