# 2006 iTest Problems/Problem 28

## Problem

The largest prime factor of $999999999999$ is greater than $2006$. Determine the remainder obtained when this prime factor is divided by $2006$.

## Solution

Note that $999999999999 = 10^{12} - 1$. This expression can be factored with difference of squares and sum/difference of cubes. \begin{align*} 10^{12} - 1 &= (10^6 + 1)(10^6 - 1) \\ &= (10^2 + 1)(10^4 - 10^2 + 1)(10^3 - 1)(10^3 + 1) \end{align*} Note that since $10^3 - 1, 10^3 + 1, 10^2 + 1$ are all less than $2006$, none of them are the wanted factors. The only option left is $10^4 - 10^2 + 1 = 9901$. By doing a prime check (or noting that if $9901$ has factors larger than 5, then the largest prime factor of the original number can not be greater than $2006$), we confirm that $9901$ is the largest prime factor of $999999999999$. The remainder when $9901$ is divided by $2006$ is $\boxed{1877}$.

## See Also

 2006 iTest (Problems) Preceded by:Problem 27 Followed by:Problem 29 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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