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Bunuel
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Kinshook
Bunuel
If n is a positive integer and \(n!\) is a multiple of \(7^{20}\) but not \(3^{20}\), then how many value can n take ?

A. 21
B. 20
C. 16
D. 11
E. 75


Are You Up For the Challenge: 700 Level Questions

Asked: If n is a positive integer and \(n!\) is a multiple of \(7^{20}\) but not \(3^{20}\), then how many value can n take ?

Highest power of 7 in 126! = 18 + 2 = 20
Highest power of 3 in 126! = 42 + 14 + 4 + 1 = 61

Hi Bunuel
Please check whether question is correct. Since there may not be any value of n such that \(n!\) is a multiple of \(7^{20}\) but not \(3^{20}\)

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Fixed. Thank you.
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So I’m getting the concept here but is there a quicker way to find N than just trial and error? I had to start with 20 and just try numbers until it fit. I’m trying to determine if there’s a more efficient way to solve this.

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hi

plz explain this
Highest power of 2 in 23! = 11 + 5 + 2 + 1 = 19 ????- how you came to this equation.

can you please share concept of this too ???
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23/2^1+23/2^2+23/2^3+23/2^4=19

So the formula is basically to figure out how many of a given factor are in n!, divide that number by that factor to 1,2,3, etc. power until it no longer will go into that number( ignore the remainder). Then find the sum of those numbers.

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hi

plz explain this
Highest power of 2 in 23! = 11 + 5 + 2 + 1 = 19 ????- how you came to this equation.

can you please share concept of this too ???
23!= 23×22×21×---------3×2×1
All the multiple of 2 will give atleast one 2 so 11
All the multiple of 4 will give you extra one 2 so 5
All the multiple of 8 will give you extra one 2 so 2
All the multiple of 16 will give you extra one 2 so 1
11+5+2+1 =19

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Alextus93
23/2^1+23/2^2+23/2^3+23/2^4=19

So the formula is basically to figure out how many of a given factor are in n!, divide that number by that factor to 1,2,3, etc. power until it no longer will go into that number( ignore the remainder). Then find the sum of those numbers.

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My suggestion is to understand how this is working because you can't do it like this when you have to find out number of 10's in 23!.
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Kinshook
Bunuel
If n is a positive integer and \(n!\) is a multiple of \(2^{20}\) but not \(3^{20}\), then how many value can n take ?

A. 21
B. 20
C. 16
D. 11
E. 10


Are You Up For the Challenge: 700 Level Questions

Asked: If n is a positive integer and \(n!\) is a multiple of \(2^{20}\) but not \(3^{20}\), then how many value can n take ?

Highest power of 2 in 23! = 11 + 5 + 2 + 1 = 19
Highest power of 2 in 24! = 12 + 6 + 3 + 1 = 22

Minimum value of n = 24

Highest power of 3 in 44! = 14 + 4 + 1 = 19
Highest power of 3 in 45! = 15 + 5 + 1 = 21

Maximum value of n = 44

Number of values of n feasible = 44-24 + 1 = 21

IMO A

hey Kinshook
how did you come up with highlighted numbers and why these numbers
thanks :)
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dave13
Kinshook
Bunuel
If n is a positive integer and \(n!\) is a multiple of \(2^{20}\) but not \(3^{20}\), then how many value can n take ?

A. 21
B. 20
C. 16
D. 11
E. 10


Are You Up For the Challenge: 700 Level Questions

Asked: If n is a positive integer and \(n!\) is a multiple of \(2^{20}\) but not \(3^{20}\), then how many value can n take ?

Highest power of 2 in 23! = 11 + 5 + 2 + 1 = 19
Highest power of 2 in 24! = 12 + 6 + 3 + 1 = 22

Minimum value of n = 24

Highest power of 3 in 44! = 14 + 4 + 1 = 19
Highest power of 3 in 45! = 15 + 5 + 1 = 21

Maximum value of n = 44

Number of values of n feasible = 44-24 + 1 = 21

IMO A

hey Kinshook
how did you come up with highlighted numbers and why these numbers
thanks :)

Hi dave13
It is done by hit and try method
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Kinshook
Bunuel
If n is a positive integer and \(n!\) is a multiple of \(2^{20}\) but not \(3^{20}\), then how many value can n take ?

A. 21
B. 20
C. 16
D. 11
E. 10


Are You Up For the Challenge: 700 Level Questions

Asked: If n is a positive integer and \(n!\) is a multiple of \(2^{20}\) but not \(3^{20}\), then how many value can n take ?

Highest power of 2 in 23! = 11 + 5 + 2 + 1 = 19
Highest power of 2 in 24! = 12 + 6 + 3 + 1 = 22

Minimum value of n = 24

Highest power of 3 in 44! = 14 + 4 + 1 = 19
Highest power of 3 in 45! = 15 + 5 + 1 = 21

Maximum value of n = 44

Number of values of n feasible = 44-24 + 1 = 21

IMO A

Bunuel can you please write the concept behind this ?
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