Mixing
Dividing every term in congruence
$begingroup$
Suppose we have the following congruence $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $10^6$).
I have read somewhere that it is possible to divide this congruence by $2^6$ to get the following:
$frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$)
Now the author of this also states that eliminating the fractions (for example multiplying by $2^6$) would result in finding $x$ modulo $5^6$ and not modulo $10^6$.
Author's words:
Take note that we do not use this technique (multiplying by $2^6$) on $frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$) so that we do not eliminate the $2^6$. Doing so would result in finding congruence for $5^6$ and not $10^6$.
So leaving this congruence in this fractional form means that we are solving it modulo $10^6$? How is that possible?
This is the original author's solution: brilliant.org
elementary-number-theory modular-arithmetic
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
$endgroup$
|
show 5 more comments
$begingroup$
Suppose we have the following congruence $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $10^6$).
I have read somewhere that it is possible to divide this congruence by $2^6$ to get the following:
$frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$)
Now the author of this also states that eliminating the fractions (for example multiplying by $2^6$) would result in finding $x$ modulo $5^6$ and not modulo $10^6$.
Author's words:
Take note that we do not use this technique (multiplying by $2^6$) on $frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$) so that we do not eliminate the $2^6$. Doing so would result in finding congruence for $5^6$ and not $10^6$.
So leaving this congruence in this fractional form means that we are solving it modulo $10^6$? How is that possible?
This is the original author's solution: brilliant.org
elementary-number-theory modular-arithmetic
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
$endgroup$
$begingroup$
The chinese remainder theorem is the key to do that. If we know the residues modulo coprime numbers $a$ and $b$, we can easily determine the residue modulo the product $ab$. I however do not understand, why the author concentrates on the fraction.
$endgroup$
– Peter
Jan 24 at 9:47
$begingroup$
@Peter I am not really interested in solving this, I am more interested in understanding what means that if we do not eliminate the denominators that we would still find a solution modulo $10^6$.
$endgroup$
– Michael Munta
Jan 24 at 9:55
1
$begingroup$
He/She apparently means the following : Since $2^6$ obviously is a divisor, the number denoted by the fraction is an integer. Suppose, we have the residue of this integer modulo $5^6$ , then we can multiply it with $2^6$ and again reduce modulo $5^6$ to get the residue of the power tower modulo $5^6$. It is however confusing that the $x$ is also divided by $2^6$. Anyway, I think that the author overcomplicates a relatively easy thing.
$endgroup$
– Peter
Jan 24 at 10:00
$begingroup$
I think he meant specifically to multiply both sides by $2^6$ to get $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $5^6$)
$endgroup$
– Michael Munta
Jan 24 at 10:47
$begingroup$
He did the division in the first place to enable usage of Euler theorem, since with modulo $10$ power it was not coprime.
$endgroup$
– Michael Munta
Jan 24 at 10:55
|
show 5 more comments
$begingroup$
Suppose we have the following congruence $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $10^6$).
I have read somewhere that it is possible to divide this congruence by $2^6$ to get the following:
$frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$)
Now the author of this also states that eliminating the fractions (for example multiplying by $2^6$) would result in finding $x$ modulo $5^6$ and not modulo $10^6$.
Author's words:
Take note that we do not use this technique (multiplying by $2^6$) on $frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$) so that we do not eliminate the $2^6$. Doing so would result in finding congruence for $5^6$ and not $10^6$.
So leaving this congruence in this fractional form means that we are solving it modulo $10^6$? How is that possible?
This is the original author's solution: brilliant.org
elementary-number-theory modular-arithmetic
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
$endgroup$
Suppose we have the following congruence $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $10^6$).
I have read somewhere that it is possible to divide this congruence by $2^6$ to get the following:
$frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$)
Now the author of this also states that eliminating the fractions (for example multiplying by $2^6$) would result in finding $x$ modulo $5^6$ and not modulo $10^6$.
Author's words:
Take note that we do not use this technique (multiplying by $2^6$) on $frac{6^{6^{6^{6^{6^{6}}}}}}{2^6} equiv frac{x}{2^6}$ (mod $5^6$) so that we do not eliminate the $2^6$. Doing so would result in finding congruence for $5^6$ and not $10^6$.
So leaving this congruence in this fractional form means that we are solving it modulo $10^6$? How is that possible?
This is the original author's solution: brilliant.org
elementary-number-theory modular-arithmetic
elementary-number-theory modular-arithmetic
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
edited Jan 24 at 11:13
Michael Munta
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
asked Jan 24 at 9:45
Michael MuntaMichael Munta
85111
85111
$begingroup$
The chinese remainder theorem is the key to do that. If we know the residues modulo coprime numbers $a$ and $b$, we can easily determine the residue modulo the product $ab$. I however do not understand, why the author concentrates on the fraction.
$endgroup$
– Peter
Jan 24 at 9:47
$begingroup$
@Peter I am not really interested in solving this, I am more interested in understanding what means that if we do not eliminate the denominators that we would still find a solution modulo $10^6$.
$endgroup$
– Michael Munta
Jan 24 at 9:55
1
$begingroup$
He/She apparently means the following : Since $2^6$ obviously is a divisor, the number denoted by the fraction is an integer. Suppose, we have the residue of this integer modulo $5^6$ , then we can multiply it with $2^6$ and again reduce modulo $5^6$ to get the residue of the power tower modulo $5^6$. It is however confusing that the $x$ is also divided by $2^6$. Anyway, I think that the author overcomplicates a relatively easy thing.
$endgroup$
– Peter
Jan 24 at 10:00
$begingroup$
I think he meant specifically to multiply both sides by $2^6$ to get $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $5^6$)
$endgroup$
– Michael Munta
Jan 24 at 10:47
$begingroup$
He did the division in the first place to enable usage of Euler theorem, since with modulo $10$ power it was not coprime.
$endgroup$
– Michael Munta
Jan 24 at 10:55
|
show 5 more comments
$begingroup$
The chinese remainder theorem is the key to do that. If we know the residues modulo coprime numbers $a$ and $b$, we can easily determine the residue modulo the product $ab$. I however do not understand, why the author concentrates on the fraction.
$endgroup$
– Peter
Jan 24 at 9:47
$begingroup$
@Peter I am not really interested in solving this, I am more interested in understanding what means that if we do not eliminate the denominators that we would still find a solution modulo $10^6$.
$endgroup$
– Michael Munta
Jan 24 at 9:55
1
$begingroup$
He/She apparently means the following : Since $2^6$ obviously is a divisor, the number denoted by the fraction is an integer. Suppose, we have the residue of this integer modulo $5^6$ , then we can multiply it with $2^6$ and again reduce modulo $5^6$ to get the residue of the power tower modulo $5^6$. It is however confusing that the $x$ is also divided by $2^6$. Anyway, I think that the author overcomplicates a relatively easy thing.
$endgroup$
– Peter
Jan 24 at 10:00
$begingroup$
I think he meant specifically to multiply both sides by $2^6$ to get $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $5^6$)
$endgroup$
– Michael Munta
Jan 24 at 10:47
$begingroup$
He did the division in the first place to enable usage of Euler theorem, since with modulo $10$ power it was not coprime.
$endgroup$
– Michael Munta
Jan 24 at 10:55
$begingroup$
The chinese remainder theorem is the key to do that. If we know the residues modulo coprime numbers $a$ and $b$, we can easily determine the residue modulo the product $ab$. I however do not understand, why the author concentrates on the fraction.
$endgroup$
– Peter
Jan 24 at 9:47
$begingroup$
The chinese remainder theorem is the key to do that. If we know the residues modulo coprime numbers $a$ and $b$, we can easily determine the residue modulo the product $ab$. I however do not understand, why the author concentrates on the fraction.
$endgroup$
– Peter
Jan 24 at 9:47
$begingroup$
@Peter I am not really interested in solving this, I am more interested in understanding what means that if we do not eliminate the denominators that we would still find a solution modulo $10^6$.
$endgroup$
– Michael Munta
Jan 24 at 9:55
$begingroup$
@Peter I am not really interested in solving this, I am more interested in understanding what means that if we do not eliminate the denominators that we would still find a solution modulo $10^6$.
$endgroup$
– Michael Munta
Jan 24 at 9:55
1
1
$begingroup$
He/She apparently means the following : Since $2^6$ obviously is a divisor, the number denoted by the fraction is an integer. Suppose, we have the residue of this integer modulo $5^6$ , then we can multiply it with $2^6$ and again reduce modulo $5^6$ to get the residue of the power tower modulo $5^6$. It is however confusing that the $x$ is also divided by $2^6$. Anyway, I think that the author overcomplicates a relatively easy thing.
$endgroup$
– Peter
Jan 24 at 10:00
$begingroup$
He/She apparently means the following : Since $2^6$ obviously is a divisor, the number denoted by the fraction is an integer. Suppose, we have the residue of this integer modulo $5^6$ , then we can multiply it with $2^6$ and again reduce modulo $5^6$ to get the residue of the power tower modulo $5^6$. It is however confusing that the $x$ is also divided by $2^6$. Anyway, I think that the author overcomplicates a relatively easy thing.
$endgroup$
– Peter
Jan 24 at 10:00
$begingroup$
I think he meant specifically to multiply both sides by $2^6$ to get $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $5^6$)
$endgroup$
– Michael Munta
Jan 24 at 10:47
$begingroup$
I think he meant specifically to multiply both sides by $2^6$ to get $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $5^6$)
$endgroup$
– Michael Munta
Jan 24 at 10:47
$begingroup$
He did the division in the first place to enable usage of Euler theorem, since with modulo $10$ power it was not coprime.
$endgroup$
– Michael Munta
Jan 24 at 10:55
$begingroup$
He did the division in the first place to enable usage of Euler theorem, since with modulo $10$ power it was not coprime.
$endgroup$
– Michael Munta
Jan 24 at 10:55
|
show 5 more comments
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$begingroup$
The chinese remainder theorem is the key to do that. If we know the residues modulo coprime numbers $a$ and $b$, we can easily determine the residue modulo the product $ab$. I however do not understand, why the author concentrates on the fraction.
$endgroup$
– Peter
Jan 24 at 9:47
$begingroup$
@Peter I am not really interested in solving this, I am more interested in understanding what means that if we do not eliminate the denominators that we would still find a solution modulo $10^6$.
$endgroup$
– Michael Munta
Jan 24 at 9:55
1
$begingroup$
He/She apparently means the following : Since $2^6$ obviously is a divisor, the number denoted by the fraction is an integer. Suppose, we have the residue of this integer modulo $5^6$ , then we can multiply it with $2^6$ and again reduce modulo $5^6$ to get the residue of the power tower modulo $5^6$. It is however confusing that the $x$ is also divided by $2^6$. Anyway, I think that the author overcomplicates a relatively easy thing.
$endgroup$
– Peter
Jan 24 at 10:00
$begingroup$
I think he meant specifically to multiply both sides by $2^6$ to get $6^{6^{6^{6^{6^{6}}}}} equiv x$ (mod $5^6$)
$endgroup$
– Michael Munta
Jan 24 at 10:47
$begingroup$
He did the division in the first place to enable usage of Euler theorem, since with modulo $10$ power it was not coprime.
$endgroup$
– Michael Munta
Jan 24 at 10:55