Is this a good recreational math problem?

It’s interesting. I’m not very number oriented but I was writing down combinations of numbers and trying to puzzle it out. I’ll try it out on my maths enthusiastic family and see if it is interesting to them too

Oooh, this is fun.

If we’re limited to 2 numbers, you can think of this as maximizing the area of a rectangle given a fixed perimeter. A square is always optimal. I remember noticing when I was a kid that the same number multiplied by itself (e.g. 7*7 = 49) is always 1 higher than if you take the number one higher and one lower and multiply them (e.g. 6*8 = 48). I didn’t know algebra at the time, but nowadays this is easy enough to prove.

But this – being able to use as many numbers as you want, as long as they add up to the same value – is different, and more interesting. I’m going to think about it for a bit.

Wow! I never knew you could use recreational and math in the same sentence.

@rojo

‘The “Mathematical Games” column in Scientific American that began in January 1957 is a legend in publishing, even though it’s been almost 30 years since the last one appeared.’

Since your proposed solution given in the details was not limited to only 2 digits and allowed repeating digits, then my proposed answer is:
2 + 2 + 2 + 2 + 2 + 2 = 12
and 2^6 = 64

(Also, in this particular case:
4 + 4 + 4 = 12
and 4^3 = 64, as well. But this seems to be a special case where 4 is a potential repeating factor.)

If you include a number more than 4 you could split it into x/2 and x/2. These multiplied give x squared over 4 which is x times x/4. This means you shouldn’t include any number more than 4. 3 times 3 is more than 2 times 2 times 2 so you are better using 3’s. 4 can be divided into 2 times 2. Therefore you should include as many 3’s as possible filling up with 2’s and 1’s.

Hmmm… yeah, 3 + 3 + 3 + 3 = 12
and 3^4 = 81

So, new answer.

Also, are we talking about only positive numbers? Because
-1 – 99 + 112 = 12
and factoring in that way can result in a very high product of the addends. That’s the right word, isn’t it? The “additive factors” are called addends, aren’t they?

@flutherother , Nicely done. Your proof, showing that x must be less than 4, is much simpler than the one I thought of. There is one correction. You should never use 1. If you have all 3’s and a 1, you should replace one 3 and the 1 with two 2’s or one 4. 2×2 is larger than 3×1.

@CWOTUS , That is the correct answer.

Ah, you did say “positive integers”, which wasn’t available for review while I was composing my revision. Never mind, then.

@LostInParadise You are right. I should have avoided using 1’s.

“I will outline the proof later”.

That makes three of us.

My proof is a bit more involved than @flutherother ‘s, but I think it is interesting in its own way, but to understand it you need to be familiar with basic calculus.

Start by taking the log of the multiplication log(a x b x c x…) = log(a) + log(b) + ...
We want to maximize the sum of the logs. Imagine that the numbers represent time intervals and the log values represent distance traveled in a time interval. We want to maximize distance traveled. Since the time intervals always add to the same number, the way to maximize distance traveled is to go as fast as possible. The speed in time interval x is log(x)/x. The best choice of x is the one that maximizes log(x)/x.

Using calculus, we can take the derivative of log(x)/x, getting (1-log(x))/x^^2. To get the maximum, set this to 0, giving 1 – log(x) = 0, log(x) = 1. x is the natural number base e, which is about 2.7. The closest integers to e are 2 and 3, so the maximum log(x)/x for an integer must be at 2 or 3.

We know that 3 is the better value from the original problem because the sum the factors of 3^^2 and 2^^3 are both 6, but 3^^2 > 2^^3, If you need to see this formally, we have log(3^^2) > log(2^^3). 2 log(3) > 3 log(2). log(3)/3 > log(2)/2

To answer the question, yes this is a good recreational math puzzle.

Sure go for it.