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My solution to first problem is

$98$

This is the strategy:

First, everyone sends a mail to $A$ ($49$ mails in total). Then $A$ sends a mail to everyone else ($49$ more mails).
Why is this optimal? To share his information, everyone must send at least an email, though if they want to know other information, they must receive at least a mail as well! So, $49+49=98$ is necessarily the minimum!

My solution to the second problem is

$96$

The strategy is:

Choose $4$ "important" people in the group $(A,B,C,D)$. The other people call one of these "important" guys, doesn't matter who. These are $50-4$ calls. Those $4$ people then share their knowledge between themselves, so that they know everything. This process requires only 4 calls. Then, they call back the "ignorants", which takes again $50-4$ mails.

My solution to first problem is

$98$

This is the strategy:

First, everyone sends a mail to $A$ ($49$ mails in total). Then $A$ sends a mail to everyone else ($49$ more mails).

Why is this optimal?

To share his information, everyone must send at least an email, though if they want to know other information, they must receive at least a mail as well! So, $49+49=98$ is necessarily the minimum!

My solution to the second problem is

$96$

A possible strategy is:

Choose $4$ "important" people in the group $(A,B,C,D)$. The other people call one of these "important" guys, doesn't matter who. These are $50-4$ calls. Those $4$ people then share their knowledge between themselves, so that they know everything. This process requires only 4 calls. Then, they call back the "ignorants", which takes again $50-4$ mails.

Why is it optimal?

It's pretty obvious that $calls(4)=4$.
When you add a member, he must make at least a call (to share his info), then he must receive at least another call (to receive the whole gossip).
Thus, $calls(n+1)=calls(n)+2$.
You can rewrite as $calls(n)=calls(4)+2(n-4)=2n-4$

My solution to first problem is

$98$

This is the strategy:

First, everyone sends a mail to $A$ ($49$ mails in total). Then $A$ sends a mail to everyone else ($49$ more mails).
Why is this optimal? To share his information, everyone must send at least an email, though if they want to know other information, they must receive at least a mail as well! So, $49+49=98$ is necessarily the minimum!

My solution to the second problem is

$96$

The strategy is:

Choose $4$ people in the group $(A,B,C,D)$. The other people send an email to one of these, doesn't matter who. This is $50-4$ mails. Those $4$ people then share their knowledge between themselves, so that they know everything. This process requires only 4 mails. Then, they send back an email to the "ignorants", which takes again $50-4$ mails.

My solution to first problem is

$98$

This is the strategy:

First, everyone sends a mail to $A$ ($49$ mails in total). Then $A$ sends a mail to everyone else ($49$ more mails).
Why is this optimal? To share his information, everyone must send at least an email, though if they want to know other information, they must receive at least a mail as well! So, $49+49=98$ is necessarily the minimum!

My solution to the second problem is

$96$

The strategy is:

Choose $4$ "important" people in the group $(A,B,C,D)$. The other people call one of these "important" guys, doesn't matter who. These are $50-4$ calls. Those $4$ people then share their knowledge between themselves, so that they know everything. This process requires only 4 calls. Then, they call back the "ignorants", which takes again $50-4$ mails.

My solution to first problem is

$98$

This is the strategy:

First, everyone sends a mail to $A$ ($49$ mails in total). Then $A$ sends a mail to everyone else ($49$ more mails).
Why is this optimal? To share his information, everyone must send at least an email, though if they want to know other information, they must receive at least a mail as well! So, $49+49=98$ is necessarily the minimum!

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

My solution to the second problem is

$96$

The strategy is:

Choose $4$ people in the group $(A,B,C,D)$. The other people send an email to one of these, doesn't matter who. This is $50-4$ mails. Those $4$ people then share their knowledge between themselves, so that they know everything. This process requires only 4 mails. Then, they send back an email to the "ignorants", which takes again $50-4$ mails.

My solution to first problem is

$98$

This is the strategy:

First, everyone sends a mail to $A$ ($49$ mails in total). Then $A$ sends a mail to everyone else ($49$ more mails).
Why is this optimal? To share his information, everyone must send at least an email, though if they want to know other information, they must receive at least a mail as well! So, $49+49=98$ is necessarily the minimum!

My solution to the second problem is

$96$

The strategy is:

Choose $4$ people in the group $(A,B,C,D)$. The other people send an email to one of these, doesn't matter who. This is $50-4$ mails. Those $4$ people then share their knowledge between themselves, so that they know everything. This process requires only 4 mails. Then, they send back an email to the "ignorants", which takes again $50-4$ mails.