The "1" Problem

The integer 1111 11111 1111 \ldots \ldots 11111 , which consists of 91 1's is :-

Divisible by 9 but not 27 Divisible by 27 A Composite Number A Prime Number

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14 solutions

Rahul Saha
Jan 24, 2014

If a number is divisible by 27 27 , it must be divisible by 9 9 as well.However, 9 91 1 9\nmid 91\cdot 1 . Therefore,it cannot be divisible by 9 9 or 27 27 . This rules out the first 2 2 options.

Prime numbers are hard to spot.Therefore let us hope that our number is composite.

11111.....1111111111 = 1 0 90 + 1 0 89 + . . . . . . . . . . . . . . . . . . . . . . + 1 11111.....1111111111=10^{90}+10^{89}+......................+1

which is

1 0 84 ( 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 10 + 1 ) + 1 0 77 ( 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 10 + 1 ) + . . . . 10^{84}(10^6+10^5+10^4+10^3+10^2+10+1)+10^{77}(10^6+10^5+10^4+10^3+10^2+10+1)+....

. . . + ( 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 10 + 1 ) ...+(10^6+10^5+10^4+10^3+10^2+10+1)

which is obviously factorizable.More generally,

The number 1111.........111111 1111.........111111 with n n 1 1 's is composite when n n is composite.

Here, 91 = 13 7 91=13\cdot 7

and hence is factorizable.For more info,check out the Wikipedia article on repunits .

For even more more more generally...

If g c d ( n , m ) = k gcd(n,m)=k , then g c d ( R n , R m ) = R k gcd(R_{n},R_{m})=R_{k} (and vice versa)

If you want a proof for this statement, ask me for that.

Samuraiwarm Tsunayoshi - 7 years, 4 months ago

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I tried it a bit myself.What do you use?Induction on k k or double induction on m , n m,n ?

Rahul Saha - 7 years, 4 months ago

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Almost the same method as you solve this question.

Samuraiwarm Tsunayoshi - 7 years, 4 months ago

Sorry, i don't understand what R position right here. What does it means anyway?

Hafizh Ahsan Permana - 7 years, 4 months ago

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R n = 111 11 n R_n = \underbrace{111 \dots 11}_\text{n}

Ben Frankel - 7 years, 4 months ago

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@Ben Frankel all right then.. awesome theories

Hafizh Ahsan Permana - 7 years, 4 months ago

Plz give me the proof

Vedant Mittal - 7 years, 4 months ago

The creator of the question picked the most disguised two digit prime, 91 91 :)

Ben Frankel - 7 years, 4 months ago

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The most disguised 2 digit composite number,you mean.

Rahul Saha - 7 years, 4 months ago

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Yes, of course. It tricked me again!

Ben Frankel - 7 years, 4 months ago

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@Ben Frankel What's the next largest looks-like-a-prime number one should know?

faraz masroor - 7 years, 4 months ago

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@Faraz Masroor Well, the reason that 91 91 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 77 77 ), or they are a square (ie 49 49 ). 91 91 is the only exception to these rules below 100.

I guess that the next smallest such composite number would be 7 17 = 119 7 \cdot 17 = 119 .

Ben Frankel - 7 years, 4 months ago

Note that the given repunit is divisible by 1111111 1111111 [i.e. 1 0 6 + 1 0 5 + 1 0 4 + 1 0 3 + 1 0 2 + 10 + 1 10^6+10^5+10^4+10^3+10^2+10+1 ]

Rahul Saha - 7 years, 4 months ago

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Yes, and it's also divisible by R 13 = 1111111111111 R_{13} = 1111111111111 for the same reason.

Ben Frankel - 7 years, 4 months ago

I misread the problem as having 911 1's.....

Xuming Liang - 7 years, 4 months ago

If You Liked This Then You Can Try This

Anand Raj - 7 years, 3 months ago
Ben Frankel
Jan 24, 2014

Let N = 1111......11111 N = 1111......11111 , having 91 decimal digits.

By the 9 divisibility rule, 9 N 9 \nmid N , and so we now have only two possible answers, prime or composite.

If N 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 N has 91 ones, and 91 = 7 13 91 = 7 \cdot 13 , and so N N is composite.

N ∉ P \boxed{N\not\in\mathbb{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.

William Cui - 7 years, 4 months ago

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Thanks, Willy.

Noah Singer - 7 years, 4 months ago

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You're welcome, Noah Singer. (do you Noah singer?)

William Cui - 7 years, 4 months ago

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.

Rahul Saha - 7 years, 4 months ago
Shamik Banerjee
Jan 24, 2014

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

Tara Liu
Jan 25, 2014

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

Joe Aliyev - 7 years, 4 months ago
Louie Tan Yi Jie
Jan 25, 2014

The integer in the question can be expressed as 1 9 ( 1 0 91 1 ) \frac{1}{9} \left(10^{91}-1\right) .

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 11 × 1 0 4 + 11 × 1 0 2 + 11 × 1 0 0 11\times 10^4+11\times 10^2+11\times 10^0 . 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. 1 9 ( 1 0 91 1 ) = n = 0 12 1111111 × 1 0 7 n \frac{1}{9} \left(10^{91}-1\right)=\sum _{n=0}^{12} 1111111\times 10^{7 n}

Since each component 1111111 × 1 0 7 n 1111111\times 10^{7 n} , where n n is a positive integer, is divisible by 1111111 1111111 , 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.

Leon Fernandes - 7 years, 4 months ago

Let me set a function so you can read this solution easier.

Let F n F_{n} be a function such that F n = 1111...1 F_{n} = 1111...1 ( n n 1's) for any n I + n \in I^{+} .

F 91 F_{91} can be factored as...

F 7 × 1 0 84 + F 7 × 1 0 77 + . . . + F 7 F_{7}\times 10^{84} + F_{7}\times 10^{77} + ... + F_{7} (split F 7 F_{7} into 13 groups, 7*13 = 91.)

Then factor 1111111 out we get ( F 7 ) ( 1 0 84 + 1 0 77 + . . . + 1 ) (F_{7})(10^{84} + 10^{77} + ... + 1) , which is a composite number .~~~

PS: You can also do F 13 × 1 0 78 + F 13 × 1 0 65 + . . . . + F 13 F_{13} \times 10^{78} + F_{13} \times 10^{65} + .... + F_{13} for experiments and stuffs and we still get a composite number.

PSS: There's a proof that for any m , n m,n such that...

  • If n m n|m , then F n F m F_{n}|F_{m} .
  • If g c d ( n , m ) = 1 gcd(n,m) = 1 , then g c d ( F n , F m ) = 1 gcd(F_{n},F_{m}) = 1 .
  • (For more generalized) If g c d ( n , m ) = k gcd(n,m) = k , then g c d ( F n , F m ) = F k gcd(F_{n},F_{m}) = F_{k} .

If you want any proofs, feel free to ask me in the comment section.

HariShankar Pv
Mar 14, 2014

(1+1+....+1+1) 91 times = 91 = 7 *13 , hence composite!

Vaibhav Agarwal
Mar 1, 2014

The no. contains 7 sets of 13 times 1...thus it is composite

Benjamin Wong
Feb 12, 2014

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

Mahbubur Rahman
Feb 7, 2014

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?

Akash D - 7 years, 4 months ago
Meet Udeshi
Jan 26, 2014

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.

Dilip DSouza
Jan 26, 2014

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.

Pranav Khandka
Jan 25, 2014

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.

Leon Fernandes - 7 years, 4 months ago

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