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I'm just earning hints as we go. After I asked about what you said, it seems as if you are not supposed to consider the seperate teams as much, but instead the gymnasts by themselves.

The answer is \(\boxed 5\) but I need to type up an explanation now if someone can help. 

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By the way. I tried this answer and it came out as incorrect. 

Okay, so let me explain this problem better so that someone can explain the answer to me.

Alrighty.

So yes, the teams are supposed to be the same size. There are \(x\) number of teams, and \(y\) number of gymnasts on each team.

EVERY gymnast must shake hands with EVERYONE else except for herself.

After this handshaking process finished, a single coach on a certain team came down and only shook the hands of the gymnasts on that certain team with \(y\) gymnasts.

In total, there were 281 handshakes. We are trying to solve for the number of handshakes that the single coach on that certain team made, and since the coach only shook the hands of the gymnasts on her own team, I guess we are solving for \(y\).

So rewriting the problem, we get: Solve for the lowest possible value of \(y\).

@Guest, so what this problem is trying to ask, is that there is one certain team, and the gymnasts from that certain team is doing handshakes with gymnasts from other teams. The "coach" is the coach from that certain team.

"There was a total of 281 handshakes. What is the fewest number of handshakes the coach could have participated in?"

Which coach?

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I honestly do not know. This is all the problem provided me with. So I'm guessing that team size does not necessarily matter, but I'm not sure.

Listfor(n, 1, 165, (99!! % 3^n) =(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 508 3731656658, 508 3731656658, 5083 7316566580, 11946 7693931463, 53124 9958120761, 114892 .......etc.

I see that the first 26 powers of 3 are factors of 99!!. Therefore, the largest integer n = 26, or:

3^26 =2,541,865,828,329 which is a factor of 99!!.

Note: 99!! is the double factorial function, i.e., 99 x 97 x 95, 93 x.............x 5 x 3 x 1.

Are the teams the same size?

But we want the SUM of the last two digits of the SUM of 8^25 + 12^25  NOT the sum of the last (single) digits of each term   .....I realize the answer may be the same, but this method is not really a true method......

\(P[\text{3rd 6 on roll 10}]=P[\text{2 6's in 9 rolls and a 6 on the 10th roll}=\\ \dbinom{9}{2}\left(\dfrac 1 6\right)^2 \left(\dfrac 5 6\right)^7 \cdot \dfrac 1 6 = \dfrac{78125}{1679616}\)

 21 divisors:1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576

so the size of the number n = \(21\)

Hope it helps! 

Sorry....the question asks for    8^(25)   and  12^(25)     NOT ^(5)     ~~

0 is the answer everyone! Thanks all! The way to do it is that we find out that the expression is divisible by 100.

The way I came up with 108 and 216 is as follows:

Take the first combination to make up a 10: 6 + 2 + 2 =10. But the first 6 comes in 3 colors: red, blue, green.

The second 2 also comes in 3 colors: red, blue, green. The same thing goes for the third 2. So, we have a combination of 3 x 3 x 3 = 27 possible ways to make up 6 + 2 + 2 =10. And this should apply to the other 3 remaining combinations of 10. So, 27 x 4 =108 ways of making up a 10.

Exactly the same reasoning applies to making up the 20s, 3 x 3 x 3 x 8 =216 ways.

That is my best guess!.

I got it right the first time!!! :( I got -8 when I went back and check my work... I guess I should never go back and check my work again?

This was what I was thinking for (a):

There are 3!*(3+6+3+3) ways so it's 90.

(3! is for the colors, 3+6+3+3 is the number of ways the sum is 10)

So how am I supposed to do it then?

But 12^5 DOES NOT END IN ZERO =248,832 !!

Also, 8^5 ............................................  =32,768 !!

Hey, SG here is a short way!

We know that the square root of 25 is 5, so every 5th power should end in the last digit.

Thus, 12^5 ends in a 0, and 8^5 also end in a zero. Now, we have two zeroes and doing some quick computation, we have 0+0=0.

thank you

(700 -  7744 /4.5 ) / 32.2 = tan (theta)

theta= - 88.2 degrees

That is the maximum....actually, the ONLY  C  for a circle with area = 400 cm^2

I'm thinking that each 3-number combination for making up a 10 should be:

3^3 x 4 =108 ways.

And each 3-number combination for making up a 20 should be:

3^3 x 8 =216 ways.