2000 PMWC Problems/Problem I2

Problem

As far as we know, the greatest prime number is $2^{6972593}-1$ . What is the remainder when $2^{6972593}-1$ is divided by $5$ ?

Solution

Since the question only asks for the remainder when this monstrous number is divided by 5, we can use some tricks to reduce this number.

Specifically, since $2^4 = 16$ , and 16 has a remainder of 1 when divided by 5, all of the $2^4$ s within $2^{6972593}$ can be divided out. Another way of explaining that step is that if you split $2^{6972593}$ into as many factors of 16 as you could and had whatever was left on the side, you could remove all of the $16$ s because multiplying a number by $16$ is multiplying it by $(3 * 5) + 1$ , and so will not change its remainder when divided by $5$ . Applying this logic to $2^{6972593}$ , we find that it simplifies to $2^1$ , leaving $2^1 - 1$ , which is clearly just $1$ .

Thus the remainder is $1$ .

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2000 PMWC Problems/Problem I2

Problem

Solution

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