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

Guest Mar 13, 2017

#2
+18356
+10

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  $${\displaystyle n!}$$

heureka  Mar 13, 2017
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#1
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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

CPhill  Mar 13, 2017
#2
+18356
+10

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  $${\displaystyle n!}$$

heureka  Mar 13, 2017
#3
+18356
+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.

heureka  Mar 13, 2017

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