Anybody know how many wins it takes to level a class to 100?
If I counted correct 5050 if you get only 1 xp point/game because that’s the amount of xp needed
But since in some games you get 2 points it may vary.
Also there are the events (class trails) where you can earn class xp and you can buy some goodies in the chop amongst others some classxp
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imho (and i currently have all classes at level 100) the fastest way to get xp is easy casual refresh teams above 6k and use a build like hero fire ruby staff/sunbird/leprechaun/firebomb ((crypt banner))… matches against opponents under 6k team score will have stats compareable to normal explore but wins give you 2xp instead of 1 like explore.
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Maths plagiarized from a google search.
Poking @Saltypatra because maths scare her. @Cyrup likes a bit of math methinks.
This question relates back to a famous mathematician, Gauss. In elementary school in the late 1700’s, Gauss was asked to find the sum of the numbers from 1 to 100. The question was assigned as “busy work” by the teacher, but Gauss found the answer rather quickly by discovering a pattern. His observation was as follows:
1 + 2 + 3 + 4 + … + 98 + 99 + 100
Gauss noticed that if he was to split the numbers into two groups (1 to 50 and 51 to 100), he could add them together vertically to get a sum of 101.
1 + 2 + 3 + 4 + 5 + … + 48 + 49 + 50
100 + 99 + 98 + 97 + 96 + … + 53 + 52 + 51
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
.
.
.
48 + 53 = 101
49 + 52 = 101
50 + 51 = 101
Gauss realized then that his final total would be 50(101) = 5050.
The sequence of numbers (1, 2, 3, … , 100) is arithmetic and when we are looking for the sum of a sequence, we call it a series. Thanks to Gauss, there is a special formula we can use to find the sum of a series:
S is the sum of the series and n is the number of terms in the series, in this case, 100.
Hope this helps!
There are other ways to solve this problem. You can, for example, memorize the formula
This is an arithmetic series, for which the formula is:
S = n[2a+(n-1)d]/2
where a is the first term, d is the difference between terms, and n is the number of terms.
For the sum of the first 100 whole numbers:
a = 1, d = 1, and n = 100
Therefore, sub into the formula:
S = 100[2(1)+(100-1)(1)]/2 = 100[101]/2 = 5050
You can also use special properties of the particular sequence you have.
An advantage of using Gauss’ technique is that you don’t have to memorize a formula, but what do you do if there are an odd number of terms to add so you can’t split them into two groups, for example “what is the sum of the first 21 whole numbers?” Again we write the sequence “forwards and backwards” but using the entire sequence.
1 + 2 + 3 + … + 19 + 20 + 21
21 + 20 + 19 + … + 3 + 2 + 1
Now if you add vertically you get
22 + 22 + 22 + … + 22 + 22 + 22 = 21(22) = 462
But this is twice the sum of the first 21 whole numbers so
1 + 2 + 3 + … + 19 + 20 + 21 = 462/2 = 231
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replies with ‘hilarious’ Sheldon Cooper meme
…but couldn’t be bothered to find one since the update is here 
tldr;
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101
…
50+51 = 101
[/quote]
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