Question

for what max value of n be the expression 10200!/504^n will be an integer?
help!!

Answer 1

10200!=10200*10199*10198*.....*2*1 right?

Answer 2

out of these you have to find out how many are divisible by 10200

Answer 3

sorry, divisible by 504*

Answer 4

so [10200/504] (where [ ] denotes greatest integer function) =20
so there are 20 numbers divisible by 504 between 1 and 10200

Answer 5

now, as you divide the numbers by 504, you will get quotient as 1,2,3....20
so you have to find out again, how many times you can divide 20! by 504
now 504=2^3 * 3^2 *7
between 1 and 20, there are two numbers divisible by 2^3, 8 and 16, two numbers divisible by 3^2, 9 and 18 and two numbers divisible by 7, 7 and 14
so 20! is divisible by 504^2

so maximum value of n can be 22

Answer 6

But answer given is 1698. How? (

Answer 7

are you really writing? @nikvist

Answer 8

\[504=2^3\cdot 3^2\cdot 7\]\[10200!=2^a\cdot 3^b\cdot 7^c\cdot X\quad,\quad X\in\mathbb{N}\]\[a=\sum\limits_{i=1}^{13}\left[\frac{10200}{2^i}\right]=10192\]\[b=\sum\limits_{i=1}^{8}\left[\frac{10200}{3^i}\right]=5094\]\[c=\sum\limits_{i=1}^{4}\left[\frac{10200}{7^i}\right]=1698\]\[10200!=2^{10192}\cdot 3^{5094}\cdot 7^{1698}\cdot X\]\[\frac{10200!}{504^n}=\frac{2^{10192}\cdot 3^{5094}\cdot 7^{1698}\cdot X}{2^{3n}\cdot 3^{2n}\cdot 7^n}\quad\Rightarrow\quad n=1698\]

Answer 9

we have to start with prime factors of 504 , which are 3^2*7*8(2^3)
Then we should go on looking for number of 2^3, 3^2 and 7's
The lowest of them will be the answer.

Answer 10

\[\left[ 10200/2 \right]+\left[ 10200/4 \right]+\left[ 102000/8 \right]+\left[ 10200/16 \right]+\left[ 10200/32 \right]+....+\left[ 10200/8192 \right]\], this gives number of 2=10192---> no. of 2^3=3397

\[\left[ 10200/3 \right]+\left[ 10200/9 \right] +....\left[ 10200/6561 \right]\]
this gives no. of 3's=5094----> no. of 3^3= 2547

same should be done with number of 7's and the lowest of these three will give value of n, which is of number of 7's=1698