Solve $left(frac mnright)^k=0.overline{x_1x_2…x_9}$ (no computers!)
I got this problem from my son and he picked it from some local math competition. It's fairly simple:
For numbers $k,m,nin N$ ($kge2)$ we know the following:
$$left(frac mnright)^k=0.overline{x_1x_2...x_9}tag{1}$$
On the right side we have an infintely repeating sequence of exactly nine digits $x_iin{0,1,2,dots9}$ ($i=1dots9)$ and these digits are not necessarily distinct. Find all possible values of expression (1).
The solution seems to be simple: we can replace the repeating sequence of digits with some number $a$:
$$a=overline{x_1x_2...x_9}$$
Relation (1) now becomes:
$$left(frac mnright)^k=frac{a}{10^9-1}$$
$$m^k(10^9-1)=an^k$$
If we assume that $m$ and $n$ are coprime then:
$$n^kmid10^9-1$$
If we are able to find prime factors of $(10^9-1)$ quickly, we are done. True, we can find a few prime factors fairly easily:
$$10^9-1=(10^3)^3-1=(10^3-1)(10^6+10^3+1)=9times111times1001001 \
=3^2times3times 37times3times333667=3^4times37times333667$$
However, the last number (333667) is a tough nut to crack. We can proceed only if we know its factors.
With some help from the computer you can easily find out that 333667 is a prime and the rest of the solution is fairly straightforward.
However, suppose that you are in a real competition - you don't have a computer or a pocket calculator. Factoring 333667 by hand is a time consuming activity and you have other problems to solve as well.
Is there a better approach?
Happy holidays :)
elementary-number-theory
|
show 5 more comments
I got this problem from my son and he picked it from some local math competition. It's fairly simple:
For numbers $k,m,nin N$ ($kge2)$ we know the following:
$$left(frac mnright)^k=0.overline{x_1x_2...x_9}tag{1}$$
On the right side we have an infintely repeating sequence of exactly nine digits $x_iin{0,1,2,dots9}$ ($i=1dots9)$ and these digits are not necessarily distinct. Find all possible values of expression (1).
The solution seems to be simple: we can replace the repeating sequence of digits with some number $a$:
$$a=overline{x_1x_2...x_9}$$
Relation (1) now becomes:
$$left(frac mnright)^k=frac{a}{10^9-1}$$
$$m^k(10^9-1)=an^k$$
If we assume that $m$ and $n$ are coprime then:
$$n^kmid10^9-1$$
If we are able to find prime factors of $(10^9-1)$ quickly, we are done. True, we can find a few prime factors fairly easily:
$$10^9-1=(10^3)^3-1=(10^3-1)(10^6+10^3+1)=9times111times1001001 \
=3^2times3times 37times3times333667=3^4times37times333667$$
However, the last number (333667) is a tough nut to crack. We can proceed only if we know its factors.
With some help from the computer you can easily find out that 333667 is a prime and the rest of the solution is fairly straightforward.
However, suppose that you are in a real competition - you don't have a computer or a pocket calculator. Factoring 333667 by hand is a time consuming activity and you have other problems to solve as well.
Is there a better approach?
Happy holidays :)
elementary-number-theory
Well, you need to somehow know at least that $333667$ is squarefree - otherwise, for its factor $p^2$, $1/p^2$ would have purely periodic expansion of length $9$. The problem at this point is actually equivalent to checking if $333667$ is squarefree.
– Wojowu
2 days ago
@Wojowu Yes, that's the key point. If we know that 333667 is squarefree, we are done. But I have no idea how to prove it quickly.
– Oldboy
2 days ago
Testing squarefree-ness is a known problem in computational complexity with no known polynomial-time algorithm. This doesn't exactly tell us there is no way to solve this problem easily, since $333667$ might have some magical properties which make it simpler, but I am being a little skeptical here...
– Wojowu
2 days ago
@JohnDouma A digit can be zero, I have clarified this.
– Oldboy
2 days ago
There was no need to clarify. I just misunderstood. That's why I deleted the comment.
– John Douma
2 days ago
|
show 5 more comments
I got this problem from my son and he picked it from some local math competition. It's fairly simple:
For numbers $k,m,nin N$ ($kge2)$ we know the following:
$$left(frac mnright)^k=0.overline{x_1x_2...x_9}tag{1}$$
On the right side we have an infintely repeating sequence of exactly nine digits $x_iin{0,1,2,dots9}$ ($i=1dots9)$ and these digits are not necessarily distinct. Find all possible values of expression (1).
The solution seems to be simple: we can replace the repeating sequence of digits with some number $a$:
$$a=overline{x_1x_2...x_9}$$
Relation (1) now becomes:
$$left(frac mnright)^k=frac{a}{10^9-1}$$
$$m^k(10^9-1)=an^k$$
If we assume that $m$ and $n$ are coprime then:
$$n^kmid10^9-1$$
If we are able to find prime factors of $(10^9-1)$ quickly, we are done. True, we can find a few prime factors fairly easily:
$$10^9-1=(10^3)^3-1=(10^3-1)(10^6+10^3+1)=9times111times1001001 \
=3^2times3times 37times3times333667=3^4times37times333667$$
However, the last number (333667) is a tough nut to crack. We can proceed only if we know its factors.
With some help from the computer you can easily find out that 333667 is a prime and the rest of the solution is fairly straightforward.
However, suppose that you are in a real competition - you don't have a computer or a pocket calculator. Factoring 333667 by hand is a time consuming activity and you have other problems to solve as well.
Is there a better approach?
Happy holidays :)
elementary-number-theory
I got this problem from my son and he picked it from some local math competition. It's fairly simple:
For numbers $k,m,nin N$ ($kge2)$ we know the following:
$$left(frac mnright)^k=0.overline{x_1x_2...x_9}tag{1}$$
On the right side we have an infintely repeating sequence of exactly nine digits $x_iin{0,1,2,dots9}$ ($i=1dots9)$ and these digits are not necessarily distinct. Find all possible values of expression (1).
The solution seems to be simple: we can replace the repeating sequence of digits with some number $a$:
$$a=overline{x_1x_2...x_9}$$
Relation (1) now becomes:
$$left(frac mnright)^k=frac{a}{10^9-1}$$
$$m^k(10^9-1)=an^k$$
If we assume that $m$ and $n$ are coprime then:
$$n^kmid10^9-1$$
If we are able to find prime factors of $(10^9-1)$ quickly, we are done. True, we can find a few prime factors fairly easily:
$$10^9-1=(10^3)^3-1=(10^3-1)(10^6+10^3+1)=9times111times1001001 \
=3^2times3times 37times3times333667=3^4times37times333667$$
However, the last number (333667) is a tough nut to crack. We can proceed only if we know its factors.
With some help from the computer you can easily find out that 333667 is a prime and the rest of the solution is fairly straightforward.
However, suppose that you are in a real competition - you don't have a computer or a pocket calculator. Factoring 333667 by hand is a time consuming activity and you have other problems to solve as well.
Is there a better approach?
Happy holidays :)
elementary-number-theory
elementary-number-theory
edited 2 days ago
asked 2 days ago
Oldboy
7,1191832
7,1191832
Well, you need to somehow know at least that $333667$ is squarefree - otherwise, for its factor $p^2$, $1/p^2$ would have purely periodic expansion of length $9$. The problem at this point is actually equivalent to checking if $333667$ is squarefree.
– Wojowu
2 days ago
@Wojowu Yes, that's the key point. If we know that 333667 is squarefree, we are done. But I have no idea how to prove it quickly.
– Oldboy
2 days ago
Testing squarefree-ness is a known problem in computational complexity with no known polynomial-time algorithm. This doesn't exactly tell us there is no way to solve this problem easily, since $333667$ might have some magical properties which make it simpler, but I am being a little skeptical here...
– Wojowu
2 days ago
@JohnDouma A digit can be zero, I have clarified this.
– Oldboy
2 days ago
There was no need to clarify. I just misunderstood. That's why I deleted the comment.
– John Douma
2 days ago
|
show 5 more comments
Well, you need to somehow know at least that $333667$ is squarefree - otherwise, for its factor $p^2$, $1/p^2$ would have purely periodic expansion of length $9$. The problem at this point is actually equivalent to checking if $333667$ is squarefree.
– Wojowu
2 days ago
@Wojowu Yes, that's the key point. If we know that 333667 is squarefree, we are done. But I have no idea how to prove it quickly.
– Oldboy
2 days ago
Testing squarefree-ness is a known problem in computational complexity with no known polynomial-time algorithm. This doesn't exactly tell us there is no way to solve this problem easily, since $333667$ might have some magical properties which make it simpler, but I am being a little skeptical here...
– Wojowu
2 days ago
@JohnDouma A digit can be zero, I have clarified this.
– Oldboy
2 days ago
There was no need to clarify. I just misunderstood. That's why I deleted the comment.
– John Douma
2 days ago
Well, you need to somehow know at least that $333667$ is squarefree - otherwise, for its factor $p^2$, $1/p^2$ would have purely periodic expansion of length $9$. The problem at this point is actually equivalent to checking if $333667$ is squarefree.
– Wojowu
2 days ago
Well, you need to somehow know at least that $333667$ is squarefree - otherwise, for its factor $p^2$, $1/p^2$ would have purely periodic expansion of length $9$. The problem at this point is actually equivalent to checking if $333667$ is squarefree.
– Wojowu
2 days ago
@Wojowu Yes, that's the key point. If we know that 333667 is squarefree, we are done. But I have no idea how to prove it quickly.
– Oldboy
2 days ago
@Wojowu Yes, that's the key point. If we know that 333667 is squarefree, we are done. But I have no idea how to prove it quickly.
– Oldboy
2 days ago
Testing squarefree-ness is a known problem in computational complexity with no known polynomial-time algorithm. This doesn't exactly tell us there is no way to solve this problem easily, since $333667$ might have some magical properties which make it simpler, but I am being a little skeptical here...
– Wojowu
2 days ago
Testing squarefree-ness is a known problem in computational complexity with no known polynomial-time algorithm. This doesn't exactly tell us there is no way to solve this problem easily, since $333667$ might have some magical properties which make it simpler, but I am being a little skeptical here...
– Wojowu
2 days ago
@JohnDouma A digit can be zero, I have clarified this.
– Oldboy
2 days ago
@JohnDouma A digit can be zero, I have clarified this.
– Oldboy
2 days ago
There was no need to clarify. I just misunderstood. That's why I deleted the comment.
– John Douma
2 days ago
There was no need to clarify. I just misunderstood. That's why I deleted the comment.
– John Douma
2 days ago
|
show 5 more comments
2 Answers
2
active
oldest
votes
You want to find whether there exists a prime $p$ such that $p^2mid n$, where $n=333667$. Suppose that such $p$ exists. Then, we know that
$$pleq sqrt{n}<578.$$
It is easily seen that $p>11$, so
$$13leq pleq 577.$$
However, if $p>67$, then $$frac{n}{p^2}leq frac{n}{71^2}<66.$$
Thus, $n$ must have a prime divisor $q<66$ such that $qmid n$ (noting that $n$ is not a perfect square per TonyK's comment under Ross Millikan's answer). Therefore, $n$ must have a prime divisor that is inclusively between $13$ and $67$: $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, and $67$. We can easily rule out $37$ as $n-1$ is divisible by $111=3cdot 37$. This leaves $13$ primes to deal with.
There will be some cumbersome computations. It is not too difficult (but a little bit tedious) to find the square root or the cubic root of $n$ by hand (the cubic root of $n$ is used to obtain $67$ when I say that if $p>67$ then there exists a prime divisor $q<66$). And then you have to divide $n$ by $13$ primes. This is doable, but not very nice.
A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done.
– Ross Millikan
2 days ago
add a comment |
To prove $333667$ is squarefree, you just have to show it has no prime factor smaller than $sqrt[3]{333667} approx 69$ The small ones can be done by divisibility rules, say $2,3,5,7,11$. That leaves $14$ to try, which is not too bad. You might even know the variants on the classic test for $7$ that you double the last digit and subtract it from the rest of the number. This is based on the fact that $21$ is a multiple of $7$. For $13$ you can note that $39$ is a multiple of $13$ and multiply the last digit by $4$ and add to the rest of the number. For $17$ you can use $51$. That gets you the next few. It would be a few minutes, but if you are quick with arithmetic much less than $10$.
4
Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.)
– TonyK
2 days ago
add a comment |
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2 Answers
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2 Answers
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You want to find whether there exists a prime $p$ such that $p^2mid n$, where $n=333667$. Suppose that such $p$ exists. Then, we know that
$$pleq sqrt{n}<578.$$
It is easily seen that $p>11$, so
$$13leq pleq 577.$$
However, if $p>67$, then $$frac{n}{p^2}leq frac{n}{71^2}<66.$$
Thus, $n$ must have a prime divisor $q<66$ such that $qmid n$ (noting that $n$ is not a perfect square per TonyK's comment under Ross Millikan's answer). Therefore, $n$ must have a prime divisor that is inclusively between $13$ and $67$: $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, and $67$. We can easily rule out $37$ as $n-1$ is divisible by $111=3cdot 37$. This leaves $13$ primes to deal with.
There will be some cumbersome computations. It is not too difficult (but a little bit tedious) to find the square root or the cubic root of $n$ by hand (the cubic root of $n$ is used to obtain $67$ when I say that if $p>67$ then there exists a prime divisor $q<66$). And then you have to divide $n$ by $13$ primes. This is doable, but not very nice.
A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done.
– Ross Millikan
2 days ago
add a comment |
You want to find whether there exists a prime $p$ such that $p^2mid n$, where $n=333667$. Suppose that such $p$ exists. Then, we know that
$$pleq sqrt{n}<578.$$
It is easily seen that $p>11$, so
$$13leq pleq 577.$$
However, if $p>67$, then $$frac{n}{p^2}leq frac{n}{71^2}<66.$$
Thus, $n$ must have a prime divisor $q<66$ such that $qmid n$ (noting that $n$ is not a perfect square per TonyK's comment under Ross Millikan's answer). Therefore, $n$ must have a prime divisor that is inclusively between $13$ and $67$: $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, and $67$. We can easily rule out $37$ as $n-1$ is divisible by $111=3cdot 37$. This leaves $13$ primes to deal with.
There will be some cumbersome computations. It is not too difficult (but a little bit tedious) to find the square root or the cubic root of $n$ by hand (the cubic root of $n$ is used to obtain $67$ when I say that if $p>67$ then there exists a prime divisor $q<66$). And then you have to divide $n$ by $13$ primes. This is doable, but not very nice.
A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done.
– Ross Millikan
2 days ago
add a comment |
You want to find whether there exists a prime $p$ such that $p^2mid n$, where $n=333667$. Suppose that such $p$ exists. Then, we know that
$$pleq sqrt{n}<578.$$
It is easily seen that $p>11$, so
$$13leq pleq 577.$$
However, if $p>67$, then $$frac{n}{p^2}leq frac{n}{71^2}<66.$$
Thus, $n$ must have a prime divisor $q<66$ such that $qmid n$ (noting that $n$ is not a perfect square per TonyK's comment under Ross Millikan's answer). Therefore, $n$ must have a prime divisor that is inclusively between $13$ and $67$: $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, and $67$. We can easily rule out $37$ as $n-1$ is divisible by $111=3cdot 37$. This leaves $13$ primes to deal with.
There will be some cumbersome computations. It is not too difficult (but a little bit tedious) to find the square root or the cubic root of $n$ by hand (the cubic root of $n$ is used to obtain $67$ when I say that if $p>67$ then there exists a prime divisor $q<66$). And then you have to divide $n$ by $13$ primes. This is doable, but not very nice.
You want to find whether there exists a prime $p$ such that $p^2mid n$, where $n=333667$. Suppose that such $p$ exists. Then, we know that
$$pleq sqrt{n}<578.$$
It is easily seen that $p>11$, so
$$13leq pleq 577.$$
However, if $p>67$, then $$frac{n}{p^2}leq frac{n}{71^2}<66.$$
Thus, $n$ must have a prime divisor $q<66$ such that $qmid n$ (noting that $n$ is not a perfect square per TonyK's comment under Ross Millikan's answer). Therefore, $n$ must have a prime divisor that is inclusively between $13$ and $67$: $13$, $17$, $19$, $23$, $29$, $31$, $37$, $41$, $43$, $47$, $53$, $59$, $61$, and $67$. We can easily rule out $37$ as $n-1$ is divisible by $111=3cdot 37$. This leaves $13$ primes to deal with.
There will be some cumbersome computations. It is not too difficult (but a little bit tedious) to find the square root or the cubic root of $n$ by hand (the cubic root of $n$ is used to obtain $67$ when I say that if $p>67$ then there exists a prime divisor $q<66$). And then you have to divide $n$ by $13$ primes. This is doable, but not very nice.
edited 2 days ago
answered 2 days ago
Zvi
4,960430
4,960430
A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done.
– Ross Millikan
2 days ago
add a comment |
A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done.
– Ross Millikan
2 days ago
A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done.
– Ross Millikan
2 days ago
A cube root is not good enough because we could have $333667=p^2q$ for $p,q$ prime. But if we show there is no prime factor smaller than the cube root we are done.
– Ross Millikan
2 days ago
add a comment |
To prove $333667$ is squarefree, you just have to show it has no prime factor smaller than $sqrt[3]{333667} approx 69$ The small ones can be done by divisibility rules, say $2,3,5,7,11$. That leaves $14$ to try, which is not too bad. You might even know the variants on the classic test for $7$ that you double the last digit and subtract it from the rest of the number. This is based on the fact that $21$ is a multiple of $7$. For $13$ you can note that $39$ is a multiple of $13$ and multiply the last digit by $4$ and add to the rest of the number. For $17$ you can use $51$. That gets you the next few. It would be a few minutes, but if you are quick with arithmetic much less than $10$.
4
Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.)
– TonyK
2 days ago
add a comment |
To prove $333667$ is squarefree, you just have to show it has no prime factor smaller than $sqrt[3]{333667} approx 69$ The small ones can be done by divisibility rules, say $2,3,5,7,11$. That leaves $14$ to try, which is not too bad. You might even know the variants on the classic test for $7$ that you double the last digit and subtract it from the rest of the number. This is based on the fact that $21$ is a multiple of $7$. For $13$ you can note that $39$ is a multiple of $13$ and multiply the last digit by $4$ and add to the rest of the number. For $17$ you can use $51$. That gets you the next few. It would be a few minutes, but if you are quick with arithmetic much less than $10$.
4
Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.)
– TonyK
2 days ago
add a comment |
To prove $333667$ is squarefree, you just have to show it has no prime factor smaller than $sqrt[3]{333667} approx 69$ The small ones can be done by divisibility rules, say $2,3,5,7,11$. That leaves $14$ to try, which is not too bad. You might even know the variants on the classic test for $7$ that you double the last digit and subtract it from the rest of the number. This is based on the fact that $21$ is a multiple of $7$. For $13$ you can note that $39$ is a multiple of $13$ and multiply the last digit by $4$ and add to the rest of the number. For $17$ you can use $51$. That gets you the next few. It would be a few minutes, but if you are quick with arithmetic much less than $10$.
To prove $333667$ is squarefree, you just have to show it has no prime factor smaller than $sqrt[3]{333667} approx 69$ The small ones can be done by divisibility rules, say $2,3,5,7,11$. That leaves $14$ to try, which is not too bad. You might even know the variants on the classic test for $7$ that you double the last digit and subtract it from the rest of the number. This is based on the fact that $21$ is a multiple of $7$. For $13$ you can note that $39$ is a multiple of $13$ and multiply the last digit by $4$ and add to the rest of the number. For $17$ you can use $51$. That gets you the next few. It would be a few minutes, but if you are quick with arithmetic much less than $10$.
answered 2 days ago
Ross Millikan
292k23197371
292k23197371
4
Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.)
– TonyK
2 days ago
add a comment |
4
Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.)
– TonyK
2 days ago
4
4
Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.)
– TonyK
2 days ago
Well, you also have to show that $333667$ is not a perfect square. (But that's easy, because a perfect square can't end in $7$.)
– TonyK
2 days ago
add a comment |
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Well, you need to somehow know at least that $333667$ is squarefree - otherwise, for its factor $p^2$, $1/p^2$ would have purely periodic expansion of length $9$. The problem at this point is actually equivalent to checking if $333667$ is squarefree.
– Wojowu
2 days ago
@Wojowu Yes, that's the key point. If we know that 333667 is squarefree, we are done. But I have no idea how to prove it quickly.
– Oldboy
2 days ago
Testing squarefree-ness is a known problem in computational complexity with no known polynomial-time algorithm. This doesn't exactly tell us there is no way to solve this problem easily, since $333667$ might have some magical properties which make it simpler, but I am being a little skeptical here...
– Wojowu
2 days ago
@JohnDouma A digit can be zero, I have clarified this.
– Oldboy
2 days ago
There was no need to clarify. I just misunderstood. That's why I deleted the comment.
– John Douma
2 days ago