Three 3's

FIrst. divide both of them..

$\frac{ N^{3} - 3 }{ N -3 }$

which is equal to

$N^{2}$ + 3N + 9 $\frac{24}{ N - 3 }$

then for the quotient to be an integer, the remainder should be transformed to an integer..

thus, $\frac{24}{ N - 3 }$ => 24 = N - 3 => N = 27

We know that $N^3 - 3^3 = N^3 - 27 = (N-3)(N^2 +3N +3^2)$ .

So $N-3 | N^3 - 27$ and we are told that $N-3 | N^3 - 3$ so, $N-3$ divides any liner combination of $N^3 - 27$ and $N^3 - 3$ , in particular $N - 3 | 24$ .

Now, obviously the largest number that divides 24 is 24 itself, and that means that $N = 27$ .

Note that $N-3$ evenly divides $(N-3)(N^2+3x-9)=N^3-27$ .

Hence, $N-3$ must evenly divide $(N^3-3)-(N^3-27)=24$ .

The largest number that divides $24$ is $24$ , so $N-3$ is at most $24$ . Hence, $N$ is at most $27$ .

If we have $N=27$ , then $N-3=27-3=24$ and $N^3-3=27^3-3=19683-3=19680$ . $\frac{19680}{24}=820$ , which is an integer.

Hence, the maximum possible value of $N$ such that $N-3$ evenly divides $N^3-3$ is $\boxed{27}$ .

If $N-3$ evenly divides $N^{3}-3$ , and $N$ is an integer, then

$(N-3) * c =N^{3}-3$ for some integer $c$ .

If we divide $N^{3}-3$ by $N-3$ using long division, we get

$N^{2} + 3N + 9 + \frac{24}{N-3}$

Clearly $N^{2} + 3N + 9$ is an integer since $N$ is an integer, and so we only need to worry about $\frac{24}{N-3}$ being an integer, which is true when $24$ is divisible by $N-3$ . The largest factor of $24$ is $24$ , so $N-3=24$ .

Hence $\boxed{N=27}$

Once we use synthetic division to compute $\frac{N^3-3}{N-3}$ , we get $N^3+3N+9+\frac{24}{N-3}$

Here we can see that $N$ is maximized when $\frac{24}{N-3}$ is minimized(for integers) So we let $\frac{24}{N-3}=1$

Thus we get the largest integer $N$ for which $N-3$ evenly divides $N^3-3$ as $\boxed{27}$

Let us say that some $k$ satisfies the conditions. By the Euclidean Algorithm:

$(k-3)(k^2 + 3k + 9) + 24 = k^3 - 3$

Therefore, if $(k-3)|k^3 - 3, (k-3)|24$ . The largest divisor of $24$ is $24$ . Therefore, our required number is $27$ .

$N^{3}$ - $3$ = $(N-3) \times (N^{2}+3N +9)$ + $24$

Clearly when N=27 then RHS divides LHS with highest quotient

We have to find $N$ such that $\frac{N^3-3}{N-3} =k$ , where $k$ is an integer. Now $\frac{N^3-3}{N-3} = \frac{N^3-27+24}{N-3} = \frac{N^3-3^3+24}{N-3} = \frac{N^3-3^3}{N-3}+\frac{24}{N-3}$ . And $N^3-3^3=(N-3)(N^2+3N+9)$ . So the equation reduces to $N^2+3N+9+\frac{24}{N-3}$ is integer if $\frac{24}{N-3}$ is an integer. So maximum value of N turns out to be 27 because if $N \geq 27$ the fraction becomes proper and hence is not integer for any value of N

$N-3 | N^3-3 <=> N^3 -3 = 0$ mod $N-3$

$<=> N^3 = 3$ mod $N-3$

Substitute N-3 with M to get: $(M+3)^3 = 3$ mod $M$

<=> $M^3 + 9M^2 + 27M + 27 = 3$ mod $M$

<=> $27 = 3$ mod $M$

The maximum value of $M$ for which this holds is $M=24$ , which gives $N=27$ as solution to the problem.

The problem gives us the problem N^{3}-3/N-3. We can simplify this using synthetic division. [example link] (http://www.purplemath.com/modules/synthdiv.htm)

Through synthetic division we can simplify this even more and we get that the quotient is N^{2}+3N+9+24/N-3. Every part of this expression will give an integer value except for the 24/N-3. For it to be the maximum integer value, we set 24/N-3 = 1 because if it equals less than one it will not be an integer value. We get N=27, which is the answer.

Given an integer n and positive integer d , there exist unique integers q and r such that, $n = q \cdot d + r$ and $0 \leq r < d$ . Where q is the quotient, d is the divisor, and r is the remainder.

So we can think of the problem as $(N^3 -3) = q \cdot (N - 3) \quad \Rightarrow \quad \frac {(N^3-3)}{(N - 3)} = q \quad$ where $\quad q \in \mathbb{Z} \\$

This means we have to find a value for N such that $N-3$ is a factor of $N^3-3$ . However, there is more than one value that satisfies that condition, and what the largest value is, is not clear from the expression.

$\mbox{}$

Our next step is to do some polynomial long division.

\begin{aligned} \renewcommand\arraystretch{2} \begin{array}{r@{\hskip\arraycolsep}c@{\hskip\arraycolsep}l*2r} &\underline{\mspace{20mu}N^2 + 3N + 9} \mspace{30mu} \\ N - 3& \Big )\mspace{10mu} N^3 - 3 \mspace{80mu} \\ & \underline{-(N^3 - 3N^2)} \mspace{50mu} \\ & 3N^2 - 3 \mspace{73mu} \\ & \underline{-(3N^2 - 9N)} \mspace{50mu} \\ & 9N - 3 \mspace{80mu} \\ & \underline{-(9N-27)} \mspace{63mu} \\ & 24 \mspace{70mu} \\ \end{array} \end{aligned} \\ \mbox{}

Now our expression looks like:

$N^2 + 3N + 9 + {24 \over N - 3} = q \\$

Let $\quad k = N^2 +3N + 9 \quad$ where $\quad k \in \mathbb{Z} \quad$

We can do this since integers are closed under multiplication and addition.

$\mbox{}$

Now we have the expression:

$k + {24 \over N -3 } = q \quad \\$

Both k and q are integers. All that is left is to find the largest integer N that will still produce an integer from $24 \over N -3$ .

To reduce a fraction to an integer, the denominator has to be a factor of the numerator. The largest factor of the numerator is 24 . So to solve the problem, we have to find N such that N-3 is equal to 24 .

$\\ \mspace{1mu} \\ N - 3 = 24 \quad \Rightarrow \quad \therefore N = 27$