What is the value of n for greatest integer function to equal 2012?

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SUMMARY

The value of n for which the greatest integer function equals 2012 is determined by the equation $\displaystyle [\frac{n}{1!}]+[\frac{n}{2!}]+[\frac{n}{3!}]+...+[\frac{n}{10!}] = 2012$. The function f(n) is non-decreasing, with f(1000) < 2012 < f(2000). A binary search or bisection method can efficiently find the exact value of n within 10 iterations, given that the difference between 2000 and 1000 is less than 210.

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Calculate Natural no. $n$ for which $\displaystyle [\frac{n}{1!}]+[\frac{n}{2!}]+[\frac{n}{3!}]+...+[\frac{n}{10!}] = 2012$

where $[x] = $ Greatest Integer function
 
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The function $f(n)=\left\lfloor\dfrac{n}{1!}\right\rfloor+\dots+\left\lfloor\dfrac{n}{10!}\right\rfloor$ is non-decreasing and f(1000) < 2012 < f(2000). Since $2000-1000<2^{10}$, you can find n for which f(n) = 2012 in 10 iterations using binary search, or bisection method.
 

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