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Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

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\(20! = 2^{18}*3^8*5^4*7^4*11^1*13^1*17^1*19^1\)

Now, If p = 2 then k = 18 and p*k = 36
Now, If p = 3 then k = 8 and p*k = 24
Now, If p = 5 then k = 4 and p*k = 20

Now we see a decreasing trend in values of p*k

i.e. \(p*k_{max.} = 36\)

Answer: Option A
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ElciaM
Given that p is prime and p^k is a factor of 20!,
20! has primes 2,3,5,7,11,13,17,19
and for p^k to be a factor of 20! it should be any one of 2^2, 2^3, 2^4, 3, 3^2, 5,7,11,13,17,19 where the powers represent k ie., k=1,2,3,4
Hence pk=9*4=36 (other options don't give the maximum value of pk)

Answer: A

Hi ElciaM

Given that p = Prime, you can NOT choose the value of p = 9 :)
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Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question

I think the problem wholly is wrong or I am having some difficulty understanding . Bunuel can you have a look please .:

So 20! apparently consists of eight 3's (2 from 18,1 from each of 15,12,6,3 and 2 from 9 ) . So apparently 3^9 is a factor of 20! . Again 11^1 is also a factor . So this makes pk to be 99 .

I just realized if we take 2 and not 3 , then it will be far more. So am I thinking in the correct direction experts?


Hi MrSengupta

Please note the fact that \(p^k\) is a factor of 20!

3^9 is NOT factor of 20! [Highlighted the mistake]

You can NOT choose p from someplace and k from some other factor. p^k must be factor of 20! so p and k both must belong to the same factor of 20!

I hope this help!!! :)
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Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question

I think the problem wholly is wrong or I am having some difficulty understanding . Bunuel can you have a look please .:

So 20! apparently consists of eight 3's (2 from 18,1 from each of 15,12,6,3 and 2 from 9 ) . So apparently 3^9 is a factor of 20! . Again 11^1 is also a factor . So this makes pk to be 99 .

I just realized if we take 2 and not 3 , then it will be far more. So am I thinking in the correct direction experts?


Hi MrSengupta

Please note the fact that \(p^k\) is a factor of 20!

3^9 is NOT factor of 20! [Highlighted the mistake]

You can NOT choose p from someplace and k from some other factor. p^k must be factor of 20! so p and k both must belong to the same factor of 20!

I hope this help!!! :)

Yes thanks a lot . That helped .
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Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question

for the given two conditions ;
a) P has to be prime and
b) find largest value of pk
the value of p has to be the least prime value i.e 2
so total factors of 20! at p =2 is ;
20/2+20/4+20/8+20/16 = 10+5+2+1 ; 18 = k
and p*k = 18*2 ; 36
OPTION A ; :)
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Primes that are factors of 20! include 2,3,5,7,11,13,17 and 19.

The highest powers of these primes that are also factors of 20! include 2^18, 3^8, 5^4, 7^2, 11^1, 13^1, 17^1 and 19^1

Any of these can be considered as p^k, but the one giving highest value of pk is 2^18.
In this case, pk = 2 x 18 = 36.

In all other cases, pk is less than 36.

Hence, answer is A.

Kudos if you liked the explanation.

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How did you decide that the minimum value will occur at the smallest prime factor?
I understand one way to do it is to observe the trend, but wanted to check if there is some other way also.

Archit3110
Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question

for the given two conditions ;
a) P has to be prime and
b) find largest value of pk
the value of p has to be the least prime value i.e 2
so total factors of 20! at p =2 is ;
20/2+20/4+20/8+20/16 = 10+5+2+1 ; 18 = k
and p*k = 18*2 ; 36
OPTION A ; :)
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Gylmitul
How did you decide that the minimum value will occur at the smallest prime factor?
I understand one way to do it is to observe the trend, but wanted to check if there is some other way also.

Archit3110
Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question

for the given two conditions ;
a) P has to be prime and
b) find largest value of pk
the value of p has to be the least prime value i.e 2
so total factors of 20! at p =2 is ;
20/2+20/4+20/8+20/16 = 10+5+2+1 ; 18 = k
and p*k = 18*2 ; 36
OPTION A ; :)



Hi Gylmitul
You have to maximize \(p^{k}\)

Thus in order to do that you have to choose the smallest prime factor but with the biggest k.

\(2^{18}\) is the best choice in order to maximize \(p^{k}\)
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Gylmitul
How did you decide that the minimum value will occur at the smallest prime factor?
I understand one way to do it is to observe the trend, but wanted to check if there is some other way also.

Archit3110
Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question

for the given two conditions ;
a) P has to be prime and
b) find largest value of pk
the value of p has to be the least prime value i.e 2
so total factors of 20! at p =2 is ;
20/2+20/4+20/8+20/16 = 10+5+2+1 ; 18 = k
and p*k = 18*2 ; 36
OPTION A ; :)

Hi Gylmitul

One can NOT be sure about it in a short span of time (which we get while solving problem in test) so it's recommended that we see a pattern of how the values of the required function are changing rather than just staying with one set of values. :)
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Gylmitul
well the question has asked for maximum value of p*k ; where p is prime integer and k is factor of 20 ! at p^k ;
determining that at which value of p will we get max k ; its basically the lower the value of divisor more is the share we get eg when 20/2 we get 10 parts and when 20/10 we get 2 parts.
similarly when we are finding the factors for 20! for a prime integer p ; the highest factors would be determined at lowest value p i.e 2 in this case..factors would be k
in this case
..
for this question we do need to determine the trend for atleast 2 different values of p .. you may do it for first two primes i.e 2 &3 and you would realise that p*k value is decreasing with every prime integer for 20! ..
once thats done then p*k can be found as done in explanation above...
hope this helps :)


Gylmitul
How did you decide that the minimum value will occur at the smallest prime factor?
I understand one way to do it is to observe the trend, but wanted to check if there is some other way also.

Archit3110
Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question

for the given two conditions ;
a) P has to be prime and
b) find largest value of pk
the value of p has to be the least prime value i.e 2
so total factors of 20! at p =2 is ;
20/2+20/4+20/8+20/16 = 10+5+2+1 ; 18 = k
and p*k = 18*2 ; 36
OPTION A ; :)
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Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

KUDOS if you appreciate this question
For reference, here is 20! expanded: 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

The largest prime number that is a factor of 20! is 19. So p could be 19; however, in that case, k would be 1, and the value of pk would be 19(1) = 19. Now let’s see if we can find a value for pk that is greater than 19.

If p is any prime number from 11 to 17, inclusive, the value of k would still be 1, and the value of pk would then be less than 19.

If p = 7, then k = 2 (since 7 and 14 are factors of 20!). However, the value of pk is 7(2) = 14, which is less than 19.

If p = 5, then k = 4 (since 5, 10, 15 and 20 are factors of 20!). The value of pk is 5(4) = 20, which is greater than 19. However, 20 is not one of the answer choices, so we must continue.

If p = 3, then k = 8 (since 3, 6, 9, 12, 15, and 18 are factors of 20!. Notice that 9 and 18 have 2 factors of 3 while the other factors have 1 factor of 3 each). The value of pk is 3(8) = 24. Again, this is not one of the answer choices, so we must continue.

If p = 2, then k = 18 (since 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20 are factors of 20!. Notice that 4, 12, and 20 have 2 factors of 2, 8 has 3 factors of 2, and 16 has 4 factors of 2, while the other factors have 1 factor of 2 each). The value of pk is 2(18) = 36. Since we’ve exhausted all the possibilities, the largest value of pk is 36.

Answer: A

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Ravixxx
If p is a prime number, k is a positive integer, and \(p^{k}\) is a factor of \(20!\), what is the largest possible value of \(pk\) ?

(A) 36
(B) 48
(C) 53
(D) 64
(E) 72

2 will have
2->10
4->5
8->2
16->1
Total= 18
Hence 2*18=36
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