PMO problem

$1\times 3\times 5\times ....\times 35\\ =\frac { 1\times 2\times 3\times 4\times 5\times ......\times 35\times 36 }{ 2\times 4\times 6\times 8\times .....\times 36 } \\ =\frac { 36! }{ { 2 }^{ 18 }\times 18! }$

Now, power of 3 in 36! is $\left\lfloor \frac { 36 }{ 3 } \right\rfloor +\left\lfloor \frac { 36 }{ 9 } \right\rfloor +\left\lfloor \frac { 36 }{ 27 } \right\rfloor \\ =12+4+1\\ =17$

And the power of 3 in 18! is $\left\lfloor \frac { 18 }{ 3 } \right\rfloor +\left\lfloor \frac { 18 }{ 9 } \right\rfloor \\ =6+2\\ =8$

So, $1\times 3\times 5\times ....\times 35\\$ can be written as $\frac { k\times { 3 }^{ 17 } }{ m\times { 3 }^{ 8 } }$ for some constants k & m and in simple power of 3 it can be further reduced to $A\times { 3 }^{ 9 }$ where mA=k

I'm gonna give a general solution to this.

First we find the first and last numbers divisible by $3$ ( $3$ and $33$ here) which are formed by multiplying $3$ by $1$ and $11$

This is an A.P of $1$ to $11$ with $d \ = \ 2$ Thus, there are $n \ = \ 6$ terms in the sequence. Now, we find the smallest power $x$ of $3$ greater than $1$ ( $2$ here) and find all of it's odd multiples, except the ones containing any multiple of $3$ in the sequence $y$ and write this as $(x \ - \ 1) \times y$ ( which is equal to $1$ for $3^{2}$ and $2$ for $3^{3}$ here). We go on till there are no more powers of $3$ in the sequence. Now sum these and add it to the original $n$ i.e:- $6 \ + \ 2 \ + \ 1 \ = \boxed{9}$ here.

This can be used for any huge sequence for $3$ or any other odd number except one( probably not even).

the largest power of 3 for 35! is 15, deleting the even numbers of the set, gives us 6 powers of 3, thus 15 -6 = 9 largest remaining powers of 3 for the given set.

$11$ is the largest number for which $11 \cdot 3 \le 35$ .

Therefore powers of three are enclosed within this series:

$(1\cdot 3)(3\cdot 3)(5\cdot 3)...(11\cdot 3)$

Now we need to find the largest power of 3 which can divide the above series.

Since we have already simplified the sum, we just need to count the threes from above series. By doing that, we find out that that the largest power of $3$ is $6$ .

Now, the simplified series (after removing the powers of 3 which we have counted):

$1\cdot 3\cdot 5...11$

But we are not done yet as some terms still contain powers of three. We separate these terms and rewrite the series:

$3\cdot 9$

Since there are only two terms, we can count the largest power $3$ which can divide the above product which is three.

We add the two powers $6 + 3$ . So the answer is $9$ .

The multiples of 3 are the terms whose index gives a remainder of 2 when divided by 3. There are 18 terms total, so there are 6 terms which have a factor of 3. Of those, 2 have a factor of 9, and 1 of those has a factor of 27. Thus we have 6 + 2(2-1) + 1(3-2) = 6 + 2 + 1 = 9.

However, the question is a bit ambiguous. I put 3^9 = 19683 as my answer at first. Maybe it's just me though.

Trying an easier (but lazier) approach: Among 1, 3, 5, ... , 33 and 35, there are six numbers that are divisible by 3: 3,9,15,21,27, and 33.

With 9=3^2 , and 27=3^3, and the other four numbers having 3 in a single multiplicity, then 3+2+1+1+1+1 = 9.