Agnis, being the casanova that he is, bought 1000 roses to give out the the girls in his school on Valentines Day. To the first girl, he gave her 1 rose. To each subsequent girl, as he professes his undying like of them, he gave out (strictly) more roses then he did to the previous.
What is the most number of girls that Agnis could give roses to?
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Lol ! @Satvik Golechha
By the way , I am waiting for your 1000 followers problem. :P
Using the formula for the sum of the Arithmetic Progression to get the maximum value of n:
$S_n= \frac{n}2(2a_1+(n1)d)$
$Where: S_n=1000$ $a_1=1$ $d=1$
We get:
$2000=n(2(1)+n1) \rightarrow n^2+n2000=0$
Using the quadratic formula to solve for the positive value of n:
$n=\frac{1+\sqrt{1(4)(2000)(1)}}{2(1)}$ $n=\frac{1+\sqrt{8001}}{2}$ $n \approx 44.74 \ldots= \boxed{44}$
Since we are talking to a person (n should be an integer), n can't be 44.74, so the maximum number of girls is 44.
Why must this give the maximum? He certainly can't give it to 43.7 girls.
Note that there is an arithmetic error in your solution.
This is a bit similar to Infinite , but easy
A little Python simulation will do just fine.
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FYI In true D&D rules, the max level is 20. There is not level 42 wizard :)
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Incorrect; An experiened kaboobly dooist has the capacity to exceed the standard limit. Downvoted.
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@Calvin Lin – It was a song first. A beautiful, beautiful song.
https://youtube.com/watch?v=ACiA1TX0tvA
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Keep in mind that Agnis has to give out an integral number of roses to an integral number of girls, and he must give strictly more roses at each subsequent girl. To give roses to the most girls, Agnis would have to distribute the roses in fashion of $1, 2, 3 ,4... n$ . So, Agnis is giving out $1+2+3+4+...+n$ roses. So essentially, we're looking for the largest triangular number $T_n$ such that $T_n = \frac{n(n+1)}{2} < 1000$ , where $n$ is our answer, which happens to be $\boxed{44}$ .