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What is the largest integer $n$ such that $7^n$ divides $1000!$ ?

Guest Mar 13, 2017

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 #2
avatar+18777 
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What is the largest integer $n$ such that $7^n$ divides $1000!$ ?

 

In mathematics, Legendre's formula gives an expression for the exponent of the largest power of a prime p that divides the factorial  \({\displaystyle n!} \)

 

see: https://en.wikipedia.org/wiki/Legendre%27s_formula

 


laugh

heureka  Mar 13, 2017
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 #1
avatar+79867 
+5

I'll take a stab at this one......

 

Notice that 1000!  has   1000 / 7  =  142  numbers that are divisible by at least one  7

 

And 1000!   has    1000/49  = 20 numbers that are divisible by  7^2  [ 20 additional 7s]

 

And it has    1000/343   =  2   numbers that are divisible by 7^3  [ 2 additional 7s ]

 

So.......the largest n such that 7^n will divide 1000!   =    (142 + 20 + 2)  = 164

 

 

 

cool cool cool

CPhill  Mar 13, 2017
 #2
avatar+18777 
+10
Best Answer

What is the largest integer $n$ such that $7^n$ divides $1000!$ ?

 

In mathematics, Legendre's formula gives an expression for the exponent of the largest power of a prime p that divides the factorial  \({\displaystyle n!} \)

 

see: https://en.wikipedia.org/wiki/Legendre%27s_formula

 


laugh

heureka  Mar 13, 2017
 #3
avatar+18777 
+10

What is the largest integer $n$ such that $7^n$ divides $1000!$ ?

 

1. We calculate 1000 in base 7:

\(\begin{array}{rcll} 1000_{10} &=& 2626_7 \\ \end{array} \)

 

2. The sum of the digits in base 7 is: \(\begin{array}{rcll} 2+6+2+6 = 16 \end{array} \)

 

3. \(n =\ ?\)

\(\begin{array}{|rcll|} \hline n &=&\frac{1000-(\text{sum of the standard base-p digits of 1000})}{7-1} \\ n &=&\frac{1000-16}{6} \\ n &=&164 \\ \hline \end{array} \)

 

\(7^{164} \) will be divide 1000! and \(7^{164} \) is one prime factor of 1000! for prime = 7.\(\)

 

laugh

heureka  Mar 13, 2017

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