Pascal's Triangle!(Easy Isn't It)

Probability Level 3

The number of odd numbers in the $81st$ 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 $\large {(A+B)!}$ .

Details:

The sum of the positions refers to the sum of the position numbers of these numbers. For example, If these numbers are at positions 1 , 2 , and 1729 then the sum is 1732
The ith row and jth entry is ${ i \choose j }$ . Note that the first row is actually row 0, and the first entry is the 0th entry.

The answer is 81.

1 solution

Sualeh Asif
Jan 1, 2015

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 $81= 64+16+1$

The number of subsets are: $0,1,16,64,17(16+1),65(64+1),80(16+64),81(1+16+64)$ $B = 324$

Thus, $A+B= 324+8= 332$

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

$\lfloor {\frac {332}{5}}\rfloor +\lfloor {\frac {332}{25}}\rfloor+\lfloor {\frac {332}{125}}\rfloor$

$=66+13+2 = 81$

Now we can conclude that the number of trailing zeroes in $332!$ are $\boxed {81}$

Pascal's Triangle!(Easy Isn't It)

The answer is 81.

1 solution

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