Multiples of 67 in a constructed array of arrays? A triangle-shaped table is constructed such the first row is "1, 3, 5, 7, ..., 99"

Each subsequent row has the following properties:


One less element than the row above
Each element is the sum of the two elements above it


Like so:

1   3   5    7    9    11 ....
  4   8   12   16   20 ....
    ...............


This continues until the last row with one element.

How many elements are divisible by 67?

Fewest lines, chars, or whatever may win.  Depends on ingenuity.

EDIT: the answer is 17 but you have to prove it.

ADDENDUM:
I have three-line python code for this but try to beat it:

a = [[2*i+1 for i in range(50)]]
while len(a[-1])&gt;1:a.append([a[-1][i]+a[-1][i+1] for i in range(len(a[-1])-1)])
print len([i for b in a for i in b if i%67==0])
Haskell - 86 chars

98 → 92: Use list comprehensions and sum instead of filter and length.

92 → 86: Convert nested list comprehensions into one, and change ==0 to &lt;1 (thanks st0le).

f s=zipWith(+)s$tail s
main=print$sum[1|r&lt;-take 50$iterate f[1,3..99],n&lt;-r,n`mod`67&lt;1]


Ungolfed:

initialRow :: [Integer]
initialRow = [1,3..99]

nextRow :: (Num a) =&gt; [a] -&gt; [a]
nextRow xs = zipWith (+) xs (tail xs)

triangle :: [[Integer]]
triangle = takeWhile (not . null) $ iterate nextRow initialRow

main = print $ length $ filter ((== 0) . (`mod` 67)) $ concat triangle