Is $1111111111111111111111111111111111111111111111111111111$ ($55$ $1$'s) a composite number?
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This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited 18 hours ago
Asaf Karagila♦
303k32429761
asked yesterday
user69284user69284
10216
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user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited 18 hours ago
Asaf Karagila♦
303k32429761
asked yesterday
user69284user69284
10216
New contributor
user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
yesterday
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
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– Jeppe Stig Nielsen
yesterday
3
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This type of number is called a “repunit”.
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– Dan
yesterday
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
16 hours ago
add a comment |
$begingroup$
This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited 18 hours ago
Asaf Karagila♦
303k32429761
asked yesterday
user69284user69284
10216
New contributor
user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = frac{1}9(999...) = frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?
sequences-and-series algebra-precalculus geometric-series geometric-progressions
sequences-and-series algebra-precalculus geometric-series geometric-progressions
edited 18 hours ago
Asaf Karagila♦
303k32429761
asked yesterday
user69284user69284
10216
New contributor
user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 hours ago
Asaf Karagila♦
303k32429761
asked yesterday
user69284user69284
10216
New contributor
user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 hours ago
Asaf Karagila♦
303k32429761
edited 18 hours ago
Asaf Karagila♦
303k32429761
edited 18 hours ago
Asaf Karagila♦
303k32429761
303k32429761
asked yesterday
user69284user69284
10216
New contributor
user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
user69284user69284
10216
asked yesterday
user69284user69284
10216
10216
New contributor
user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
user69284 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
yesterday
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
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– Jeppe Stig Nielsen
yesterday
3
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
yesterday
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
16 hours ago
add a comment |
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
yesterday
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
yesterday
3
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
yesterday
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
16 hours ago
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
yesterday
$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
yesterday
1
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
yesterday
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
yesterday
3
3
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
yesterday
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
yesterday
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
16 hours ago
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
16 hours ago
add a comment |
5 Answers
5
active
oldest
votes
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The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered yesterday
Ekesh KumarEkesh Kumar
90428
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1
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Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
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– user69284
yesterday
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Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
yesterday
12
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"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
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– Chris
yesterday
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
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– Chris
yesterday
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
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– boboquack
yesterday
|
show 1 more comment
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You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered yesterday
ChrisChris
4381513
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1
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If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
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– potestasity
yesterday
5
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@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
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– David Robinson
yesterday
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@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
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– Chris
yesterday
2
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@Chris If you had 55 2's you simply would divide by 2.
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– rus9384
yesterday
1
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@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
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– Random832
yesterday
|
show 6 more comments
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More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered yesterday
heropupheropup
63.6k762102
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1
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Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
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– orion
17 hours ago
add a comment |
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This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered yesterday
DanijelDanijel
826517
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If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
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– gnasher729
yesterday
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I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
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– Acccumulation
yesterday
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This is the best answer here imo
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– goblin
22 hours ago
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@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
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– Danijel
16 hours ago
add a comment |
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The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
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6
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Can you clarify how you obtained this?
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– Pedro A
yesterday
3
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I think if the OP searched for the factorization alone, he had used a calculator in the first place
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– Nico Haase
yesterday
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@PedroA Several repunit factorizations are available here
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– LegionMammal978
yesterday
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I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
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– Daniel Schepler
yesterday
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Or Maple or any other computer algebra system.
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– David G. Stork
yesterday
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered yesterday
Ekesh KumarEkesh Kumar
90428
$endgroup$
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
yesterday
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
yesterday
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
yesterday
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
yesterday
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
yesterday
|
show 1 more comment
$begingroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered yesterday
Ekesh KumarEkesh Kumar
90428
$endgroup$
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
yesterday
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
yesterday
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
yesterday
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
yesterday
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
yesterday
|
show 1 more comment
$begingroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered yesterday
Ekesh KumarEkesh Kumar
90428
$endgroup$
The number is composite.
We have
begin{align*}
underbrace{11ldots111}_{55 text{ times }} = frac{1}{9} cdot (10^{55} - 1) \
= frac{1}{9} cdot ((10^{5})^{11} - 1) \
end{align*}
Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).
answered yesterday
Ekesh KumarEkesh Kumar
90428
edited yesterday
answered yesterday
Ekesh KumarEkesh Kumar
90428
answered yesterday
Ekesh KumarEkesh Kumar
90428
answered yesterday
Ekesh KumarEkesh Kumar
90428
90428
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
yesterday
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
yesterday
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
yesterday
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
yesterday
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
yesterday
|
show 1 more comment
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
yesterday
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
yesterday
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
yesterday
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
yesterday
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
yesterday
1
1
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
yesterday
$begingroup$
Can you please elaborate on the last step? How did you get k and concluded it to be an integer? Sorry, I am terrible at maths.
$endgroup$
– user69284
yesterday
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
yesterday
$begingroup$
Check the edit. I have found a better (and easier to understand) solution.
$endgroup$
– Ekesh Kumar
yesterday
12
12
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
yesterday
$begingroup$
"As a result, we see that the quantity is divisible by 99999" - this isn't right, surely. If it is divisible by 99999 then it is also divisible by 9 and as we all know something is divisible by 9 if the sum of the digits is divisible by 9. The sum of the digits of the original number is obviously 55 so it is not divisible by 9.
$endgroup$
– Chris
yesterday
19
19
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
yesterday
$begingroup$
Oh, I think I see the confusion now. You have your factor of $1/9$ multiplying your number so in fact I believe you mean to say that the original number is divisible by 11111
$endgroup$
– Chris
yesterday
1
1
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
yesterday
$begingroup$
@Dan That is true, but since we have a factor of $1/9$, we get $(1/9)cdot(10-1)=1$ divides $11dots1$ which isn't very helpful.
$endgroup$
– boboquack
yesterday
|
show 1 more comment
$begingroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered yesterday
ChrisChris
4381513
$endgroup$
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
yesterday
5
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
yesterday
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
yesterday
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
yesterday
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
yesterday
|
show 6 more comments
$begingroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered yesterday
ChrisChris
4381513
$endgroup$
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
yesterday
5
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
yesterday
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
yesterday
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
yesterday
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
yesterday
|
show 6 more comments
$begingroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered yesterday
ChrisChris
4381513
$endgroup$
You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.
Similarly you can see trivially see that 11111 is a factor of the number.
answered yesterday
ChrisChris
4381513
answered yesterday
ChrisChris
4381513
answered yesterday
ChrisChris
4381513
answered yesterday
ChrisChris
4381513
4381513
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
yesterday
5
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
yesterday
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
yesterday
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
yesterday
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
yesterday
|
show 6 more comments
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
yesterday
5
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
yesterday
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
yesterday
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
yesterday
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
yesterday
1
1
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
yesterday
$begingroup$
If I may add: this becomes clearer if you think of splitting the number into equally-sized chunks. Thus you can conclude that any number whose representation consists of a non-prime number of digits is composite.
$endgroup$
– potestasity
yesterday
5
5
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
yesterday
$begingroup$
@potestasity I think you mean "a non-prime number of identical digits"? Otherwise you may have accidentally destroyed most of the world's prime numbers. And of course it does not matter if the number of digits is composite, if the digit is itself not 1, e.g. 22222.
$endgroup$
– David Robinson
yesterday
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
yesterday
$begingroup$
@DavidRobinson: While I agree with your initial point I don't follow your last sentence. If you had 55 2s instead of 55 1s the logic still works doesn't it?
$endgroup$
– Chris
yesterday
2
2
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
yesterday
$begingroup$
@Chris If you had 55 2's you simply would divide by 2.
$endgroup$
– rus9384
yesterday
1
1
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
yesterday
$begingroup$
@Chris The point is that it works for 53 2's as well, because it's (53 1's) * 2.
$endgroup$
– Random832
yesterday
|
show 6 more comments
$begingroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered yesterday
heropupheropup
63.6k762102
$endgroup$
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
17 hours ago
add a comment |
$begingroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered yesterday
heropupheropup
63.6k762102
$endgroup$
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
17 hours ago
add a comment |
$begingroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered yesterday
heropupheropup
63.6k762102
$endgroup$
More explicitly,
$$begin{align*}
frac{10^{55} - 1}{9}
&= frac{(10^5)^{11} - 1}{9} \
&= frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + cdots + 10 + 1)}{9} \
&= frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + cdots + 1)}{9} \
&= (10^4 + 10^3 + cdots + 1)(10^{50} + 10^{45} + cdots + 1).
end{align*}$$
The first factor is $11111$, which in turn is $41 cdot 271$, and the second factor has as its smallest prime factor $1321$.
answered yesterday
heropupheropup
63.6k762102
answered yesterday
heropupheropup
63.6k762102
answered yesterday
heropupheropup
63.6k762102
answered yesterday
heropupheropup
63.6k762102
63.6k762102
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
17 hours ago
add a comment |
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
17 hours ago
1
1
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
17 hours ago
$begingroup$
Also, splitting $5times 11$ the other way around, you get that 11111111111 is also a factor, and it doesn't share any factor with 11111, so this way, you can divide out both of them (if you wanted to get closer to full factorization instead of just proving it's composite). The remaining number is still large, though.
$endgroup$
– orion
17 hours ago
add a comment |
$begingroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered yesterday
DanijelDanijel
826517
$endgroup$
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
yesterday
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
yesterday
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
22 hours ago
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
16 hours ago
add a comment |
$begingroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered yesterday
DanijelDanijel
826517
$endgroup$
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
yesterday
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
yesterday
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
22 hours ago
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
16 hours ago
add a comment |
$begingroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered yesterday
DanijelDanijel
826517
$endgroup$
This number can be written as $$x=sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,qinmathbb N$. Then,
$$x=sum_{k=1}^{pq}a^{k-1} = sum_{l=1}^p sum_{m=1}^q a^{q(l-1)+(m-1)}=
left(sum_{l=1}^p a^{q(l-1)}right)left(sum_{m=1}^q a^{(m-1)}right)$$
Since, $55$ is composite, it follows that $sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2cdot3$) as $$x = 111000 + 111= 1001cdot111$$ or $$x=110000 + 1100 + 11 = 10101cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
answered yesterday
DanijelDanijel
826517
answered yesterday
DanijelDanijel
826517
answered yesterday
DanijelDanijel
826517
answered yesterday
DanijelDanijel
826517
826517
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
yesterday
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
yesterday
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
22 hours ago
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
16 hours ago
add a comment |
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
yesterday
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
yesterday
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
22 hours ago
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
16 hours ago
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
yesterday
$begingroup$
If a natural number can be represented as two or more repetitions of the same digit sequence other than "1" then it is composite. If you have a composite number a*b of 1's then you have a repeats of (b 1's).
$endgroup$
– gnasher729
yesterday
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
yesterday
$begingroup$
I think the math is nicer if you do $sum_{k=0}^{54}10^k$.
$endgroup$
– Acccumulation
yesterday
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
22 hours ago
$begingroup$
This is the best answer here imo
$endgroup$
– goblin
22 hours ago
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
16 hours ago
$begingroup$
@Acccumulation perhaps it is. I just wanted to write $55$ without writing $55-1$. Writing $k-1$ looks less bad to me.
$endgroup$
– Danijel
16 hours ago
add a comment |
$begingroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
yesterday
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
yesterday
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
yesterday
$begingroup$
I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
$endgroup$
– Daniel Schepler
yesterday
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
yesterday
add a comment |
$begingroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
yesterday
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
yesterday
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
yesterday
$begingroup$
I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
$endgroup$
– Daniel Schepler
yesterday
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
yesterday
add a comment |
$begingroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
$endgroup$
The prime factorization of the number is:
$$41 cdot 271 cdot 1321 cdot 21649 cdot 62921 cdot 513239 cdot 83251631 cdot 1300635692678058358830121$$
obtained by FactorInteger in Mathematica.
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
edited yesterday
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
answered yesterday
David G. StorkDavid G. Stork
10.9k31432
10.9k31432
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
yesterday
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
yesterday
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
yesterday
$begingroup$
I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
$endgroup$
– Daniel Schepler
yesterday
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
yesterday
add a comment |
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
yesterday
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
yesterday
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
yesterday
$begingroup$
I got the same thing just enteringfactor((10^55-1)/9);into Maxima.
$endgroup$
– Daniel Schepler
yesterday
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
yesterday
6
6
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
yesterday
$begingroup$
Can you clarify how you obtained this?
$endgroup$
– Pedro A
yesterday
3
3
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
yesterday
$begingroup$
I think if the OP searched for the factorization alone, he had used a calculator in the first place
$endgroup$
– Nico Haase
yesterday
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
yesterday
$begingroup$
@PedroA Several repunit factorizations are available here
$endgroup$
– LegionMammal978
yesterday
$begingroup$
I got the same thing just entering
factor((10^55-1)/9); into Maxima.$endgroup$
– Daniel Schepler
yesterday
$begingroup$
I got the same thing just entering
factor((10^55-1)/9); into Maxima.$endgroup$
– Daniel Schepler
yesterday
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
yesterday
$begingroup$
Or Maple or any other computer algebra system.
$endgroup$
– David G. Stork
yesterday
add a comment |
user69284 is a new contributor. Be nice, and check out our Code of Conduct.
user69284 is a new contributor. Be nice, and check out our Code of Conduct.
user69284 is a new contributor. Be nice, and check out our Code of Conduct.
user69284 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
It is listed under geometric progression and geometric mean because that's how you get from $111ldots$ to $frac{1}{9}(10^{55} - 1)$
$endgroup$
– Ekesh Kumar
yesterday
1
$begingroup$
You can either do five blocks of eleven 1s: $$11111111111 11111111111 11111111111 11111111111 11111111111 \ = 11111111111times 100000000001000000000010000000000100000000001$$ or you can do eleven blocks of five 1s: $$11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 11111 \ = 11111times 100001000010000100001000010000100001000010000100001$$ It turns out, in this case, that all these factors are themselves composite.
$endgroup$
– Jeppe Stig Nielsen
yesterday
3
$begingroup$
This type of number is called a “repunit”.
$endgroup$
– Dan
yesterday
$begingroup$
@JeppeStigNielsen. And this shows that for a number consisting of $n$ ones (in any basis!) can only be prime if $n$ is prime. This is the primary background of Mersenne primes.
$endgroup$
– md2perpe
16 hours ago