1 1 1 1 … … 1 1 1 1 1 , which consists of 91 1's is :-
The integer
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If g c d ( n , m ) = k , then g c d ( R n , R m ) = R k (and vice versa)
If you want a proof for this statement, ask me for that.
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I tried it a bit myself.What do you use?Induction on k or double induction on m , n ?
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Almost the same method as you solve this question.
Sorry, i don't understand what R position right here. What does it means anyway?
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R n = n 1 1 1 … 1 1
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@Ben Frankel – all right then.. awesome theories
Plz give me the proof
The creator of the question picked the most disguised two digit prime, 9 1 :)
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The most disguised 2 digit composite number,you mean.
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Yes, of course. It tricked me again!
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@Ben Frankel – What's the next largest looks-like-a-prime number one should know?
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@Faraz Masroor – Well, the reason that 9 1 is good to know is because all of the other composite numbers below 100 either end in 0, 2, 4, 5, 6, 8, or the sum of their digits is 3, or they have a repeated digit (ie 7 7 ), or they are a square (ie 4 9 ). 9 1 is the only exception to these rules below 100.
I guess that the next smallest such composite number would be 7 ⋅ 1 7 = 1 1 9 .
Note that the given repunit is divisible by 1 1 1 1 1 1 1 [i.e. 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 1 0 + 1 ]
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Yes, and it's also divisible by R 1 3 = 1 1 1 1 1 1 1 1 1 1 1 1 1 for the same reason.
I misread the problem as having 911 1's.....
Let N = 1 1 1 1 . . . . . . 1 1 1 1 1 , having 91 decimal digits.
By the 9 divisibility rule, 9 ∤ N , and so we now have only two possible answers, prime or composite.
If N were prime, it would be a special kind of prime called a Repunit prime. It is known that a Repunit can only possibly be prime if the amount of digits that it has is prime. N has 91 ones, and 9 1 = 7 ⋅ 1 3 , and so N is composite.
N ∈ P
Also we know that the two possible answers are prime or composite because every number has to be prime or composite except 1 and we know that 1111111111.... is not equal to 1.
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Thanks, Willy.
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You're welcome, Noah Singer. (do you Noah singer?)
If (one of) the first 2 options were correct,then that would automatically imply that option 3 is correct.So the first 2 options make no sense since there can only be 1 answer.
91 = 7 * 13 The integer n , which consists of 91 1's, can be looked at as 13 continuous arrangements of 7 1's. That is to say, n = 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111 1111111. So, it has to be completely divisible by 1111111 and so. composite.
111...11111 (91 1's) = 1111111 * 1000000100000010000001000000100000010000001000000100000010000001000000100000010000001
Since there are 91 1's, the sum of the digits is 91, which is not divisible by 3. Therefore, the integer isn't divisible by 3, which eliminates two of the possible answers.
All that remains is to figure out whether the integer is prime or composite. Notice that 91 is divisible by 7, so the integer would be divisible by 1111111 (the integer would be equal to 1111111 * 100000010000001....) Therefore, the integer is a composite number.
nice
The integer in the question can be expressed as 9 1 ( 1 0 9 1 − 1 ) .
A number is divisible by 9 if the sum of the digits are evenly divisible by 9. Here the sum of the digits is 91, which is not divisible by 9. Hence the integer is not divisible by 9.
If the integer is not divisible by 9 it is also not divisible by 27.
Now, for exemplification, the number 111111 can be broken into 1 1 × 1 0 4 + 1 1 × 1 0 2 + 1 1 × 1 0 0 . This number is clearly divisible by 11 because each of the components are divisible by 11.
Similarly, the number consisting of 91 ones can be broken into 13 strings of 7 ones, i.e. 9 1 ( 1 0 9 1 − 1 ) = n = 0 ∑ 1 2 1 1 1 1 1 1 1 × 1 0 7 n
Since each component 1 1 1 1 1 1 1 × 1 0 7 n , where n is a positive integer, is divisible by 1 1 1 1 1 1 1 , we can conclude that the integer can be divided by 1111111. Hence the integer is a composite number.
My method of solving was that I realised that since 91 1's can be broken into 13 strings of 7 1's then clearly 1111111 divides it.
Let me set a function so you can read this solution easier.
Let F n be a function such that F n = 1 1 1 1 . . . 1 ( n 1's) for any n ∈ I + .
F 9 1 can be factored as...
F 7 × 1 0 8 4 + F 7 × 1 0 7 7 + . . . + F 7 (split F 7 into 13 groups, 7*13 = 91.)
Then factor 1111111 out we get ( F 7 ) ( 1 0 8 4 + 1 0 7 7 + . . . + 1 ) , which is a composite number .~~~
PS: You can also do F 1 3 × 1 0 7 8 + F 1 3 × 1 0 6 5 + . . . . + F 1 3 for experiments and stuffs and we still get a composite number.
PSS: There's a proof that for any m , n such that...
If you want any proofs, feel free to ask me in the comment section.
(1+1+....+1+1) 91 times = 91 = 7 *13 , hence composite!
The no. contains 7 sets of 13 times 1...thus it is composite
It cannot be divisible by 9 cuz its digits add up to 91, which is not divisible by 9, and certainly not divisible by 27. Not divisible by numbers 2 through 12, but guess composite anyway, because the prime density gets lower and lower as the range gets higher and higher, wait, it involves ln but i forgot aaaaaa
As there are 91 1's so the sum of the number is 91.As 91 is not divided by 3 or 9 so the number is not divided by 27 . Now we have 2 options :prime or composite. the number is divided by a number which contain 13 1's and another number that contains 7 1's because 13*7=91. For example: a number that contains 6 1's is divisible by the number that contains 2 1's and 3 1's. so we have only one option. the answer is definitely composite.
Let no: of 1's be n
if n is a multiple of 3 * (i.e.,n=3 p)** then, the number will be divisible by 3
if n is 1 more than a multiple of 3 (i.e.,n=3*p + 1) then, the number will be divisible by 11
Here n=91 91=90+1=3(30)+1
Therefore, the number will be divisible by 11.
Hence, It is a Composite number.
How do you say that the number will be divisible by 11?
This number will be divisible by 1111111(7 times '1') and 1111111111111(13 times '1'). It is not divisible by 9 because sum of digits is not divisible by nine.
The clue was 91, which is 13x7. i.e. if you write the number 1,111,111 (7 digits) 13 times, you get the number in question.
Call this 91-digit number y. Call the number 1,111,111 "x".
y = x + x * 10^7 + x * 10^14 + ... + x * 10^84. Clearly divisible by x. Thus y is not prime, and is composite.
clearly 1111111111.............111111111111111(91 times) is divisible by 11 so it is a composite number . since 1+1+1+1+1+..............1+1+1(91 times)=91 and 91 is not divisible by 9 or 27 so only one option is left i.e. it is a composite number. easy peasy
Actually 11111....11111 91 times is not divisible by 11. divisibility test for 11 : sum of digits at odd places = 46 sum of digits at even places = 45 11 does not divide (46 - 45 ) and hence it does not divide 1111...........111.
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If a number is divisible by 2 7 , it must be divisible by 9 as well.However, 9 ∤ 9 1 ⋅ 1 . Therefore,it cannot be divisible by 9 or 2 7 . This rules out the first 2 options.
Prime numbers are hard to spot.Therefore let us hope that our number is composite.
1 1 1 1 1 . . . . . 1 1 1 1 1 1 1 1 1 1 = 1 0 9 0 + 1 0 8 9 + . . . . . . . . . . . . . . . . . . . . . . + 1
which is
1 0 8 4 ( 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 1 0 + 1 ) + 1 0 7 7 ( 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 1 0 + 1 ) + . . . .
. . . + ( 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 1 0 + 1 )
which is obviously factorizable.More generally,
The number 1 1 1 1 . . . . . . . . . 1 1 1 1 1 1 with n 1 's is composite when n is composite.
Here, 9 1 = 1 3 ⋅ 7
and hence is factorizable.For more info,check out the Wikipedia article on repunits .