Easy to count?
77 one’s 1 1 1 1 1 1 1 1 … 1
Is this number prime or composite?
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11 solutions
Nice proof! Notice that it remains valid if you choose a non-prime number of 1's, just replacing 77 with a factorized number n=a*b. Today I learned something really nice, thank you!!!
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Nice! If we have composite number of 1's, then it is a composite number. If we have a prime number of 1's, can we conclude that it would be a prime number?
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Not necessarily. Take 111, a composite number but prime number of 1's
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@Yash Jain – I'd even speculate that 11 is the only prime number made up exclusively of 1's. I suspect I'm wrong but I haven't yet found any other prime of this nature.
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@Malcolm Rich – On further investigation it appears that there are a few but they are quite rare. 1,111,111,111,111,111,111 is the smallest one after 11.
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@Malcolm Rich – I found this sequence OEIS A004023 which goes like 2, 19, 23, 317, 1031, ... For every n in this sequence, the number with n ones is a prime number. The sequence seems to be quite random, with no discernible pattern.
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@Pranshu Gaba – I saw that sequence too. Looking for patterns in the prime factorisation I've seen that, at least one of the prime factors takes the form (p * n) + 1, where p is the number of 1's in the number concerned.
Nice final step.
👍👍..... 77 times😁
Nicely done.
The fact that there are 77 1's tells the whole story. 77 is a composite number, 7 * 11. That means if we divide the big number by a number with either 7 or 11 1's, there will be no remainder. Hence, the big number is composite.
Good analysis, Tom. Expanding the idea a bit in case it's not clear to someone...
111111...1 can be grouped into 11 consecutive groups of seven 1's.
e.e 1111111 1111111 1111111 1111111..... 1111111
That should make it plain to see that you can divide the original number by a string of seven ones (1111111)
111111...1 / 1111111 = (000000)10000001000000100000010000001....0000001.
The quotient is 77 digits long if you include the leading zeroes, or 71 digits long if you don't. But it doesn't really matter how long the quotient string is; it is a factor of the original number, with no remainder. Therefore the original number is composite.
Of course, the converse isn't true. For instance, 3 is prime but 111 is composite.
The exact same principle, in binary, explains why Mersenne primes necessarily have prime exponents.
Repunit primes (containing only the decimal digit 1) are that with 2 (it is 11), 19, 23, 317, 1031 1's. With your reasoning how do you explane that R(11)=21649 · 513239 is not prime. Or R(13) and many, many more. For even number of 1's it is obvious.
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You've basically answered your own question - R(11) isn't prime because it's 21649 · 513239. :)
As I noted, the converse isn't true. In this particular instance you can answer the question by realising that the number of 1s is composite, but if the number of 1s is prime then you can't draw any conclusion, so would have to test it for primality in some other way.
I think this was wrong. Let me consider a small num. Let it be 4 1's. 4 is a composite num,2*2. That means 1111 must be divisible by 2( as per ur logic) which is not true.
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Your example is easier to see, but it is incorrect. Instead of "divisible by 2", it should say "divisible by a number with 2 1's " (my logic) which is true. 1111 / 11 = 101. This proves that 1111 is composite. It suggests the general case that only a prime number string of 1's can possibly be prime. Even then, the number could be divisible by other prime numbers, so, only some of these eligible numbers may be prime themselves
Thnx. I was confused on how you guys got composite. I assumed that as 11 is prime that 111111111111...1 would be the same.
if there was an even number of 1's your logic would be correct, but 77 is odd.
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Surely, you have misunderstood something. I won't try to guess what even and oddness has to do with my comment. The fact is, 77 is not only odd, but also the product of two primes, 7 and 11. That means the digits can be collected in 7 groups of 11, or 11 groups of 7. It is clear that if you divide a group by a copy of itself, there will be no remainder. Therefore, the 77 digit number is composite.
This is wrong. We can easily rule out that 7 and 11 are NOT the divisors of this large number by Divisibliity Rules (2,3,5,7,11,13,17,19,...) of 7 and 11.
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You got something wrong. He meant to say 7 or 11 number of 1's .
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Whoops, sorry about that. I misread the solution. Yup, it's definitely correct~!
1111...1 = (1/9) * (9999...9) = (1/9) * (10^77 - 1) = ( 10^77 - 1) / ( 10 - 1) = ( 10 - 1) ( 1+10 + 100 + 1000 + ....+ 10^76 ) / ( 10 - 1) = ( 1+10 + 100 + 1000 + ....+ 10^76 ) = 11111....1 76 1's which is divisible by 11 ,(sum of geometric progression with a = 1 & r = 10 )
No, this is incorrect.
By looking at your solution, it tells us that this number is divisible by 10, which is clearly not.
Yes, @Pi Han Goh is right.
( 1 0 − 1 ) ( 1 0 7 7 − 1 ) is a GP with the first term as 1, not 10. Thus you cannot take 10 as a common factor.
ya you are right i have revised the solution
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Still doesn't look right. Note that ( 1 + 10 + 100 + ... + 10^76) has 77 1's (You've restated the number as itself, but lost a 1 along the way). It is not divisible by 11.
There are 771 digits of one so the sum of digits is 771 7+7+1=15 1+5=6 6 is divided by 3 so this whole number is divided by 3
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There are 77 digits that are 1, not 771 digits. Let me improve the clarity in the problem.
Easy to show 11111111...1 (77 1's) as a Composite Number !
11111111...1 (77 1's) can be factorised as:
(a) by grouping of 7 1's: 1111111×(10^70+10^63+10^56+...+10^14+10^7+1)
(b) by grouping of 11 1's: 11111111111×(10^66+10^55+10^44+...+10^11+1)
Any number formed by an odd number of 1's has a factor of 11, as 1010..... (10 written n times) x 11 = 1111...(1 written 2n+1 times).
How is that true? 11111 doesn't have 11 as a factor. I think your construction only works when there is an even number of 1's, eg 11 * 101 = 1111..
771 is divisible by 3 such that the given num is divisible by 3
And the original number is not divisible by 3
All series of 11's can be written by the sum of 10^0,10^1,10^2,...10^n-1 because n = 77 there are 77 of these and they just need to be separated into equal groups (because size of group only changes number of 1's (not 0's which always allow division)). The possible equal groups are of 7's or of 11's befause there are 77 of these terms total. These all have 11 or 7 1's followed by 0's and all of these terms divide by their 7 or 11 1's number.
Ex:
11111111+1111111 * 10^7 +1111111 * 10^14...
Consists of
10^0,10^1,10^2...
1+10+100...
Can we sort out the use of apostrophes please?
Clearly, this number is divisible by 77.
No, this number is not divisible by 77. If it's divisible by 77, then it must be divisible by both 7 and 11. We can show none of these are true by divisibility rules of 7 and 11 .
38 1's at odd positions 1,3,.......77 and 38 1's at even positions 2,4,6,.........76 add up to equal sums, and according to the divisibility rule for 11, the number is divisible by 11. Hence composite by rule of alternating sums being divisible by 11. qed
Nope, check your working again, the alternating sum is not divisible by 11, so the number is not divisible by 11.
Any number whos digits sum to a multiple of 3 is divisible by 3. Hence (1+1+1+...+1)=771, and (7+7+1)=15, which is a multiple of 3, hence the number is a composite number
77 1s sum to 77, not 771.
subtraction of sum for digits at odd and even number position is 0, which is divisible by 11.
You're close, but your reasoning is incorrect. There are an odd number of 1's, so the alternating sum of these digits gives 1, not 0, so it is not divisible by 11.
77 1’s 1 1 1 1 1 1 … 1
= 1 0 7 6 + 1 0 7 5 + 1 0 7 4 … + 1 0 1 + 1
= 1 + 1 0 1 + 1 0 2 … + 1 0 7 5 + 1 0 7 6
= 1 0 − 1 1 ( 1 0 7 7 − 1 ) (Sum of a GP with a=1, r=10, n=77)
= 1 0 − 1 ( 1 0 7 7 − 1 ) × ( 1 0 7 − 1 ) ( 1 0 7 − 1 )
= 1 0 7 − 1 ( ( 1 0 7 ) 1 1 − 1 ) × 1 0 − 1 ( 1 0 7 − 1 ) (Rearranging the terms)
= ( 1 + 1 0 7 + 1 0 1 4 … + 1 0 7 0 ) ( 1 + 1 0 + 1 0 2 ⋯ + 1 0 6 ) (Product of two GPs)
Now the number is in the form of product of 2 GPs, which means product of two finite numbers other than 1 and the number itself.