nth tern of the sequence [on hold]
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if $a_0=3$, $a_1=3^3$,..., $a_n=3^{a_{n-1}}$ for all positive n. Then what is the last digit of $a_{1000}$
discrete-mathematics
asked yesterday
mathmania12
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put on hold as off-topic by Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber♦ yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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if $a_0=3$, $a_1=3^3$,..., $a_n=3^{a_{n-1}}$ for all positive n. Then what is the last digit of $a_{1000}$
discrete-mathematics
asked yesterday
mathmania12
6
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mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber♦ yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
4
What have you tried? Have you, say, written out the sequence of last digits starting with $3$ for $a_0$? Maybe you can spot a pattern.
– lulu
yesterday
3
Have you just computed the last digit for the first few terms? You should notice a pattern. A spreadsheet with mod(3^n,10) will make it easy
– Ross Millikan
yesterday
add a comment |
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if $a_0=3$, $a_1=3^3$,..., $a_n=3^{a_{n-1}}$ for all positive n. Then what is the last digit of $a_{1000}$
discrete-mathematics
asked yesterday
mathmania12
6
New contributor
mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
if $a_0=3$, $a_1=3^3$,..., $a_n=3^{a_{n-1}}$ for all positive n. Then what is the last digit of $a_{1000}$
discrete-mathematics
discrete-mathematics
asked yesterday
mathmania12
6
New contributor
mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
mathmania12
6
New contributor
mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
mathmania12
6
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mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
mathmania12
6
asked yesterday
mathmania12
6
6
New contributor
mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
mathmania12 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber♦ yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber♦ yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack D'Aurizio, lulu, Servaes, Mark, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
4
What have you tried? Have you, say, written out the sequence of last digits starting with $3$ for $a_0$? Maybe you can spot a pattern.
– lulu
yesterday
3
Have you just computed the last digit for the first few terms? You should notice a pattern. A spreadsheet with mod(3^n,10) will make it easy
– Ross Millikan
yesterday
add a comment |
4
What have you tried? Have you, say, written out the sequence of last digits starting with $3$ for $a_0$? Maybe you can spot a pattern.
– lulu
yesterday
3
Have you just computed the last digit for the first few terms? You should notice a pattern. A spreadsheet with mod(3^n,10) will make it easy
– Ross Millikan
yesterday
4
4
What have you tried? Have you, say, written out the sequence of last digits starting with $3$ for $a_0$? Maybe you can spot a pattern.
– lulu
yesterday
What have you tried? Have you, say, written out the sequence of last digits starting with $3$ for $a_0$? Maybe you can spot a pattern.
– lulu
yesterday
3
3
Have you just computed the last digit for the first few terms? You should notice a pattern. A spreadsheet with mod(3^n,10) will make it easy
– Ross Millikan
yesterday
Have you just computed the last digit for the first few terms? You should notice a pattern. A spreadsheet with mod(3^n,10) will make it easy
– Ross Millikan
yesterday
add a comment |
1 Answer
1
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The answer is $boxed{7 text{ } }$. An explanation is provided below.
We will look for a pattern. First, we examine powers of $3$ modulo $10$.
The last digit of $3^{1} = 3$ is just $3$.
The last digit of $3^{2} = 9$ is $9$.
The last digit of $3^{3} = 27$ is $7$.
The last digit of $3^{4} = 81$ is $1$.
The last digit of $3^{5} = 243$ is $3$.
So, we see that the last digit of powers of $3$ repeat in cycles of $4$ (this means that the last digit of, for example, $3^{10}$ will equal the last digit of $3^{6}$, which equals the last digit of $3^{2}$, which is $9$).
Now, we can look for a pattern in our sequence ${a_{n}}$ using the pattern we found above.
- The last digit of $a_{0} = 3$ is just $3$.
The last digit of $a_{1} = 27$ is $7$.
The last digit of $a_{2} = 3^{27}$ equals the last digit of $3^{3}$, which is $7$.
Therefore, we can keep on computing powers, and we will see that the last digit remains at $7$.
answered yesterday
Ekesh
3514
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The last digit repeats in cycles of $4$, not $3$, right? Also, I do not see how this proves that the last digit will remain $7$.
– Servaes
yesterday
Oops, yes. I made a typo. I just corrected it.
– Ekesh
yesterday
We can't work mod $10$ since $3^{10}notequiv 1pmod{!10}.,$ But mod $20$ does the trick: $begin{align}\ {rm Note}qquadquad a_1 &= 3^{large 3} = 7+20.\ \ {rm so} bmod 20!: a_n &= 7+20k\ \ Rightarrow a_{n+1} &= 3^{large a_n}! = 3^{large 7+20k}! = 3^{large 3}(3^{large 4})^{large 1+5k}! equiv 7(1)^{1+5k}! equiv 7end{align}$
– Bill Dubuque
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The answer is $boxed{7 text{ } }$. An explanation is provided below.
We will look for a pattern. First, we examine powers of $3$ modulo $10$.
The last digit of $3^{1} = 3$ is just $3$.
The last digit of $3^{2} = 9$ is $9$.
The last digit of $3^{3} = 27$ is $7$.
The last digit of $3^{4} = 81$ is $1$.
The last digit of $3^{5} = 243$ is $3$.
So, we see that the last digit of powers of $3$ repeat in cycles of $4$ (this means that the last digit of, for example, $3^{10}$ will equal the last digit of $3^{6}$, which equals the last digit of $3^{2}$, which is $9$).
Now, we can look for a pattern in our sequence ${a_{n}}$ using the pattern we found above.
- The last digit of $a_{0} = 3$ is just $3$.
The last digit of $a_{1} = 27$ is $7$.
The last digit of $a_{2} = 3^{27}$ equals the last digit of $3^{3}$, which is $7$.
Therefore, we can keep on computing powers, and we will see that the last digit remains at $7$.
answered yesterday
Ekesh
3514
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Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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The last digit repeats in cycles of $4$, not $3$, right? Also, I do not see how this proves that the last digit will remain $7$.
– Servaes
yesterday
Oops, yes. I made a typo. I just corrected it.
– Ekesh
yesterday
We can't work mod $10$ since $3^{10}notequiv 1pmod{!10}.,$ But mod $20$ does the trick: $begin{align}\ {rm Note}qquadquad a_1 &= 3^{large 3} = 7+20.\ \ {rm so} bmod 20!: a_n &= 7+20k\ \ Rightarrow a_{n+1} &= 3^{large a_n}! = 3^{large 7+20k}! = 3^{large 3}(3^{large 4})^{large 1+5k}! equiv 7(1)^{1+5k}! equiv 7end{align}$
– Bill Dubuque
yesterday
add a comment |
up vote
0
down vote
The answer is $boxed{7 text{ } }$. An explanation is provided below.
We will look for a pattern. First, we examine powers of $3$ modulo $10$.
The last digit of $3^{1} = 3$ is just $3$.
The last digit of $3^{2} = 9$ is $9$.
The last digit of $3^{3} = 27$ is $7$.
The last digit of $3^{4} = 81$ is $1$.
The last digit of $3^{5} = 243$ is $3$.
So, we see that the last digit of powers of $3$ repeat in cycles of $4$ (this means that the last digit of, for example, $3^{10}$ will equal the last digit of $3^{6}$, which equals the last digit of $3^{2}$, which is $9$).
Now, we can look for a pattern in our sequence ${a_{n}}$ using the pattern we found above.
- The last digit of $a_{0} = 3$ is just $3$.
The last digit of $a_{1} = 27$ is $7$.
The last digit of $a_{2} = 3^{27}$ equals the last digit of $3^{3}$, which is $7$.
Therefore, we can keep on computing powers, and we will see that the last digit remains at $7$.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The last digit repeats in cycles of $4$, not $3$, right? Also, I do not see how this proves that the last digit will remain $7$.
– Servaes
yesterday
Oops, yes. I made a typo. I just corrected it.
– Ekesh
yesterday
We can't work mod $10$ since $3^{10}notequiv 1pmod{!10}.,$ But mod $20$ does the trick: $begin{align}\ {rm Note}qquadquad a_1 &= 3^{large 3} = 7+20.\ \ {rm so} bmod 20!: a_n &= 7+20k\ \ Rightarrow a_{n+1} &= 3^{large a_n}! = 3^{large 7+20k}! = 3^{large 3}(3^{large 4})^{large 1+5k}! equiv 7(1)^{1+5k}! equiv 7end{align}$
– Bill Dubuque
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
The answer is $boxed{7 text{ } }$. An explanation is provided below.
We will look for a pattern. First, we examine powers of $3$ modulo $10$.
The last digit of $3^{1} = 3$ is just $3$.
The last digit of $3^{2} = 9$ is $9$.
The last digit of $3^{3} = 27$ is $7$.
The last digit of $3^{4} = 81$ is $1$.
The last digit of $3^{5} = 243$ is $3$.
So, we see that the last digit of powers of $3$ repeat in cycles of $4$ (this means that the last digit of, for example, $3^{10}$ will equal the last digit of $3^{6}$, which equals the last digit of $3^{2}$, which is $9$).
Now, we can look for a pattern in our sequence ${a_{n}}$ using the pattern we found above.
- The last digit of $a_{0} = 3$ is just $3$.
The last digit of $a_{1} = 27$ is $7$.
The last digit of $a_{2} = 3^{27}$ equals the last digit of $3^{3}$, which is $7$.
Therefore, we can keep on computing powers, and we will see that the last digit remains at $7$.
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The answer is $boxed{7 text{ } }$. An explanation is provided below.
We will look for a pattern. First, we examine powers of $3$ modulo $10$.
The last digit of $3^{1} = 3$ is just $3$.
The last digit of $3^{2} = 9$ is $9$.
The last digit of $3^{3} = 27$ is $7$.
The last digit of $3^{4} = 81$ is $1$.
The last digit of $3^{5} = 243$ is $3$.
So, we see that the last digit of powers of $3$ repeat in cycles of $4$ (this means that the last digit of, for example, $3^{10}$ will equal the last digit of $3^{6}$, which equals the last digit of $3^{2}$, which is $9$).
Now, we can look for a pattern in our sequence ${a_{n}}$ using the pattern we found above.
- The last digit of $a_{0} = 3$ is just $3$.
The last digit of $a_{1} = 27$ is $7$.
The last digit of $a_{2} = 3^{27}$ equals the last digit of $3^{3}$, which is $7$.
Therefore, we can keep on computing powers, and we will see that the last digit remains at $7$.
answered yesterday
Ekesh
3514
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Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
answered yesterday
Ekesh
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered yesterday
Ekesh
3514
answered yesterday
Ekesh
3514
3514
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ekesh is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The last digit repeats in cycles of $4$, not $3$, right? Also, I do not see how this proves that the last digit will remain $7$.
– Servaes
yesterday
Oops, yes. I made a typo. I just corrected it.
– Ekesh
yesterday
We can't work mod $10$ since $3^{10}notequiv 1pmod{!10}.,$ But mod $20$ does the trick: $begin{align}\ {rm Note}qquadquad a_1 &= 3^{large 3} = 7+20.\ \ {rm so} bmod 20!: a_n &= 7+20k\ \ Rightarrow a_{n+1} &= 3^{large a_n}! = 3^{large 7+20k}! = 3^{large 3}(3^{large 4})^{large 1+5k}! equiv 7(1)^{1+5k}! equiv 7end{align}$
– Bill Dubuque
yesterday
add a comment |
The last digit repeats in cycles of $4$, not $3$, right? Also, I do not see how this proves that the last digit will remain $7$.
– Servaes
yesterday
Oops, yes. I made a typo. I just corrected it.
– Ekesh
yesterday
We can't work mod $10$ since $3^{10}notequiv 1pmod{!10}.,$ But mod $20$ does the trick: $begin{align}\ {rm Note}qquadquad a_1 &= 3^{large 3} = 7+20.\ \ {rm so} bmod 20!: a_n &= 7+20k\ \ Rightarrow a_{n+1} &= 3^{large a_n}! = 3^{large 7+20k}! = 3^{large 3}(3^{large 4})^{large 1+5k}! equiv 7(1)^{1+5k}! equiv 7end{align}$
– Bill Dubuque
yesterday
The last digit repeats in cycles of $4$, not $3$, right? Also, I do not see how this proves that the last digit will remain $7$.
– Servaes
yesterday
The last digit repeats in cycles of $4$, not $3$, right? Also, I do not see how this proves that the last digit will remain $7$.
– Servaes
yesterday
Oops, yes. I made a typo. I just corrected it.
– Ekesh
yesterday
Oops, yes. I made a typo. I just corrected it.
– Ekesh
yesterday
We can't work mod $10$ since $3^{10}notequiv 1pmod{!10}.,$ But mod $20$ does the trick: $begin{align}\ {rm Note}qquadquad a_1 &= 3^{large 3} = 7+20.\ \ {rm so} bmod 20!: a_n &= 7+20k\ \ Rightarrow a_{n+1} &= 3^{large a_n}! = 3^{large 7+20k}! = 3^{large 3}(3^{large 4})^{large 1+5k}! equiv 7(1)^{1+5k}! equiv 7end{align}$
– Bill Dubuque
yesterday
We can't work mod $10$ since $3^{10}notequiv 1pmod{!10}.,$ But mod $20$ does the trick: $begin{align}\ {rm Note}qquadquad a_1 &= 3^{large 3} = 7+20.\ \ {rm so} bmod 20!: a_n &= 7+20k\ \ Rightarrow a_{n+1} &= 3^{large a_n}! = 3^{large 7+20k}! = 3^{large 3}(3^{large 4})^{large 1+5k}! equiv 7(1)^{1+5k}! equiv 7end{align}$
– Bill Dubuque
yesterday
add a comment |
4
What have you tried? Have you, say, written out the sequence of last digits starting with $3$ for $a_0$? Maybe you can spot a pattern.
– lulu
yesterday
3
Have you just computed the last digit for the first few terms? You should notice a pattern. A spreadsheet with mod(3^n,10) will make it easy
– Ross Millikan
yesterday