# Equal Averages - Neal's Solution

Thank you to Neal for sending in his thoughts about Equal Averages

I think the answer is 33.

Reasoning. Start with a,b,100,c,100+a in increasing order.
This covers median and range constraints
To get a mode of 100, either b=100 or c=100

So consider 2 cases:
1) Try b=100:
So the numbers are
a,100,100,c,100+a
But the mean is 100 also
So a+100+100+c+100+a=500
so c=200-2a
But also, 100<=c<=100+a
so 100<=200-2a<=100+a
this gives 34<=a<=50
i.e. a set of 17 groups
from 34,100,100,132,134
to 50,100,100,100,150

2) try b=100
So the numbers are
a,b,100,100,100+a
progressing similar to above
The mean is (a+b+100+100+100+a)/5 = 100
so b=200-2a
However, a<=b<=100
so a<=200-2a<=100
so 50<=a<=66
This gives a range of 17 answers, from 50,100,100,100,150 up to 66,68,100,100,166

Finally. combine the two cases

(a,100,100,200-2a,100+a)  for 34<=a<=50
and
(a,200-2a,100,100,100+a) for 50<=a<66
Which is 33 groups (because one group (50,100,100,100,150) is shared by both cases

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