Thirty x Thirty x Thirty and so on

Since, ${ 30 }^{ x }={ 2 }^{ x }{ 3 }^{ x }{ 5 }^{ x }$ .We need to find the power of $5$ in $345!$ ,i.e. $=\left\lfloor \frac { 345 }{ 5 } \right\rfloor +\left\lfloor \frac { 345 }{ 25 } \right\rfloor +\left\lfloor \frac { 345 }{ 125 } \right\rfloor +\left\lfloor \frac { 345 }{ 625 } \right\rfloor +...\\ =69+13+2+0+0+0+....\\ =\boxed{84}$

The Power of 2 in 345! =

[345/2] + [345/4] + [345/8] + [345/16] +[345/32] +[345/64] +[345/128] +

[345/256] + [345/512] + ............ = 172 + 86 + 43 + 21 + 10 + 5 + 2 + 1 + 0 + 0 + .......

= 340

The power of 3 in 345! =

[345/3] + [345/9] +[345/27] +[345/81] +[345/243] +[345/729] + .... =

115 + 38 + 12 + 4 + 1 + 0 + 0 + ..... = 170

The power of 5 in 345! =

[345/5] +[345/25] +[345/125] +[345/625] + ... = 69 + 13 + 2 + 0 + 0 + ..... = 84

Then

345! = (2^340)(3^170)(5^84)(7^a)(11^b)(13^c) .... .............. (1)

30^x = (2^x)(3^x)(5^x) ................... (2)

From (1), (2)

x = ............., -4, -3, -2, -1, 0, 1, 2, 3, 4, ........., 84

Then, the maximum value of x = 84

This can be expressed as 345!/(2 * 3 * 5)^x which then can be expressed as 345!/(2^x * 3^x * 5^x) We can then find the number of times 2, 3, or 5 becomes a factor in 345! The least number of times becomes x. Because 5 is the biggest and there are all in the same exponent of 1 we can simply consider 5 as there will be more 3s or 2s in the numerator

The number of 5s in the expression 345! or 1 x 2 x 3 x ..... x 345 can be shown by: Numbers Divisible by 5 + Numbers Divisible by 25 + Numbers Divisible by 125 The reason is that 25 = 5 x 5 and you can have 2 5s there but one is already counted when we got the numbers divisible by 5, similarly the numbers divisible by 125 would have 5 x 5 x 5 but 2 of the 5s are counted by divisibility of 5 and 25 so we need to add only 1 more for every number divisible by 125.

Numbers divisible by 5 = round down (345/5) = 69 Numbers divisible by 25 = round down (345/25) = 13 Numbers divisible by 125 = round down (345/25) = 2 69 + 13 + 2 = 84

30 = 3 x 10. Since dividing by 30 only requires removing one '3', and there are sufficient '3's in 345! It does not affect the final answer.

Next we consider the number of 10's. Since dividing by 30 removes a '10' or getting rid of a zero, The number of trailing zeroes of 345! is the answer we want.

We then notice the following pattern:

A multiple of 5 produces one trailing zero (since there are sufficient 2's in the product) , a multiple of 25 produce 2, a multiple of 125 produce 3 and so on.

Thus 345 contains

2 + 2 + 3 + 2 + 3 + 2 + 2 + 3 + 2 + 3 + 2 + 2 + 4 + 2 + 3 + 2 + 2 + 3 + 2 + 3 + 2 + 2 + 3 + 2 + 4 + 2 + 2 + 3 + 2 + 3 + 2 + 2 + 3 + 2 + 1 = 84,

and that's the answer we wanted.