NotThatSmart

"Hey, it's still possible to do it without the function" - famous last words 😂

Thanks, CPhill!

~NTS

Euler's Totient Function! The savior!

Let's look at 1200 first.

First, let me explain the process. 

We must prime factorize every number first. For 1200, we have

\(1200 = 12 * 5^2 * 2^2 = 2^4 * 3 * 5^2\)

Now, to summarize what we do, we essentially

\(\frac{\text{number we start with}}{\text{product of the prime factors}} \cdot \text{product of every prime factor -1}\)

So for 1200, we have \(\phi{1200} = 1200 / ( 2 * 3 * 5) * (2-1)(3-1) (5-1) = 40 (1)(2)(4) = 320\)

Meaning the answer is 320. 

Now, let's do the same for the other two numbers. 

First off, 181. It's a prime number, so the only prime factor is 181. Thus, we have

\(\phi 181 = \frac{181}{181}\cdot (181-1) = 180\). So 180 is the answer. 

Now, let's also do 212. We have that \(212 = 2^2*53\)

Thus, using Euler's Totient Function, we have

\(\phi212 = \frac{212}{2*53} \cdot (2-1)(53-1) = 2*52 = 104\) so 104 is the answer. 

Thus, the 3 answers are

\(\phi1200 = 320\\ \phi 181 = 180\\ \phi 212 = 104\)

Thanks! :)

Let's just count. 

We have

\((1,1) \\\to (13, 19)\\(\text{2 steps in between})\\\to (1, 19)\\\text{(3 steps in between)}\\\to(1, 3)\\\to(15,23)\\ \text{(2 steps in between)}\\\to(3, 23)\\\text{(4 steps in between)}\\\to(3, 3)\\\to (17,25)\\\text{(3 steps in between)}\\\to (1, 25)\\\text{(5 steps in between)}\\\to(1, 1)\)

Adding all of these up, we find that there are 28 points. 

Thanks! :)

The answer is \(244224 \pmod {64} \equiv 0\)

mmm...good question. I can't answer this question completely yet, but can you tell me what types of robots. 

One robot has the board Cytron Maker Rasberry Pi, which is for younger people

Here's a link on A* path finding

https://en.wikipedia.org/wiki/A*_search_algorithm

Here' what ai said

Good luck reading it...sorry for the small font. 

Now, I'll elaborate later, but I'm m,ore of a math guy....

Thanks! :)

Alright. I'm not the best with these problems, but let me give it a shot, 

First, let's use the Law of Cosines. According to the law, we have

\(BC^2 = AB^2 + AC^2 - 2(AB* AC) cos ( BAC)\)

Now, we know a lot of the information given already, so we want to find cos BAC. Thus, we have

\(8^2=3^2+7^2-2(3*7) \cdot cos(BAC)\)

Now, isolating BAC, we have that

\(64=16 \cdot cos(BAC)\\ cos(BAC)=4\)

Now, why not apply Law of Cosines again? Let's do it! :)

This time, let's use the equation

\(DE^2 = AE^2 + AD^2 - 2(AE * AD) cos (BAC) \)

We figured out and know every term in the right side of the equation, so plugging in all the information, we have

\(DE^2 = 9^2+21^2-2(9*21)*4\)

mmm...let's check something out.

This simplifies to \(DE^2=-990 \), which defintely IS NOT possible. 

Now, disclaimer though....it's kinda late at night, and my brain is fried. my solution is probably faulty. 

...

Let's set up for a really cool function. 

First, note that \(191=191*1\)

Now, we use Euler's Totient Function. We have

\(\phi{191} = 191/ ( 191) * (191-1) = 190\)

The answer is 190

Thanks! :)

First, let's note that if a number has an odd number of divisors, then it must be a perfect square. 

It also must be a multiple of 8.

So we need a perfect square divisble by 8. 

First, we have

\(4*4=16\)

However, 16 only has 5 factors, with \(1,2,4,8,16\)

Next, we have \(8*8=64\) . However, 64 only has 7 factors, with \(1, 2, 4, 8, 16, 32, 64\)

144 doesn't work, as it has way too much. 

However, we note that \(16*16=256\)

256 has exactly 9 divisors, of \(1, 2, 4, 8, 16, 32, 64, 128,256\)

Thus, the smallest possible number is 256. 

So 256 is our answer. 

Thanks! :)

First, let's note that if a number has an odd number of factors, then it must be a perfect square. 

In the problem, it also must be divisible by 8. 

Thus first few that satsify this number is \(16,64,144\)

However, none of them work, as 16 and 64 have too few factors, and 144 has too many. 

The next number is 256. 

Note that 256 has exactly 9 factors of \(1, 2, 4, 8, 16, 32, 64, 128,256\)

Of these 9 factors, 2 is the only prime divisor!

So out of the 9, only 1 is prime

So 1 is our answer.

Thanks! :)

Let's set up for a really cool function. 

First, note that

\(1200 = 12 * 5^2 * 2^2 = 2^4 * 3 * 5^2\)

Now, we use Euler's Totient Function. We have

\(\phi{1200} = 1200 / ( 2 * 3 * 5) * (2-1)(3-1) (5-1) = 40 (1)(2)(4) = 320\)

The answer is 320

Thanks! :)

*700th answer