All exclamation marks - Part 2

Number Theory Level 1

Which of the following is a divisor of the number 10 ! + 1 10! + 1 ?

See Part 1 and Part 3 .

10 12 13 11

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Chung Kevin by Chung Kevin , 45, Australia

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3 solutions

By Wilson's Theorem we know that

( 11 − 1 ) ! ≡ − 1 ( m o d 11 ) ⟹ 10 ! + 1 ≡ 0 ( m o d 11 ) . (11 - 1)! \equiv -1 \pmod{11} \Longrightarrow 10! + 1 \equiv 0 \pmod{11}.

Looking at the other options, we know that 10 ! + 1 10! + 1 has a last digit of 1 , 1, and hence is not divisible by either 10 10 or 12. 12. And for 13 , 13, again by Wilson's Theorem we know that

12 ! ≡ − 1 ( m o d 13 ) ⟹ 12 ∗ 11 ∗ 10 ! ≡ − 1 ( m o d 13 ) . 12! \equiv -1 \pmod{13} \Longrightarrow 12*11*10! \equiv -1 \pmod{13}.

Now if it we were to assume that 10 ! ≡ − 1 ( m o d 13 ) 10! \equiv -1 \pmod{13} then we would have

12 ∗ 11 ∗ ( − 1 ) ≡ − 1 ( m o d 13 ) ⟹ − 132 ≡ − 1 ( m o d 13 ) . 12*11*(-1) \equiv -1 \pmod{13} \Longrightarrow -132 \equiv -1 \pmod{13}.

But − 132 ≡ − 2 ( m o d 13 ) , -132 \equiv -2 \pmod{13}, so by contradiction we know that 10 ! + 1 10! + 1 is not divisible by 13. 13.

Thus of the given options, only 11 \boxed{11} is a divisor of 10 ! + 1. 10! + 1.

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Or you can just do this: 12 ⋅ 11 ⋅ 10 ! ≡ − 1 ( m o d 13 ) ( − 2 ) ( − 1 ) 10 ! ≡ 12 ( m o d 13 ) 10 ! ≡ 6 ( m o d 13 ) \begin{aligned}12\cdot11\cdot10! &\equiv & -1 \pmod{13} \\ (-2) (-1) 10! & \equiv& 12 \pmod{13} \\ 10!& \equiv& 6 \pmod{13} \end{aligned}

Bonus question : By only using this list of primes as a reference, is it possible to show that ( 10 ! + 1 ) (10! + 1) only has two distinct prime factors?

I got it right instinctively. Probably got lucky.

Vinícius Melo - 6 years ago
Hadia Qadir
Jul 22, 2015

10 + 1 =3628801 which is visi: Or you can just do this:

Bonus question: By only using this list of primes as a reference, is it possible to show that ( 10 +1) only has two distinct prime factors?

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice bonus question :)

Chung Kevin - 5 years, 10 months ago

Simple, 10! Ends in 00, which is a perfect multiple of 10 and when when divided by 11 gives 1 as remainder, so when 1 is added it will be a perfect multi ple of 11

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

So you're saying that 10 ! 10! gives a remainder of 1 1 when divided by 11 11 ? Which means 10 ! + 1 10! + 1 gives a remainder of 1 + 1 = 2 1+1=2 when divided by 11 11 . Your solution made no sense. The proper way to solve this is via Wilson's Theorem .

100 also ends up with two zeroes but does not yield a remainder 1 when divided by 11. So how can you be so sure about it will give a remainder 1 when divided by 11. Also you can also see that crieria for divisibility by 11 doesnt match anyway with your answer.

Shubham Tuteja - 5 years, 12 months ago

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