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As the question explicitly asks about factors and we see no way to apply divisibility rules, we'll start by prime factorization. This is a Precise approach.
1365 = 1300 + 65 = 5(260 + 13) = 5*273 = 5*3*91=5*3*7*13 Now we'll look to our answers to help us out. p=1 definitley works so (E) is eliminated. if p = 7 were true than 200! would need to have at least 7 of each of 3,5,7,13. Since 7*13 = 91 then 200! has all multiples of 13 smaller or equal to 91 (and certainly also those for 3,5,7.) So (C) works and (D) is eliminated. Could p = 15? 15*13 =195 which is still smaller than 200! So the first 15 multiples of 13 are in 200! and (B) works. (C) is eliminated. Could p = 16? 16*13 = 195+13=208 so 200! does not have 16 different multiples of 13. However, before eliminating (A) we should look out for the square trap - since 169 = 13*13 than it has two factors of 13 and so 200! has 16 total 13s in its factorization - one for each of the 15 multiples and one more for 13^2. (B) is eliminated.
Re: If p is an integer and 1365^p is a factor of 200
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21 May 2018, 14:57
1
200! has 200/5 + 200/25 + 200/125 zeroes ~= 49 1365 -> since the answer options are not close -> Approximating as (1000)^[p] -> (10)^[3p] So 3p<49 p~<16
Re: If p is an integer and 1365^p is a factor of 200
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24 May 2018, 10:48
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pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?
A. 16 B. 15 C. 7 D. 1 E. 0
1365 = 5 x 273 = 5 x 3 x 91 = 5 x 3 x 7 x 13
Since 13 is the largest prime factor of 1365, we see that the maximum value of p will be based on the number of 13’s that are factors of 200!.
We see that 13 is a factor of 200!, and, in fact, every one of the multiples of 13 less than 200 is also a factor of 200!. Since 13 x 15 = 195, we see that there are 15 multiples of 13 that are factors of 200!. A But since 169 = 13^2, we see that there is an additional factor of 13, so there are 16 factors of 13 in 200! Thus, the maximum value of p is 16.
Re: If p is an integer and 1365^p is a factor of 200
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01 Jun 2018, 09:44
ScottTargetTestPrep wrote:
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?
A. 16 B. 15 C. 7 D. 1 E. 0
1365 = 5 x 273 = 5 x 3 x 91 = 5 x 3 x 7 x 13
Since 13 is the largest prime factor of 1365, we see that the maximum value of p will be based on the number of 13’s that are factors of 200!.
We see that 13 is a factor of 200!, and, in fact, every one of the multiples of 13 less than 200 is also a factor of 200!. Since 13 x 15 = 195, we see that there are 15 multiples of 13 that are factors of 200!.A But since 169 = 13^2, we see that there is an additional factor of 13, so there are 16 factors of 13 in 200! Thus, the maximum value of p is 16.
No, you are not missing anything. The formula you mentioned is the another way to solve this question.
What ScottTargetTestPrep meant was since we will have every factor of 13 less than 200 being a factor of 200! because 200! when expanded will include all those factors. Based on that we calculated there are 15 factors of 13 that are less than 200. All of those when divide 200! will give us an integer.
Now, we do have 169 also as one of it's factor less than 200. But this 169 is \(13^2\), that means we have two 13's in 169. So, we will add 1 to 15 and hence get the total number of 13s = 16.
Re: If p is an integer and 1365^p is a factor of 200
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03 Jun 2018, 18:48
hisho wrote:
ScottTargetTestPrep wrote:
pushpitkc wrote:
If p is an integer and \(1365^p\) is a factor of \(200!\), what is the maximum possible value of p?
A. 16 B. 15 C. 7 D. 1 E. 0
1365 = 5 x 273 = 5 x 3 x 91 = 5 x 3 x 7 x 13
Since 13 is the largest prime factor of 1365, we see that the maximum value of p will be based on the number of 13’s that are factors of 200!.
We see that 13 is a factor of 200!, and, in fact, every one of the multiples of 13 less than 200 is also a factor of 200!. Since 13 x 15 = 195, we see that there are 15 multiples of 13 that are factors of 200!.A But since 169 = 13^2, we see that there is an additional factor of 13, so there are 16 factors of 13 in 200! Thus, the maximum value of p is 16.
I'm following the below rule strictly, do you suggest a modification for it?
When we said there are 15 multiples of 13 less than 200, we included every multiple of 13 from 13 to 195. Thus 169 is also included. However, the difference between 169 and all the other multiples is that 169 = 13^2, where all the other multiples of 13 are 13^1 * k.
Thus, the reason why we said that 169 has an additional factor of 13 is because 169 as two prime factors of 13 as opposed to just 1.
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