Every positive power of $5$ appears in the last digits of bigger power of $5$
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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 '18 at 4:56
BrianHBrianH
658
$endgroup$
add a comment |
$begingroup$
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 '18 at 4:56
BrianHBrianH
658
$endgroup$
$begingroup$
Why is thecontest-mathtag used? Please edit the question to add that context.
$endgroup$
– Shaun
Dec 24 '18 at 20:48
$begingroup$
@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
$endgroup$
– BrianH
Dec 24 '18 at 20:51
1
$begingroup$
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.
$endgroup$
– Shaun
Dec 24 '18 at 20:56
add a comment |
$begingroup$
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 '18 at 4:56
BrianHBrianH
658
$endgroup$
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 '18 at 4:56
BrianHBrianH
658
asked Dec 22 '18 at 4:56
BrianHBrianH
658
edited Dec 24 '18 at 20:57
BrianH
asked Dec 22 '18 at 4:56
BrianHBrianH
658
asked Dec 22 '18 at 4:56
BrianHBrianH
658
asked Dec 22 '18 at 4:56
BrianHBrianH
658
658
$begingroup$
Why is thecontest-mathtag used? Please edit the question to add that context.
$endgroup$
– Shaun
Dec 24 '18 at 20:48
$begingroup$
@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
$endgroup$
– BrianH
Dec 24 '18 at 20:51
1
$begingroup$
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.
$endgroup$
– Shaun
Dec 24 '18 at 20:56
add a comment |
$begingroup$
Why is thecontest-mathtag used? Please edit the question to add that context.
$endgroup$
– Shaun
Dec 24 '18 at 20:48
$begingroup$
@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
$endgroup$
– BrianH
Dec 24 '18 at 20:51
1
$begingroup$
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.
$endgroup$
– Shaun
Dec 24 '18 at 20:56
$begingroup$
Why is the
contest-math tag used? Please edit the question to add that context.$endgroup$
– Shaun
Dec 24 '18 at 20:48
$begingroup$
Why is the
contest-math tag used? Please edit the question to add that context.$endgroup$
– Shaun
Dec 24 '18 at 20:48
$begingroup$
@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
$endgroup$
– BrianH
Dec 24 '18 at 20:51
$begingroup$
@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
$endgroup$
– BrianH
Dec 24 '18 at 20:51
1
1
$begingroup$
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.
$endgroup$
– Shaun
Dec 24 '18 at 20:56
$begingroup$
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.
$endgroup$
– Shaun
Dec 24 '18 at 20:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 '18 at 5:21
angryavianangryavian
40.7k23380
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add a comment |
$begingroup$
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 '18 at 5:26
bangzhengbangzheng
212
$endgroup$
$begingroup$
Could you explain to me how you reached the last two lines?
$endgroup$
– BrianH
Dec 22 '18 at 5:39
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 '18 at 5:21
angryavianangryavian
40.7k23380
$endgroup$
add a comment |
$begingroup$
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 '18 at 5:21
angryavianangryavian
40.7k23380
$endgroup$
add a comment |
$begingroup$
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 '18 at 5:21
angryavianangryavian
40.7k23380
$endgroup$
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 '18 at 5:21
angryavianangryavian
40.7k23380
answered Dec 22 '18 at 5:21
angryavianangryavian
40.7k23380
answered Dec 22 '18 at 5:21
angryavianangryavian
40.7k23380
answered Dec 22 '18 at 5:21
angryavianangryavian
40.7k23380
40.7k23380
add a comment |
add a comment |
$begingroup$
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 '18 at 5:26
bangzhengbangzheng
212
$endgroup$
$begingroup$
Could you explain to me how you reached the last two lines?
$endgroup$
– BrianH
Dec 22 '18 at 5:39
add a comment |
$begingroup$
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 '18 at 5:26
bangzhengbangzheng
212
$endgroup$
$begingroup$
Could you explain to me how you reached the last two lines?
$endgroup$
– BrianH
Dec 22 '18 at 5:39
add a comment |
$begingroup$
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 '18 at 5:26
bangzhengbangzheng
212
$endgroup$
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 '18 at 5:26
bangzhengbangzheng
212
answered Dec 22 '18 at 5:26
bangzhengbangzheng
212
answered Dec 22 '18 at 5:26
bangzhengbangzheng
212
answered Dec 22 '18 at 5:26
bangzhengbangzheng
212
212
$begingroup$
Could you explain to me how you reached the last two lines?
$endgroup$
– BrianH
Dec 22 '18 at 5:39
add a comment |
$begingroup$
Could you explain to me how you reached the last two lines?
$endgroup$
– BrianH
Dec 22 '18 at 5:39
$begingroup$
Could you explain to me how you reached the last two lines?
$endgroup$
– BrianH
Dec 22 '18 at 5:39
$begingroup$
Could you explain to me how you reached the last two lines?
$endgroup$
– BrianH
Dec 22 '18 at 5:39
add a comment |
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$begingroup$
Why is the
contest-mathtag used? Please edit the question to add that context.$endgroup$
– Shaun
Dec 24 '18 at 20:48
$begingroup$
@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
$endgroup$
– BrianH
Dec 24 '18 at 20:51
1
$begingroup$
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.
$endgroup$
– Shaun
Dec 24 '18 at 20:56