CalculatorUser

I am that guest just answered above, not the one who posted the question.

I would like to say this website is literally broken. When I posted as a guest, it said "internal server error"

There are tremondous amounts of other glitches, too.
 

Hint: Less eighteen means minus 18

Only use the hint below if you ARE SUPER STUCK and you can't solve it at all.

Here is a hint for 1.

If the n is divisible by 12, that means it is also divisible by 3 and 4.

That also means it is divisible by 3 and 2

What number between 1 and 11 is divisible by 3 and 2?

YEAH! What happened to him? His profile pic changed too!
 

Is he okay? I hope he is!

Rom, if you are reading this, you got a whole forum of people who support you!

I actually used guess and check in this problem XD XD, I only understood the format meaning of modulus I actually barely even know how to apply it.

Yes I hope Rom is okay, its a strange pattern that is going on with him.

Don't care about the white squares, just care about the grey squares.

We can fill the horizontal grey row with 1 2 3 4 5

The remaining four grey squares will contain 4 numbers from the list 1 2 3 4 5

Since is 2 (mod 6)

we just guess and check

1  + 2  + 3 + 4 + 5 = 15

That is the sum of the center horizontal row.

We then can guess and check the sum of the vertical row plus 15 to see if it fits 2 mod ( 6)

Lets say the center square is 1?

Then 29 / 6 remainder is 5

center square is 2?

Then 28 / 6 remainder is 4

Center square is 3?

then 27/6 remainder is 3

Center square is 4? 

then 26/6 remainder is 2

Center squar eis 5?

25/6 remainder is 1

We see by guess and checking that the answer is 4, four should go into the center square

wow. thats just weird.

Ummmmm, I guess I will just simplify???

\(z=\frac{{\sqrt{3}}^2-2{\sqrt{2}}^2}{\sqrt{3}-2\sqrt{2}}\)

Then simplify squared terms

\(z=\frac{3-2*2}{\sqrt{3}-2\sqrt{2}}\)

Simplify further

\(z=\frac{-1}{\sqrt{3}-2\sqrt{2}}\)

Now, lets see if \(\sqrt{3}-2\sqrt{2}\), the denominator, is larger than 1.

round sqrt(3) and sqrt(2) into, 1.7 and 1.4, respectively.

1.7 - 2(1.4) = -1.1

So we have \(z=\frac{-1}{-1.1}\)

\(\frac{1}{1.1}\)

The floor function of that is 0, so that should be the answer???