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To find last two digits of $2^{100}$

I have just learned about modular arithmetic and I wanted to solve this problem. I only know about equivalence classes and about $a=b \pmod n$. I have also learned about multiplication and addition of classes. Can someone explain to me step by step on how to apply modular arithmetic to this problem?

Thanks

Martin Sleziak
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Gathdi
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    See http://math.stackexchange.com/questions/607829/get-the-last-two-digits-of-16100-and-17100 – lab bhattacharjee Jul 31 '16 at 07:52
  • @labbhattacharjee i need to know why and how of method and not the mechanical procedure – Gathdi Jul 31 '16 at 07:53
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    Gathdi, just try it out. Experiment what happens with exponents smaller than $100$. You will find a period. It is not very short, but not excessively long either. After you see what happens, you can then take a look at it, and see what modular arithmetic is necessary to justify the result. – Jyrki Lahtonen Jul 31 '16 at 07:57
  • @JyrkiLahtonen, +1 – Andreas Caranti Jul 31 '16 at 11:17
  • @JyrkiLahtonen What do you mean by period? – Gathdi Jul 31 '16 at 14:58
  • That you will find an integer $k$ such that for all (large enough) integer $n$ the last two digits of $2^n$ and $2^{n+k}$ are the same. – Jyrki Lahtonen Jul 31 '16 at 15:15
  • Generally speaking, this website is not for giving lectures. If you don't know how to do something as obvious as modular exponentiation even the normal slow way, then your instructors suck. Given that you say you learned about "equivalence classes" without knowing how to do the basic arithmetic, that pretty much confirms it. Your instructors are awful, look somewhere else to learn this subject. – DanielV Jul 31 '16 at 18:45
  • @DanielV Can you provide some link to learn abou all this from beginning which has lots of examples – Gathdi Aug 01 '16 at 03:53

6 Answers6

11

Method 1(using fast power algorithm):

$2^0 \equiv 1 \pmod {100}$

$2^1 \equiv 2 \pmod {100}$

$2^2 \equiv 4 \pmod {100}$

$2^4 \equiv 16 \pmod {100}$

$2^8 \equiv 56 \pmod {100}$

$2^{16} \equiv 36 \pmod {100}$

$2^{32} \equiv 96 \pmod {100}$

$2^{64} \equiv 16 \pmod {100}$

Since $100 = 4 + 32 + 64$

$2^{100} \equiv 2^{4} 2^{32} 2^{64}\equiv 16 \times 96 \times 16 \equiv 76 \pmod {100}$

Method 2(using minimal cycle to improve):

Since $2^{22} \equiv 4 \pmod {100}$, $2^{100} \equiv 2^{22 \times 4 + 12} \equiv 4^4 \times 2^{12} \equiv 2^{20} \equiv 16 \times 96 \equiv 76 \pmod{100} $

Last but not least: Since $2^{20} \equiv 76 \pmod {100}$, $76 \times 76 \equiv 76 \pmod{100}$, $2^{100} = 2^{20 \times 5} \equiv 76^5 \equiv 76 \pmod{100} $ by induction.

Zau
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$$2^{100} \equiv 4^{50} \equiv 0 \pmod {4}$$ $$2^{100} \equiv 1024^{10} \equiv (-1)^{10} \equiv 1 \pmod {25}$$

So what number less than $100$ is $0 \pmod 4$ and $1 \pmod {25}$ ?

DanielV
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$2^{100}\bmod{100}=$

$2^{10\cdot10}\bmod{100}=$

$(2^{10})^{10}\bmod{100}=$

$1024^{10}\bmod{100}=$

$24^{10}\bmod{100}=$

$24^{3\cdot3+1}\bmod{100}=$

$(24^{3})^{3}\cdot24\bmod{100}=$

$13824^{3}\cdot24\bmod{100}=$

$24^{3}\cdot24\bmod{100}=$

$13824\cdot24\bmod{100}=$

$24\cdot24\bmod{100}=$

$576\bmod{100}=$

$76\pmod{100}$

barak manos
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Use the Chinese remainder theorem: $$\mathbf Z/100\mathbf Z\simeq \mathbf Z/4\mathbf Z\times\mathbf Z/25\mathbf Z.$$

Now $2^{100}\equiv 0\mod 4$, and $\;2^{100}=(2^{20})^5\equiv 1^5=1\mod 25\;$ by Euler's theorem.

So we have to solve the system of linear congruences: $\;\begin{cases} x\equiv 0\mod4,\\x\equiv 1\mod 25.\end{cases}$

From the Bézout's relation $\;25-6\cdot 4=1$, we deduce the solutions: $$x\equiv 0\cdot 24-1\cdot 6\cdot 4=-24\equiv 76\mod 100.$$

Bernard
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$2^{10}=1024$

$2^{100}=1024^{10}$

$2^{11} \equiv48$

$ 2^{12}\equiv96\equiv-2^2$

$ 2^{100}\equiv2^{12*8+2^4}\equiv-{2^2}^8*2^4$

$ =-2^{20}=-2^2*256=-24\equiv76 \pmod {100}$

another method

By euler's function, $2^{\phi(25)}=2^{20}\equiv1 \pmod{25}$

$2^{100}\equiv1,26,51,76 \pmod{100}$

$2^{100}$ is multiple of 4, so $76$.

Takahiro Waki
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${\rm mod}\ 25\!:\ \ \ \ \color{#0a0}{2^{\large 10}} = 1024\equiv \color{#0a0}{\bf -1}\,\Rightarrow$ $\,\color{#c00}{2^{\large 90}}\equiv (\color{#0a0}{2^{\large 10}})^{\large 9}\equiv (\color{#0a0}{\bf -1})^{\large 9}\equiv \color{#c00}{\bf -1}$

${\rm mod}\ 100\!:\,\ 2^{\large 100} = 2^{\large 10} \color{#c00}{2^{\large 90}} = 2^{\large 10}(\color{#c00}{\bf -1}+25n) \equiv \color{#c00}{\bf -}2^{10}\!+0\equiv -1024\equiv -24 $

Bill Dubuque
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  • All mental arithmetic, no use of Euler/Fermat or CRT etc (but of course you should learn those methods because they will be essential for more general problems). – Bill Dubuque Jul 31 '16 at 14:54