Back to Basics I

Let $m=n-11$ . Now $m$ must divide $n^3+83=(m+11)^3+83=m^3+33m^2+363m+1414$ . Thus $m$ must divide 1414; the largest such $m$ is $m=1414$ . Finally, $n=m+11=\boxed{1425}$

We can rewrite $n^{3}+83$ as

$n^{3}+83=(n^{2}+an+b)(n-11)+c$ $n^{3}+83=n^{3}-11n^{2}+an^{2}-11an+bn-11b+c$ $n^{3}+83=n^{3}+(a-11)n^{2}+(b-11a)n-11b+c$

From the above equation we can get the system:

$\begin{cases} a-11=0 \\ b-11a=0 \\ c-11b=83 \\ \end{cases}$

We can then solve $a,b,c$ one by one to get:

$a=11\\ b=121\\ c=1414$

Since the polynomial has a factor of $n-11$ in it, we only need to check for the largest value of $n$ such that $n-11|1414$

This means that the greatest possible value for $n$ is when $n-11=1414$ , since the quotient $\frac{1414}{n-11}=1$ . If we increase $n$ any further, the quotient will be less than one, and thus never again an integer.

So our greatest possible value for $n$ is

$n-11=1414$ $n=\boxed{1425}$

According to the remainder theorem, if ${ n }^{ 3 }+83$ is divided by $n - 11$ , then the remainder will be ${ (11)} ^ {3} + 83 = 1331 + 83 = 1414$

If $n - 11$ divides ${ n }^{ 3 }+83$ , then we must somehow force the remainder to become zero, since by definition, when one number divides another, the remainder is zero. The only way to do this is by setting the divided, $n - 11$ , equal to $1414$ . This works because when the dividend equals the remainder in a division problem, what really happens is the quotient increases by $1$ and the remainder becomes zero.

For example, $63 / 7$ can be written as $8$ with a remainder of $7$ , but that is really just $9$ with no remainder.

If we set $n - 11$ , equal to $1414$ ., then $n = 1425$

We can use factor theorem to divide $n^3+83$ by $n-11$ and then try to maximize $n$ keeping the remainder 0.

$\begin{matrix} 11 & | & 1 & 0 & 0 & 83 \\ & | & \underline { } & \underline { 1 } 1 & \underline { 121 } & \underline { 1331 } \\ & & 1 & 11 & 121 & 1414 \end{matrix}$

(I'm sorry for bad rendering)

So now we can rewrite

$\frac { { n }^{ 3 }+83 }{ n-11 } ={ n }^{ 2 }+11n+121+\frac { 1414 }{ n-11 }$

Since we want the remainder to be 0, we want to find an $n$ such that $n-11|1414$ . So $n-11$ is a factor of 1414. Since we want $n$ to be as large as possible, let's assign the value of $n-11$ as 1414. Hence

$n-11=1414\\ n=1425$

Since

$\frac{n^3+83}{n-11} = n^2+11n+121+\frac{1414}{n-11}$

the reminder vanishes if and only if the last fraction is a integer. The largest n for which it happens is when $n-11=1414 \rightarrow n = 1425$

$N^3 + 83 = (n^3 -1331) + 1414$ Since n -11 obviously divides the term in the bracket, we have to find largest value of n satisfying $n-11|1414$ Therefore max value of n is 1425.

its just too easy.by remainder theorem put n-11=0(to find the remainder) n=11 putting we get 11 cube +83 that is 1414 so it should be the remainder unless n is less than or equal to 1414 .now if n is a factor of 1414 then the remainder would be zero. the highest factor of number 1414 is 1414.so n-11=1414(for highest value) n=1425 too easy just some logic and basic maths follow me if u like this solution