wow someone's in an inspired mood ;p
Solution:
So we want solutions to 3a+2b+4c=100.
Let's focus on the simple case when a=0. 100=50*2 hence there are 26 different ammounts of four that can fit here. (4*0,4*1...4*25)
when
:
3a=100-2b-4c
3a=2(50-b-2c) hence a must by even.
Therefore the values
are only between 1 and 16 inclusive. Let's analyse each case in a descending order:
let 100-3a=x
The possible x values are hence 4,4+6,4+6*2.... 4+6*16. Next to each we will write down how many four's fit there.
4 - 0 or 1- 2 four's
10 - 0,1,2 - 3 four's
16 - 0,1,2,3,4 - 5 four's
22 - 0,1,2,3,4,5 - 6 four's
28 - 0,1,2,3,4,5,6,7 - 8 four's
.
.
.
You can see that the pattern is that the difference between consecutive count's of four is such: first is we jump by 1, then by 2, then by 1.. then by 2 again etc. (this can be easily proven, but i will skip it as my post is long enough, ill explain in a later post)
hence the total sum is:
+(3+6+9+12...))
So it is the sum of two arithmetic sequences, now all we need is the last term:
because
ranges from 1 to 16. there are 16 terms alltogether and hence 8 terms in each sum. So the sum is:
)+(3+6...3+3(7)))
+(3+6+9...24))
Basically add that sum to 26 (the thing we found when a=0)
edit: 200 posts
:D
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