Answer:
Prime factorization: 50 = 2 x 5 x 5
Answer:
The number of possible choices of my team and the opponents team is
Step-by-step explanation:
selecting the first team from n people we have 
possibility and choosing second team from the rest of n-1 people we have 
As { A, B} = {B , A}
Therefore, the total possibility is 
Since our choices are allowed to overlap, the second team is
Possibility of choosing both teams will be
![\frac{n(n-1)}{2} * \frac{n(n-1)}{2} \\\\= [\frac{n(n-1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%20%20%2A%20%20%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%20%20%5C%5C%5C%5C%3D%20%5B%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
We now have the formula
1³ + 2³ + ........... + n³ =![[\frac{n(n+1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
1³ + 2³ + ............ + (n-1)³ = ![[x^{2} \frac{n(n-1)}{2}] ^{2}](https://tex.z-dn.net/?f=%5Bx%5E%7B2%7D%20%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%20%5E%7B2%7D)
=![\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] = [\frac{n(n-1)}{2}]^{3}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dn-1%5C%5CE%5C%5Ci%3D1%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%20%5B%5Cfrac%7Bn%28n-1%29%7D%7B2%7D%5D%5E%7B3%7D)
Step-by-step explanation:
Since it remains only 1 sweet, we can subtract it from the total and get the amount of sweets distributed (=1024).
As all the sweets are distributed equally, we must divide the number of distributed sweets by all its dividers (excluding 1024 and 1, we'll see later why):
1) 512 => 2 partecipants
2) 256 => 4 partecipants
3) 128 => 8 partecipants
4) 64 => 16 partecipants
5) 32 => 32 partecipants
6) 16 => 64 partecipants
7) 8 => 128 partecipants
9) 4 => 256 partecipants
10) 2 => 512 partecipants
The number on the left represents the number of sweets given to the partecipants, and on the right we have the number of the partecipants. Note that all the numbers on the left are dividers of 1024.
Why excluding 1 and 1024? Because the problem tells us that there remains 1 sweet. If there was 1 sweet for every partecipant, the number of partecipants would be 1025, but that's not possible as there remains 1 sweet. If it was 1024, it wouldn't work as well because the sweets are 1025 and if 1 is not distributed it goes again against the problem that says all sweets are equally distributed.
I think the answer is B. Darnell does not know how many rides he has already used. He knows he has 12 bus rides left on the bus pass.
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