How to show that \; 11^{n} -6\; is divisible by 5 Without induction
idontknow Dec 2015 1,076 166 Earth Jan 1, 2019 #1 How to show that \(\displaystyle \; 11^{n} -6\; \) is divisible by \(\displaystyle 5\) Without induction
How to show that \(\displaystyle \; 11^{n} -6\; \) is divisible by \(\displaystyle 5\) Without induction
idontknow Dec 2015 1,076 166 Earth Jan 1, 2019 #3 Just posted to work without induction Last edited: Jan 1, 2019
topsquark Math Team May 2013 2,510 1,049 The Astral plane Jan 1, 2019 #4 idontknow said: How to show that \(\displaystyle \; 11^{n} -6\; \) is divisible by \(\displaystyle 5\) Without induction Click to expand... This seems to only work for odd n... -Dan
idontknow said: How to show that \(\displaystyle \; 11^{n} -6\; \) is divisible by \(\displaystyle 5\) Without induction Click to expand... This seems to only work for odd n... -Dan
M [email protected] Oct 2009 942 365 Jan 1, 2019 #5 topsquark said: This seems to only work for odd n... -Dan Click to expand... It doesn't. It works for all n. Can you check again and perhaps post why you don't think it works? I'm curious whether I did something wrong. Reactions: 1 person
topsquark said: This seems to only work for odd n... -Dan Click to expand... It doesn't. It works for all n. Can you check again and perhaps post why you don't think it works? I'm curious whether I did something wrong.
romsek Math Team Sep 2015 2,958 1,673 USA Jan 1, 2019 #6 $11^n - 6 \pmod{5} = $ $(2\cdot 5 + 1)^n - 6 \pmod{5} = $ $\left(\sum \limits_{k=0}^n~\dbinom{n}{k}(2\cdot 5)^k\right) - 6 \pmod{5} = $ $1 - 1 \pmod{5} = $ $0 \pmod{5}$ Reactions: 1 person
$11^n - 6 \pmod{5} = $ $(2\cdot 5 + 1)^n - 6 \pmod{5} = $ $\left(\sum \limits_{k=0}^n~\dbinom{n}{k}(2\cdot 5)^k\right) - 6 \pmod{5} = $ $1 - 1 \pmod{5} = $ $0 \pmod{5}$
topsquark Math Team May 2013 2,510 1,049 The Astral plane Jan 1, 2019 #7 Pffl!! :furious: I was using 11n, not \(\displaystyle 11^n\). Sorry about that. :cry: -Dan Reactions: 1 person
skipjack Forum Staff Dec 2006 21,473 2,466 Jan 1, 2019 #8 [email protected] said: It works for all n. Click to expand... No, it works for every non-negative integer $n$. For such $n$, x - 1 is a factor of the polynomial x$^n$ - 1. Hence 11$^n$ - 1 is a multiple of 10, and so 11$^n$ - 6 is divisible by 5. Reactions: 1 person
[email protected] said: It works for all n. Click to expand... No, it works for every non-negative integer $n$. For such $n$, x - 1 is a factor of the polynomial x$^n$ - 1. Hence 11$^n$ - 1 is a multiple of 10, and so 11$^n$ - 6 is divisible by 5.
greg1313 Forum Staff Oct 2008 8,008 1,174 London, Ontario, Canada - The Forest City Jan 1, 2019 #9 \(\displaystyle (5+6)^n-6\rightarrow5^n\cdots6^n-6\rightarrow5^n-6(6^{n-1}-1)\) Six to any positive power is a number with six as the last digit. I will leave it to the reader to finish up. Last edited: Jan 2, 2019 Reactions: 1 person
\(\displaystyle (5+6)^n-6\rightarrow5^n\cdots6^n-6\rightarrow5^n-6(6^{n-1}-1)\) Six to any positive power is a number with six as the last digit. I will leave it to the reader to finish up.
greg1313 Forum Staff Oct 2008 8,008 1,174 London, Ontario, Canada - The Forest City Jan 2, 2019 #10 Of course, one could apply the same sort of idea to $11^n$ ...