CoolStuffYT

Cool, hey!

No

They are real adults who are SMART and SUPPORTIVE

So yeah...

g(x) = 4f(x)

Using the function f(x), can you find g(x)? Multiply everything by 4.

You are very welcome!

:P

Yes, but the sides must be whole numbers

To make the area as big as possible, we should make each side as close to each other as possible. For example, for perimeter 20, a 5x5 rectangle has more area than a 3x7 because 5 and 5 are close together.

With this, we can easily find that 7+8+7+8, or a 7x8 rectangle, is the most compact rectangle with whole number dimensions.

The area of this is 56 square units.

You are very welcome!

:P

I'll give you a hint for #2: this becomes \(\int_8^9x^2\).

You are very welcome!

:P

I'm part of an AoPS class, and I've seen other people spam AoPS questions.

I've never done this personally, but I've had people literally joking in my class about web2.0calc.

One person would say, "You know there's like an answer key to AoPS questions online?"

Another would say, "Yeah! web2.0calc is epic!"

And they just start laughing.

Divisibility rules of 8 and 9 are easiest, so let's find that first.

Modulo may be a bit hard for now, so I'll show you a more rudimentary way.

The list of numbers that when divided by 9 give a remainder of 7 are 7, 16, 25, 34, 43, 52

Same for 8 that give 4: 4, 12, 20, 28, 36, 44, 52

We see that 52 meets both of these requirements! This will happen every cycle of 72, because the LCM of 8 and 9 is 72.

(in other words, 52, 124, 196, 268, 340, ...)

Now, the three digit numbers are 124, 196, 268, etc.

We still haven't used the divisible by 7 with remainder of 1 rule!

Those numbers are 1, 8, 15, 22, etc.

If we test every single number that we had before (124, 196, 268, 340, 412, 484, 556, 628, 772, 854, 916, 988), we can find that

484 can be divided by 7 to give a remainder of 1. So can 988.

Therefore, there are two numbers - 484 and 988.

There is a faster way to do this - with modulo, but try looking at that for yourself! Search up "modulo" online and start looking at the math sites.

You are very welcome!

:P