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.
This is a problem from the worksheet of the class Putnam Seminar at CMU. Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
asked Dec 22 at 4:56
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.
This is a problem from the worksheet of the class Putnam Seminar at CMU. Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
asked Dec 22 at 4:56
BrianH
517
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.
Why is thecontest-mathtag used? Please edit the question to add that context.
– Shaun
Dec 24 at 20:48
@Shaun Well what I'm I suppose to add? It was a problem on the worksheet from my Putnam class. It feel like saying that is just distracting
– BrianH
Dec 24 at 20:51
1
Well, the fact that it's from your Putnam class. Don't worry about potential distractions. Context is strongly encouraged by many users here and, typically, it can be very helpful for some users for any number of reasons.
– Shaun
Dec 24 at 20:56
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.
This is a problem from the worksheet of the class Putnam Seminar at CMU. Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
asked Dec 22 at 4:56
BrianH
517
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.
This is a problem from the worksheet of the class Putnam Seminar at CMU. Please give hints towards the right direction and not the full solutions. Thanks!!
number-theory contest-math
number-theory contest-math
asked Dec 22 at 4:56
BrianH
517
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BrianH is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 22 at 4:56
BrianH
517
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edited Dec 24 at 20:57
asked Dec 22 at 4:56
BrianH
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asked Dec 22 at 4:56
BrianH
517
asked Dec 22 at 4:56
BrianH
517
517
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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
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Why is thecontest-mathtag used? Please edit the question to add that context.
– Shaun
Dec 24 at 20:48
@Shaun Well what I'm I suppose to add? It was a problem on the worksheet from my Putnam class. It feel like saying that is just distracting
– BrianH
Dec 24 at 20:51
1
Well, the fact that it's from your Putnam class. Don't worry about potential distractions. Context is strongly encouraged by many users here and, typically, it can be very helpful for some users for any number of reasons.
– Shaun
Dec 24 at 20:56
add a comment |
Why is thecontest-mathtag used? Please edit the question to add that context.
– Shaun
Dec 24 at 20:48
@Shaun Well what I'm I suppose to add? It was a problem on the worksheet from my Putnam class. It feel like saying that is just distracting
– BrianH
Dec 24 at 20:51
1
Well, the fact that it's from your Putnam class. Don't worry about potential distractions. Context is strongly encouraged by many users here and, typically, it can be very helpful for some users for any number of reasons.
– Shaun
Dec 24 at 20:56
Why is the
contest-math tag used? Please edit the question to add that context.– Shaun
Dec 24 at 20:48
Why is the
contest-math tag used? Please edit the question to add that context.– Shaun
Dec 24 at 20:48
@Shaun Well what I'm I suppose to add? It was a problem on the worksheet from my Putnam class. It feel like saying that is just distracting
– BrianH
Dec 24 at 20:51
@Shaun Well what I'm I suppose to add? It was a problem on the worksheet from my Putnam class. It feel like saying that is just distracting
– BrianH
Dec 24 at 20:51
1
1
Well, the fact that it's from your Putnam class. Don't worry about potential distractions. Context is strongly encouraged by many users here and, typically, it can be very helpful for some users for any number of reasons.
– Shaun
Dec 24 at 20:56
Well, the fact that it's from your Putnam class. Don't worry about potential distractions. Context is strongly encouraged by many users here and, typically, it can be very helpful for some users for any number of reasons.
– Shaun
Dec 24 at 20:56
add a comment |
2 Answers
2
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oldest
votes
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 Dec 22 at 5:21
angryavian
38.7k23180
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 Dec 22 at 5:26
bangzheng
211
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Could you explain to me how you reached the last two lines?
– BrianH
Dec 22 at 5:39
add a comment |
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2 Answers
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2 Answers
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active
oldest
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active
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votes
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 Dec 22 at 5:21
angryavian
38.7k23180
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 Dec 22 at 5:21
angryavian
38.7k23180
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 Dec 22 at 5:21
angryavian
38.7k23180
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 Dec 22 at 5:21
angryavian
38.7k23180
answered Dec 22 at 5:21
angryavian
38.7k23180
answered Dec 22 at 5:21
angryavian
38.7k23180
answered Dec 22 at 5:21
angryavian
38.7k23180
38.7k23180
add a comment |
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 Dec 22 at 5:26
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
Dec 22 at 5:39
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 Dec 22 at 5:26
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
Dec 22 at 5:39
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 Dec 22 at 5:26
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 Dec 22 at 5:26
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 Dec 22 at 5:26
bangzheng
211
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answered Dec 22 at 5:26
bangzheng
211
answered Dec 22 at 5:26
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
Dec 22 at 5:39
add a comment |
Could you explain to me how you reached the last two lines?
– BrianH
Dec 22 at 5:39
Could you explain to me how you reached the last two lines?
– BrianH
Dec 22 at 5:39
Could you explain to me how you reached the last two lines?
– BrianH
Dec 22 at 5:39
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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Why is the
contest-mathtag used? Please edit the question to add that context.– Shaun
Dec 24 at 20:48
@Shaun Well what I'm I suppose to add? It was a problem on the worksheet from my Putnam class. It feel like saying that is just distracting
– BrianH
Dec 24 at 20:51
1
Well, the fact that it's from your Putnam class. Don't worry about potential distractions. Context is strongly encouraged by many users here and, typically, it can be very helpful for some users for any number of reasons.
– Shaun
Dec 24 at 20:56