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Difference between revisions of "2000 PMWC Problems/Problem I2"

(Created page with "==Problem== As far as we know, the greatest prime number is <math>2^{6972593}-1</math>. What is the remainder when <math>2^{6972593}-1</math> is divided by <math>5</math>? ==Sol...")
 
(→‎Solution)
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==Solution==
 
==Solution==
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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.
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Specifically, since <math>2^4 = 16</math>, and 16 has a remainder of 1 when divided by 5, all of the <math>2^4</math>s within <math>2^(6972593)</math> can be divided out. Another way of explaining that step is that if you split <math>2^(6972593)</math> into as many factors of 16 as you could and had whatever was left on the side, you could remove all of the <math>16</math>s because multiplying a number by <math>16</math> is multiplying it by <math>(3 * 5) + 1</math>, and so will not change its remainder when divided by <math>5</math>. Applying this logic to <math>2^(6972593)</math>, we find that it simplifies to <math>2^1</math>, leaving <math>2^1 - 1</math>, which is clearly just <math>1</math>.
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Thus the remainder is <math>1</math>.
  
 
==See Also==
 
==See Also==

Revision as of 11:50, 23 December 2019

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 $1$.

See Also

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