Without using Fermat's Little Theorem, how can I quickly solve $2014 ^{2015} \pmod {11}$?
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3Note that $2013$ is divisible by $11$. – André Nicolas Apr 28 '15 at 23:25
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5Just note $2014=11\times 183+1$ – Ángel Mario Gallegos Apr 28 '15 at 23:25
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2I suppose you started by computing $2014^1\pmod{11},$ right? What did you get? – bof Apr 28 '15 at 23:25
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${\rm mod}\ 11!:\ 2014\equiv 4-1+0-2\equiv 1\ \ $ – Bill Dubuque Apr 28 '15 at 23:27
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@bof I did compute 2014^1 (mod 11), and I got 1. But I am not sure what to do with the 2015 exponent. Is there a way to simplify 2014^2015? – User Apr 28 '15 at 23:30
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1@OmarN $,2014\equiv 1,\Rightarrow, 2014^n\equiv 1^n\equiv 1\ $ by the Congruence Power Rule. $\ \ $ – Bill Dubuque Apr 28 '15 at 23:30
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Can you compute $1^{2015}\pmod{11}$? – bof Apr 28 '15 at 23:31
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11^2015 (mod 11) = 1. Does that mean 2014^2015 mod 11 = 1? Is it really as simple as just doing one number at a time? – User Apr 28 '15 at 23:33
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I'm not sure what you mean by "doing one number at a time", but it really is that simple. See Bill Dubuque's comments. – bof Apr 28 '15 at 23:36
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Marios observation above + binomial theorem : every term except the last will be 0 mod 11, and the last is always a number smaller than 11. Then the procedure can be iterated if needed. – mathreadler Apr 28 '15 at 23:46
3 Answers
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This can be computed in a few seconds of mental arithmetic as follows. First we cast out $11$'s to compute $2014\pmod{11}$ as its alternating digit sum:
${\rm mod}\ 11\!:\ \color{#0a0}{10\equiv -1}\,\Rightarrow\, \color{#c00}{2014}\equiv 2(\color{#0a0}{10})^3\!+\!\color{#0a0}{10}\!+\!4\equiv 2(\color{#0a0}{-1})^3\!\color{#0a0}{-\!1}\!+\!4\equiv\color{#c00} 1\ $ [cast $11$'s - see below]
Therefore $\ \color{#c00}{2014}^n\equiv \color{#c00}1^n\equiv 1\ $ by the Congruence Power Rule.
Remark $\ $ The first line is a special case of casting out elevens, the analog of casting out nines, i.e. $\, \color{#0a0}{10\equiv -1}\,\Rightarrow\,f(\color{#0a0}{10})\equiv f(\color{#0a0}{-1}) = $ alternating digit sum, where $\,f(x)\,$ is the decimal radix polynomial, e.g. above $\, f(x) = 2x^3+x+4,\ f(10) = 2014.\,$ See here for further discussion.
Bill Dubuque
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In the second line, can you tell me where all those numbers are coming from? I am not sure where the numbers in the second line come from. – User Apr 28 '15 at 23:48
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@OmarN Because $,\color{#0a0}{10\equiv -1},$ we can replace $,\color{#0a0}{10},$ by $,\color{#0a0}{-1},$ whenever it occurs as an argument of a sum or product (by the linked Cogruence Sum and Product Rules). – Bill Dubuque Apr 28 '15 at 23:50
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The fundamental fact about modular arithmetic is that $$ ab \bmod n = (a\bmod n)(b\bmod n)\bmod n $$ (and the analogous fact for addition). Apply this rule 2014 times to $$ \underbrace{2014\cdot 2014\cdot 2014 \cdots 2014}_{2015\text{ factors}} $$ (and then 2014 more times in reverse to get rid of the intermediate "$\bmod 11$"s) and you get $$ 2014^{2015} \bmod 11 = (2014 \bmod 11)^{2015} \bmod 11 = 1^{2015} \bmod 11 = 1 $$
hmakholm left over Monica
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i.e., as I said, use the Congruence Power Rule, an inductive extension of the Congruence Power Rule.. – Bill Dubuque Apr 29 '15 at 00:03
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@Bill: Yes -- but it looked like the OP needed to see a more detailed explanation of why that rule works. – hmakholm left over Monica Apr 29 '15 at 00:13
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Indeed! My comment was meant to help the OP connect the dots between our answers. – Bill Dubuque Apr 29 '15 at 00:15
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@Henning Makholm, I understand your answer and it makes perfect sense. What if though, we wanted to apply the same logic to $$ 2014^{2015} \bmod 7 = (2014 \bmod 7)^{2015} \bmod 7 = 5^{2015} \bmod 7 = ? $$ How would you find 5^2015 mod 7 without using a calculator? – User Apr 29 '15 at 19:30
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@OmarN $5^6 \equiv 1 \bmod 7$, so find $2015 \bmod 6$, which is $5$, and calculate $5^5 \bmod 7$ which is $5^2\cdot5^3 \equiv 4\cdot-1 \equiv \color{red}{3} \bmod 7$. – Joffan Apr 29 '15 at 19:42
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This is the part I get lost at. Where did you get $5^{6}$ from? I don't understand how we went from $5^{2015}$ to $5^{6}$ – User Apr 29 '15 at 20:13
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1@OmarN Fermat's Little Theorem, $a^{p-1} \equiv 1 \bmod p$ (and $6=7-1$). You don't need this in the original question, but for cutting down the exponent when $a \not\equiv 1$, it's very useful. – Joffan Apr 29 '15 at 20:15
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I like Bill Dubuque's answer, but I definitely didn't use his approach to find $2014 \bmod 11$, just used a discarding version of long division:
$20 \bmod 11 = \color{red}{9} \\ \to \color{red}{9}1 \bmod 11 = \color{green}{3} \\ \to \color{green}{3}4 \bmod 11 = 1$
And then of course $2014^{2015} \equiv 1^{2015} \equiv 1 \bmod 11$.
If $2014$ had been a bigger number, I would probably have used the alternating digits rule for divisibility by $11$, $(0+4)-(2+1) = 1$, because I would have stopped to think before starting the division process :-)
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My way is $,2014\equiv 4-2+0-1\equiv 1,,$ which, to me, is simpler, and more general (it is casting out $11$'s - see my remark). – Bill Dubuque Apr 28 '15 at 23:53
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They're both equally quick here for someone who is skilled. I give a general description of the method you use in this answer. [Horner form polynomial evaluation] – Bill Dubuque Apr 29 '15 at 00:01