Every positive power of 5 appears in the last digits of bigger power of 5
Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.
Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
asked 1 hour ago
BrianH
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Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.
Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
asked 1 hour ago
BrianH
385
New contributor
BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.
Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
asked 1 hour ago
BrianH
385
New contributor
BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Problem. Show that for every positive integer $n$, there is an integer $N > n$ such that the number $5^n$ appears as the last few digits $5^N$. For example, if $n = 3$, we have $5^3 = 125$ and $5^5 = 3125$, so $N = 5$ would satisfy.
Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
number-theory contest-math
asked 1 hour ago
BrianH
385
New contributor
BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago
BrianH
385
New contributor
BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
BrianH
385
New contributor
BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago
BrianH
385
asked 1 hour ago
BrianH
385
385
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BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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Fix $n$.
The cases $n le 3$ can be handled directly.
We now assume $n > 3$.
Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$
Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.
Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.
answered 34 mins ago
angryavian
38.4k23180
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We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits
So what we need to do is to find $5^Nequiv 5^npmod{10^n}$
This can be achieved by setting $N=n+phi (2^n)$
Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$
answered 30 mins ago
bangzheng
211
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Could you explain to me how you reached the last two lines?
– BrianH
16 mins ago
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2 Answers
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2 Answers
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active
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active
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Fix $n$.
The cases $n le 3$ can be handled directly.
We now assume $n > 3$.
Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$
Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.
Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.
answered 34 mins ago
angryavian
38.4k23180
add a comment |
Fix $n$.
The cases $n le 3$ can be handled directly.
We now assume $n > 3$.
Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$
Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.
Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.
answered 34 mins ago
angryavian
38.4k23180
add a comment |
Fix $n$.
The cases $n le 3$ can be handled directly.
We now assume $n > 3$.
Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$
Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.
Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.
answered 34 mins ago
angryavian
38.4k23180
Fix $n$.
The cases $n le 3$ can be handled directly.
We now assume $n > 3$.
Let $m = lceil n log_{10} 5 rceil$ be the smallest integer such that $10^m > 5^n$. For $n > 3$, we have $m < n$.
You want to find $N$ such that $5^N = k 10^m + 5^n$ for some integer $k$. Dividing both sides by $5^n$ yields $$5^{N-n} - 1 = k 2^m/ 5^{n-m}.$$
Thus if you show you can find a large integer $q$ such that $5^q - 1$ is divisible by $2^m$ then you can choose $N$ and $k$ appropriately to conclude the proof.
Base case: For $m=2$ we have $5^1 - 1$ divisible by $2^m$. Inductive step: if $5^q-1$ is divisible by $2^m$, then $5^{2q}-1 = (5^q-1)(5^q+1)$ is divisible by $2^{m+1}$.
answered 34 mins ago
angryavian
38.4k23180
answered 34 mins ago
angryavian
38.4k23180
answered 34 mins ago
angryavian
38.4k23180
answered 34 mins ago
angryavian
38.4k23180
38.4k23180
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We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits
So what we need to do is to find $5^Nequiv 5^npmod{10^n}$
This can be achieved by setting $N=n+phi (2^n)$
Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$
answered 30 mins ago
bangzheng
211
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bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Could you explain to me how you reached the last two lines?
– BrianH
16 mins ago
add a comment |
We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits
So what we need to do is to find $5^Nequiv 5^npmod{10^n}$
This can be achieved by setting $N=n+phi (2^n)$
Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$
answered 30 mins ago
bangzheng
211
New contributor
bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you explain to me how you reached the last two lines?
– BrianH
16 mins ago
add a comment |
We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits
So what we need to do is to find $5^Nequiv 5^npmod{10^n}$
This can be achieved by setting $N=n+phi (2^n)$
Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$
answered 30 mins ago
bangzheng
211
New contributor
bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We will prove that there exists a $N$ such that $5^n$ and $5^N$ have the same last $n$ digits
So what we need to do is to find $5^Nequiv 5^npmod{10^n}$
This can be achieved by setting $N=n+phi (2^n)$
Since $5^Nequiv 5^npmod{5^n}$ and $5^Nequiv 5^npmod{2^n}$
answered 30 mins ago
bangzheng
211
New contributor
bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 30 mins ago
bangzheng
211
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answered 30 mins ago
bangzheng
211
answered 30 mins ago
bangzheng
211
211
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New contributor
bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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bangzheng is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Could you explain to me how you reached the last two lines?
– BrianH
16 mins ago
add a comment |
Could you explain to me how you reached the last two lines?
– BrianH
16 mins ago
Could you explain to me how you reached the last two lines?
– BrianH
16 mins ago
Could you explain to me how you reached the last two lines?
– BrianH
16 mins ago
add a comment |
BrianH is a new contributor. Be nice, and check out our Code of Conduct.
BrianH is a new contributor. Be nice, and check out our Code of Conduct.
BrianH is a new contributor. Be nice, and check out our Code of Conduct.
BrianH is a new contributor. Be nice, and check out our Code of Conduct.
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