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

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To find the natural number n for which the greatest integer function equals 2012, the equation f(n) = [n/1!] + [n/2!] + ... + [n/10!] must be solved. The function f(n) is non-decreasing, and it has been established that f(1000) is less than 2012 while f(2000) exceeds it. Given that the difference between 2000 and 1000 is less than 2^10, a binary search or bisection method can efficiently determine the value of n in about 10 iterations. This approach allows for precise identification of n that satisfies the equation. The solution process highlights the effectiveness of numerical methods in solving such problems.
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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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