But it's so many digits!

First note that: $12=2^2\cdot 3$ . Let $e_p (n!)$ denote the the largest power of a prime $p$ which divides $n!$ .

By Legendre's Formula , we have:

$e_2 (49!)=\sum_{i=1}^{\infty}\left\lfloor\dfrac{49}{2^i}\right\rfloor=\left\lfloor\dfrac{49}{2^1}\right\rfloor+\left\lfloor\dfrac{49}{2^2}\right\rfloor+\left\lfloor\dfrac{49}{2^3}\right\rfloor+\left\lfloor\dfrac{49}{2^4}\right\rfloor+\left\lfloor\dfrac{49}{2^5}\right\rfloor+\left\lfloor\dfrac{49}{2^6}\right\rfloor+\cdot\cdot\cdot$

$=24+12+6+3+1+0+0+0+\cdot\cdot\cdot=46$

$e_3 (49!)=\sum_{i=1}^{\infty}\left\lfloor\dfrac{49}{3^i}\right\rfloor=\left\lfloor\dfrac{49}{3^1}\right\rfloor+\left\lfloor\dfrac{49}{3^2}\right\rfloor+\left\lfloor\dfrac{49}{3^3}\right\rfloor+\left\lfloor\dfrac{49}{3^4}\right\rfloor+\left\lfloor\dfrac{49}{3^5}\right\rfloor+\left\lfloor\dfrac{49}{3^6}\right\rfloor+\cdot\cdot\cdot$

$=16+5+1+0+0+0+\cdot\cdot\cdot=22$

So we get $2^{46}\mid\mid 49! ~~~ \Longrightarrow ~ 4^{23}\mid\mid 49!$ and $3^{22}\mid\mid 49!$ . Since $23>22$ , we get: $12^{22}\mid\mid 49!$ .

Therefore, the largest power of $12$ dividing $49!$ is $\fbox{22}$ .

How many $2$ 's are there in $49!$ ?

$\lfloor \frac {49}{2} \rfloor + \lfloor \frac {49}{4} \rfloor + \lfloor \frac {49}{8} \rfloor + \lfloor \frac {49}{16} \rfloor + \lfloor \frac {49}{32} \rfloor = 46$ .

How many $3$ 's?

$\lfloor \frac {49}{3} \rfloor + \lfloor \frac {49}{9} \rfloor + \lfloor \frac {49}{27} \rfloor = 22$ .

But we know $12 = 2\cdot2\cdot3$ . We have $23$ pairs of $2$ 's that we can choose, but only $22$ $3$ 's. Thus, maximizing the power of $12$ yields ${\color{#3D99F6}{\boxed {22}}}$ .

• _12=2^{2} x 3 _
• _ highest power of 2 in 49!= [49/2]+[49/4]+[49/8]+[49/16]+[49/32]=46 _
• _ highest power of 2^{2} in 49!= 23 _
• _highest power of 3 in 49!= [49/3]+[49/9]+[49/27]=22 _
• _ therefore the maximum power of 12 in 49 ! = 22 _

First of all, observe that $12^{n} = 3^{n} \times 2^{2n}$ . Then observe that $49!$ has $48$ factors $2$ and $22$ factors $3$ . Then we want $n$ such that

$n = min(48/2, 22) = 22$

12=2 x 2 x 3

Because you get at least one 2 every even number, compared to the amount of 3's it is absolutely enormous, so we don't want to worry about that.

there is one 3 for every multiple of 3, an additional for every multiple of 9, and an additional for every multiple of 27.

[49/3]+[49/9]+[49/27]=16+5+1= 22

We express $49!$ in terms of $2^{a}3^{b}k$ for some positive integer $a,b,k$ . Now, from Legendre's theorem, we obtain

$a = \sum_{n=1}^{\infty}\lfloor \frac{49}{2^{n}}\rfloor = 46$

and

$b = \sum_{n=1}^{\infty}\lfloor \frac{49}{3^{n}}\rfloor = 22$

It follows that $49!=2^{46}3^{22}k=(2^{2}3)^{22}2^{2}k=12^{22}4k$ . Hence, the largest power of $12$ that would divide $49!$ ,must be $\boxed{22}$

Notice that $12=2\times2\times3$ , which means that we need to obtain two "2"s and a "3" from the prime factorization of $49!$ in order to be divisible by one power of 12. Therefore, we need to calculate the number of "2"s and the number of "3"s in the prime factorization of $49!$ .

The number of "2"s in the prime factorization of $49!$

= $\left\lfloor{\frac{49}2}\right\rfloor+\left\lfloor{\frac{49}4}\right\rfloor+\left\lfloor{\frac{49}8}\right\rfloor+\left\lfloor{\frac{49}{16}}\right\rfloor+\left\lfloor{\frac{49}{32}}\right\rfloor$

= $24+12+6+3+1$

= $46$

The number of "3"s in the prime factorization of $49!$

= $\left\lfloor{\frac{49}3}\right\rfloor+\left\lfloor{\frac{49}9}\right\rfloor+\left\lfloor{\frac{49}{27}}\right\rfloor$

= $16+5+1$

= $22$

There are 46 "2"s and 22 "3"s available, and since $\frac{46}2>22$ , the answer is 22.

Here are two possible approaches (using Legendre's theorem):

1) We have $v_3(49!) = \sum_{i = 1}^{\infty} \left\lfloor \frac{49}{3^i} \right\rfloor = 22$ $v_4(49!) = \frac{1}{2}v_2(49!) = \frac{1}{2}\sum_{i = 1}^{\infty} \left\lfloor \frac{49}{2^i} \right\rfloor = 23$ Hence the answer is $\boxed{22}$ .

2) We have $v_3(49!) = \frac{49 - S_3(49)}{3 - 1} = \frac{44}{2} = 22$ $v_4(49!) = \frac{1}{2}v_2(49!) = \frac{1}{2} \cdot \frac{49 - S_2(49)}{2 - 1} = \frac{46}{2} = 23$ and the answer again comes out to be $\boxed{22}$ .

No. of 2's conatined in $49!=24+12+6+3+1=46$

No. of 3's conatined in $49!=16+5+1=22$

since in 12 2's and 3's are in the ratio $2:1$ and $46>44$

thus $\boxed{22}$

We know if to find many zero, we think that it divides by 10. but it divides by 5 up to 5^n. So, it's 12 = 2^2. 3, divides by 3 up to 3^n ----> | 49/3 | + | 49/9 | + | 49/27 | = 16 + 5 + 1 = 22. ANSWER : 22

here we will count total power of three's in 49!
3 = 1 6 = 1 9 = 2 12 = 1 15 = 1 18 = 2 21 = 1 24 = 1 27 = 3 30 = 1 33 = 1 36 = 2 39 = 1 42 = 1 45 = 2 48 = 1
total = 22
now we know that 12 = 2^2*3^1
so here in one (12) no. of three's required =1
so highest power of 12 that divides 49! is 22
(here we will not consider two since two will obviously more than twice of no. of three's and here it is 46)

nice solution

12 = 2^2×3 We find the maximum power of 2 in 49! = 492+494+498+4916+4932=24+12+6+3+1=46 So maximum power of 2^2 in 49! is 23. Now we find the maximum power of 3 in 49! = 493+499+4927=16+5+1=22 ⇒49!=(22)23×3^22×some K So 22 is the maximum power of 12 that divides 49! exactly.

As a start, we need to break 12 down into its prime factors. Thus, 12 = 2 * 2 * 3.

To now find out the greatest power of 12 that can divide 49!, we need to figure out how many 3s and how many 2s does the expanded 49! contain.

To do this, we first divide 49 by 3, which gives the numerator as 16. Thus, there are 16 numbers in 49! that are directly divisible by 3. These numbers of course are the multiples of 3 like 3,6,9......48. Now, the above calculation only takes a single 3 into account. But what about numbers like 9 which contain two 3s. The above calculation only counted them as one 3. Thus, we have 49/9 = 5 more 3s in 49!. These come from multiples of 3 squared. Lastly, we have have multiple of 3^3 in 49!. To get the exact number, we divide 49/27 = 1 more 3.

How do we know when to stop? When 49 divided by 3^n gives a numerator of 0. In this case, 49/81 would give us 0 and hence 49! does not contain any more 3s.

Total number of 3s in 49! = 16+5+1 = 22 in number.

We must also check for 2s in 49!.

These are, by similar calculation as above, 49/2 = 24 in number 49/4 = 12 in number 49/8 = 6 in number 49/16 = 3 in number 49/32 =1 49/64 = 0 and hence we stop

Total numbers of 2s in 49! = 24+12+6+3+1 = 46.

Going to the start, 12 = 2 2 3. Thus, for every two 2s, we need to have only one 3. Thus, since 22 is less in number than 46/2 (Number of 2s by 2), the number of 3s become the decisive factor in this case.

Thus, the maximum power of 12 that can divide 49! is 22.

49/3 + 49/3^2 + 49/3^3 = 22

49 / 12 = 49 /(2^2 x 3)

Now let us count how many 2^2 and 3 in 49!

sources of 2 in 49 = 49/2 + 49 /4 + 49/8 + 49/16 + 49 /32 = 24 + 12 + 6 + 2 +1 = 45

sources of 3 in 49 = 49/3 + 49 / 9 + 49 / 27 = 16 + 5 + 1 = 22

This means there are 2^45 and 3 ^ 22 in 49 !

number of 12's in 49! = 2^3 x (2^2 x 3)^22

then, the highest power of 12 in 49 ! = 22

PGCD(4,3)=1 and 12=4*3 It is trivial to prove that the power of 4 in 49! is more than the power of 3 in 49! Which means the highest power of 12 that would divide 49! is the power of 3 in 49! which is easy to compute : 22

12=2x2x3....so ,the highest power of 12 is the highest power which the greatest prime factor of 12.,that is 3 possesses .. the greatest power of 3 is given by [49/3]+[49/3x3]+[49/3 3 3]= 22

This question is all about counting prime factors 2 and 3, since the number 12 contains two factors 2 and one factor 3. In order to find the highest power of 12, we count all factors 2 and 3 in $49!$ . This is the same thing as saying we count the number of factors in each of $1,2,3,\ldots,49$ and add these. It turns out there are $46$ factors 2 and $22$ factors 3 in $49!$ . Since $46>2\cdot 22$ , the number of factors limits the powers of 12 and as a result, the highest power of 21 dividing $49!$ is 22.

Write the prime factorization of all numbers from 2 to 49. From these factorizations, cross out any number that is not a 2 or a 3. Create as many {2, 2, 3} sets as possible from the remaining numbers. You will be able to make 22 such sets.

We find the maximum power of 2 in 49! = 49/2+49/4+49/8+49/16+49/32=24+12+6+3+1=46

So maximum power of 22 in 49! is 23.

Now we find the maximum power of 3 in 49! = 49/3+49/9+49/27=16+5+1=22 ⇒49!=(2^2)^23×3^22×some K

So 22 is the maximum power of 12 that divides 49! exactly.

The highest power of 12 that divides 49! is the same as the highest power of 3 that divides 49! and this is [49/3]+[49/3^2]+[49/3^3] $$ \text{ where } [x] \text{ denotes the greatest integer } \le x$$

the highest power of 12 that would devide 49! is $\left \lfloor \frac{49}{3} \right \rfloor+\left \lfloor \frac{49}{9} \right \rfloor+\left \lfloor \frac{49}{27} \right \rfloor=16+5+1=22$

Highest power of 49! is $\lfloor \frac{49}{3} \rfloor+\lfloor \frac{49}{9} \rfloor +\lfloor \frac{49}{27} \rfloor=16+5+1=\fbox{22}$