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

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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.  
 
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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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>.
  
 
Thus the remainder is <math>1</math>.
 
Thus the remainder is <math>1</math>.
  
 
==See Also==
 
==See Also==

Revision as of 11:51, 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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