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After a gymnastics meet, each gymnast shook hands once with every gymnast on every team (except herself). Afterwards, a coach came down and only shook hands with each gymnast from her own team. There were a total of 281 handshakes. What is the fewest number of handshakes the coach could have participated in?

KnockOut  Oct 21, 2018
 #1
avatar+2729 
+1

Are the teams the same size?

Rom  Oct 21, 2018
 #2
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+8

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.

KnockOut  Oct 21, 2018
 #3
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+1

deleted

Rom  Oct 21, 2018
edited by Rom  Oct 21, 2018
edited by Rom  Oct 21, 2018
 #7
avatar+406 
+8

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

KnockOut  Oct 21, 2018
 #4
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"There was a total of 281 handshakes. What is the fewest number of handshakes the coach could have participated in?"

 

Which coach?

Guest Oct 21, 2018
 #5
avatar+406 
+8

@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.

KnockOut  Oct 21, 2018
 #6
avatar+406 
+8

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 \(y\).

KnockOut  Oct 21, 2018
 #8
avatar+2729 
+1

deleted

Rom  Oct 21, 2018
edited by Rom  Oct 21, 2018
 #9
avatar+406 
+8

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. 

KnockOut  Oct 21, 2018
 #10
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+1

Every gymnast shakes hands with every other gymnast, including the gymnasts from her own team?

 

If that is correct, i think the answer is supposed to be 10 not 5.

 

 

Edit: nvm it's 5

Guest Oct 21, 2018
edited by Guest  Oct 21, 2018
 #11
avatar+1198 
+2

Solution:

(Adapted from dialogues with Lancelot Link)

 

\(\text {Note the “handshake” formula: n(n – 1)/2 = h where (n) is the number of persons and (h) is the number of handshakes. }\\ \text {In this question the number of handshakes is depicted by }\\ \left . y + \dfrac {(xy)(xy-1)}{2} = 281 \right | \text { where (x) = number of teams, (y) = number of gymnasts. }\\ \text {Using (n) as the product of (xy) gives } y + \dfrac {n(n-1)}{2} = 281\\ \text {Find the largest integer for (n), such that }\dfrac {n(n-1)}{2} \lt 281\\ \tiny \text {(If you want to be lazy, look up OEIS A000217 “Triangular” numbers and count them. Or ...)}\\ \text {To “zero” in on the largest integer, choose a few arbitrary values for (n). }\\ n=20; n(n-1)/2 = 190\\ n=25; n(n-1)/2 = 300\\ n=24; n(n-1)/2 = 276 \leftarrow \text { This one. }\\ \;\\ y + 276 = 281\\ y = 281 – 276\\ y = 5\\ \)

 

 

GA

GingerAle  Oct 21, 2018

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