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I really have no idea how to start. The only theorem considering prime numbers I know of is Fermat's little theorem and maybe its related with binomial theorem.

Any help will be appreciated.

evil999man
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4 Answers4

14

Just think through the actual number.

$10^{91}$ is a $1$ with $91$ $0$'s after it.

$10^{91}-1$ is therefore $91$ $9$'s in a row.

$\frac{10^{91}-1}{9}$ is therefore $91$ $1$'s in a row.

Due to the form of this number, $x$ $1$'s in a row will divide it, where $x$ is a divisor of $91$.

For example $1111111$ is a divisor, so is $1111111111111$.

Hence the number is not prime.

john
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    Cute. Very pretty indeed. +1! – Robert Lewis May 17 '14 at 06:48
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    Really, you're not thinking of it in terms of "the actual number" here; you're thinking of it in terms of the decimal representation. The decimal representation isn't any more "the actual number" than the expression $\frac{10^{91}-1}{9}$, and this argument wouldn't work in hexadecimal. Still, upvoted. – user2357112 May 17 '14 at 06:57
  • Perhaps knowing too much theorems reduces your strike rate of easier problems. :D – evil999man May 17 '14 at 07:02
  • Yep, poor wording but that's what I meant by thinking of the actual number. A lot of the time the initial reaction (mine too) is big, scary, clearly I'm just supposed to just throw a theorem at it. Sometimes things aren't as bad as they seem though :) – john May 17 '14 at 07:05
  • Great solution but i guess it should be 90 9's instead of 91 , elegant solution though – avz2611 May 17 '14 at 07:39
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    no, 91 9's. Try taking some simpler examples, e.g. $10^3-1=1000-1=999$ – john May 17 '14 at 07:51
  • No 91 1s is not prime it has many divisors But it is not divisible by 7 1s or 13 1s, – Neel Basu Nov 27 '16 at 18:44
11

Hint: Note that $91=7\cdot 13$, so the numerator is divisible by $10^7-1$.

If this is not obvious, it may be clearer if we let $x=10^7$. Then $10^{91}=x^{13}$, and it is a familiar fact that $x-1$ divides $x^n-1$ for any positive integer $n$.

André Nicolas
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1

Using Why $a^n - b^n$ is divisible by $a-b$?,

$$a^{91}-1=(a^{13})^7-1^7$$ is divisible by $a^{13}-1$ and by similarly by $a^7-1$

Again, using Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$,

$$(a^7-1,a^{13}-1)=a^{(7,13)}-1=a-1$$

Hence, $\displaystyle\frac{a^{13}-1}{a-1}, \frac{a^7-1}{a-1}$ are co-prime primes and divisors of $\displaystyle\frac{a^{91}-1}{a-1}$

Community
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lab bhattacharjee
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You have

$$111111....1=1+10+10^2.....+10^{90}=(10^{84}+10^{77}+10^70.....+10^7+1) (10^6+10^5+10^4+ \dots + 10^0);$$ this is product of two integers 2 or greater, so it is not prime.

ncmathsadist
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suraj
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