The number of odd numbers in the row of the pascals triangle is A
The sum of the positions of these odd numbers can be written as B .
Find the number of trailing zeroes in .
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The number of odd numbers in the n th row of the pascals triange is 2 raised to the power of the number of one's in the binary expansion of n. Thr binary expansion of 81 is 1010001 . There are 3 one's in it thus
A = 2 3 = 8
The positions of the 8 odd numbers in row 81 of Pascal’s triangle are those numbers that can be formed using a subset (possibly empty) of 81 expressed in the powers of 2. That is 8 1 = 6 4 + 1 6 + 1
The number of subsets are: 0 , 1 , 1 6 , 6 4 , 1 7 ( 1 6 + 1 ) , 6 5 ( 6 4 + 1 ) , 8 0 ( 1 6 + 6 4 ) , 8 1 ( 1 + 1 6 + 6 4 ) B = 3 2 4
Thus, A + B = 3 2 4 + 8 = 3 3 2
The final act is to find the number of trailing zeroes in 332! This is the same as finding the exponent of ten in that value. Which is same as fimding the exponents of 2 and 5 . However there are more 2's in the value than 5's .
The number of 5's in 332! are
⌊ 5 3 3 2 ⌋ + ⌊ 2 5 3 3 2 ⌋ + ⌊ 1 2 5 3 3 2 ⌋
= 6 6 + 1 3 + 2 = 8 1
Now we can conclude that the number of trailing zeroes in 3 3 2 ! are 8 1