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 $\boxed{1877}$ .

2006 iTest (Problems, Answer Key)
Preceded by: Problem 27	Followed by: Problem 29
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2006 iTest Problems/Problem 28

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Solution

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