Last 2 digits of $143^{101}$ in base 10.
I have to use Fermat's theorem or Euler's theorem but I dont know where can I started.
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Fermat's little theorem, $a^{p-1} \equiv 1 \pmod p$ for a prime $p$ – UmbQbify Jun 28 '20 at 09:51
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And thus wiki page has enough information about Euler's theorem https://en.m.wikipedia.org/wiki/Euler's_theorem – UmbQbify Jun 28 '20 at 09:54
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1Does this answer your question? How do I compute $a^b,\bmod c$ by hand? – José Carlos Santos Jun 28 '20 at 09:55
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See https://math.stackexchange.com/questions/679842/find-the-last-two-digits-of-345 – lab bhattacharjee Jun 28 '20 at 10:05
3 Answers
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Hint:
This amounts to calculating $143^{101}\bmod 100$.
Now $143\equiv 43\bmod 100$ is coprime to $100$, so we can apply Euler's theorem. As $\varphi(100)=\varphi (2^2\cdot 5^2)=2\cdot(5\cdot4)=40$, you ultimately have calculate $43^{101\bmod 40}=43^{21}\bmod 100$.
Last, you can use the fast exponentiation algorithm, which works with repeated squaring. On the other hand, the order of $43\bmod100$ is a divisor of $40$. Repeatedly squaring will show you this order is a power of $2$.
Bernard
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1But I use Euler's theorem, since I calculate the exponent modulo $\varphi(100)$! I only added remarks to make the computation by hand easier. – Bernard Jun 28 '20 at 10:44
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As we need to find last two digits, we should divide by 100. Here is the full solution.
$$Find \space a^{x} \space mod \space n$$ $$a=143,x=101,n=100$$ $$We \space also \space know \space a^{φ(n)} \equiv 1 \space (\bmod 100) \text { if n and a are co-prime}...$$ $$\text {and 100 and 143 are co-prime.}$$ $$φ(100)=φ(2^2⋅5^2)=2^1.(2-1)⋅5^1.(5-1) = 40 \text {, where 2 and 5 are prime.} $$ $$So, 143^{40} \equiv 1 (\bmod100)$$ $$So, 143^{101} \equiv 143^{40.2+21}\equiv(143^{40})^2.143^{21} \equiv 1^2.143^{21}\space (\bmod100) \space \text { ...(A)}$$
From here, I will show two ways.
Option 1: $$Now, 143^{21} = (13.11)^{21}$$ $$So,143^{21} \bmod100 \equiv (13^{21} \bmod 100).(11^{21}\bmod100)$$ $$13^3 \bmod 100 = -3, \text {as the number ends in 97}$$ $$13^{21} \bmod 100 \equiv (13^3 \bmod 100)^7 \bmod100$$ $$13^{21} \bmod 100 \equiv (-3)^7 \bmod100$$ $$13^{21} \bmod 100 \equiv -87 \bmod100$$ $$13^{21} \bmod 100 \equiv 13 \bmod100$$ $$\text{We also know } 11^{10} \bmod 100 = 1$$ $$\text {or you can find the same way I did for 13}.$$ $$11^{21} \bmod 100 \equiv [(11^{10} \bmod 100)^2.11^1] (\bmod100) \equiv 11 \bmod 100$$ $$\text {Let's go back to, } 143^{21} \bmod100 \equiv (13^{21} \bmod 100).(11^{21}\bmod100)$$ $$So, 143^{21} \bmod100 \equiv (13 \bmod 100).(11\bmod100)$$ $$So, 143^{101} \bmod100 \equiv 143^{21} \bmod100 \equiv 143 \bmod 100 = 43$$
Or from point A,
Option 2: $$\text {We know, }(x+a)^n \bmod x \equiv a^n \bmod x$$ where x, a and n are positive integers. $$So,(100+43)^{21} \bmod 100 \equiv 43^{21} \bmod 100$$ $$\text {We know or can easily check that, } 43^4 \bmod 100 = 1$$ $$So,43^{21} \bmod 100 \equiv (43^4 \bmod 100)^5.43^1 (\bmod100)$$ $$So,43^{21} \bmod 100 \equiv (1^5.43^1) \bmod100 = 43.$$
Math Lover
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I realize I have made a blunder in my proof. Apologies. Please reject the answer so I can delete it. I will send you the proof shortly. – Math Lover Jun 28 '20 at 11:31
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Now it is correct and you can give me credit :) Apologies for the last time. – Math Lover Jun 28 '20 at 14:27
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Thank you very much :) u are very kind,no need for an apology for the last time :) – carla carta Jun 28 '20 at 15:44
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Hint: Compute $143^{101} \bmod 4$ and $143^{101} \bmod 25$. These are easy because of Euler's theorem.
lhf
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We cannot break up 100 into 4 and 25 to do this. mod 4 will give 3 and mod 25 will give 18. – Math Lover Jun 28 '20 at 15:16
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1@MathLover. Use the Chinese remainder theorem to combine those two solutions into $43$ – lhf Jun 28 '20 at 15:17
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Good point. Yes, that can also be done. But as 43^4 mod 100 = 1, I felt it was easier to just do that from the point of 43^21 mod 100. – Math Lover Jun 28 '20 at 15:26