Logged in after almost two Years.
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Logged in after almost two Years.
Ohk so I logged in after two years.That doesn't matter. Here is a brainteaser that I came across a while back.Robert is giving his exams which comprises of 4 subjects lets say ABCD. The max he can score in A B and C are 50 each and he can score 100 in D.In how many ways can he score 150 ?
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Re: Logged in after almost two Years.
Hi. 
I don't know. That sounds like math.
I assume the scores are given out in integers ranging from 0 to 50 (or 0 to 100) and no fractions? And not simply pass/fail, either 0 or 50 (or 0 or 100) with nothing in between?

I don't know. That sounds like math.
I assume the scores are given out in integers ranging from 0 to 50 (or 0 to 100) and no fractions? And not simply pass/fail, either 0 or 50 (or 0 or 100) with nothing in between?
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"Imagine an ennobling of what could be" -- the New Age BS Generator site
"You are also taking my words out of context." -- Justin
"Nullius in verba" -- The Royal Society ["take nobody's word for it"]
#ANDAMOVIE
Is Trump in jail yet?
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Re: Logged in after almost two Years.
Yeah only integers are allowed.No fractions please.All the integers including 0 and maximum are allowed.
A [0,50]
B [0,50]
C [0,50]
D [0,100]
A [0,50]
B [0,50]
C [0,50]
D [0,100]
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Re: Logged in after almost two Years.
1 way with a score of 0 for D.
3 ways with a score of 1 for D. [+2 more than previous.]
6 ways ... 2 for D. [+3 more than previous.]
...
+ ? ways ... 100 for D. [+101 more than previous.]
-------------------------------------------------------------------
= the answer.
= a0 + a1 + a2 + a3 + ...... + a100
= 1 + 1 + 2 + 1 + 2 + 3 + 1 + 2 + 3 + 4 + ....
= 1x101 + 2x100 + 3x99 + ... + 51x51 + 52x50 + ... + 101x1
= 2 x [1x101 + 2x100 + ... + 50x52] + 51x51
= 2 x sum from 0 to 49 of [1+i ][101-i] + 51x51
= 2 x sum from 0 to 49 of [101 + 100xi - ixi] + 51x51
= 2 x (50x101 + 100x49x50/2 - 49x50x99/6) + 51x51
= 176,851
3 ways with a score of 1 for D. [+2 more than previous.]
6 ways ... 2 for D. [+3 more than previous.]
...
+ ? ways ... 100 for D. [+101 more than previous.]
-------------------------------------------------------------------
= the answer.
= a0 + a1 + a2 + a3 + ...... + a100
= 1 + 1 + 2 + 1 + 2 + 3 + 1 + 2 + 3 + 4 + ....
= 1x101 + 2x100 + 3x99 + ... + 51x51 + 52x50 + ... + 101x1
= 2 x [1x101 + 2x100 + ... + 50x52] + 51x51
= 2 x sum from 0 to 49 of [1+i ][101-i] + 51x51
= 2 x sum from 0 to 49 of [101 + 100xi - ixi] + 51x51
= 2 x (50x101 + 100x49x50/2 - 49x50x99/6) + 51x51
= 176,851