Find the largest possible integer value of $x$ such that $30^x$ divides $345!$ .

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$345!$ means $345$ factorial.

The answer is 84.

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By right, you should find the powers of all of the terms, just in case.

It is obvious that power of 5 is less than the power of other prime factors of 30 i.e. 2 and 3,also,you always need 5 to make 30,hence,we can just check the powers of 5,power of others will automatically fulfill.

mudit bansal
- 6 years, 4 months ago

Of course, since 5 is coprime to 2 and 3, we can just use 5 to solve for $x$ :)

Jake Lai
- 6 years, 4 months ago

No, not exactly. See this problem for why it is important to check just in case.

@Calvin Lin – Since power of each prime factor is same, we can do it directly.

Pranjal Jain
- 6 years, 4 months ago

@Pranjal Jain – RIght. My point is that fact has be to mentioned, ie you have to check the powers.

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

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This solution is not entirely clear. Your comment "We can only compute the power of 5 in 345!" makes no connection to the problem. Nor did you explain why you chose the number of powers of $5$ instead of the number of powers of $2$ 's and $3$ 's. And since you're not using $\LaTeX$ , you should specify that [x] implies the floor function of $x$ .

We can only compute the power of 5 in 345!

Gamal Sultan
- 6 years, 4 months ago

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

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Like Calvin has mentioned, you should check for all powers of prime factors of $345$ .

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.

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This solution is not clear. How did you come up with "there are sufficient '3's in 345! " and what does it mean?

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