function

**Day** · March 16th 2009, 03:03 AM

$n\in<1,1004>$
Find how many n exist when k = 7

$2n! \over n!n!2^k$

For :

$2n! \over n!n!2^{k+1}$
Result isn't integer value.

**HallsofIvy** · March 16th 2009, 09:41 AM

Originally Posted by Day

$n\in<1,1004>$
Find how many n exist when k = 7

$2n! \over n!n!2^k$

For :

$2n! \over n!n!2^{k+1}$
Result isn't integer value.

I not clear what your question is. Do you mean that n must be an integer such that $1\le n\le 1004$ such that $\frac{2n!}{n!n!2^7}$ is an integer but $\frac{2n!}{n!n!2^{8}$ is not? And is that really "2n!" and not "(2n)!"?
If so it is asking how many "n"s are such that [tex]\frac{2n!}{n!n!} has 7 factors of 2 but not 8.

**Day** · March 16th 2009, 11:04 AM

n is integer.
how many results have $2n! \over n!n!2^k$ . How many is n in this function? This is question.

All is up.

$<br /> <br /> 2n! \over n!n!2^k$ = integer

$2n! \over n!n!2^{k+1}$ isn't integer

Thread: function

LinkBack

Thread Tools

Display

function

Similar Math Help Forum Discussions

Word problem: Cost function, Revenue Function, Profit function.

Showing concavity of a function defined in terms of concave utility function

[SOLVED] Expectancy (mean) of Probability density function composed of delta/rect function

rational function, polynomial function or some other function

Approximation of a Riemann integrable function w continuous function in the L2 metric

Search Tags