I don't know why I am wrong with this problem. Here is what I did: The last two digit of $6^{2006}$ is 36. So the answer should be 1.
Find the remainder when $6^{2006}$ is divided by 35.
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Martin Sleziak
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8$6^{2006}\equiv 36^{1003}\equiv 1^{1003}\equiv 1\pmod{35}$. – user236182 May 22 '16 at 13:53
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Thanks. So the answer sheet is wrong. – May 22 '16 at 13:55
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10Your reasoning is wrong, for example $136\equiv 31 \mod 35$. – Mathematician 42 May 22 '16 at 13:55
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5It is not true that the last two digits of $6^{2006}$ are $36$, they are $56$, but that does not help you with $6^{2006} \pmod {35}$ – Ross Millikan May 22 '16 at 13:57
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3The last two digits of $6^{2006}$ are $56$, not $36$. http://www.wolframalpha.com/input/?i=6%5E(2006)+mod+100 – user236182 May 22 '16 at 13:57
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@FSK It seems that your answer sheet says $1$, and that's indeed the answer. What's wrong? – user236182 May 22 '16 at 13:59
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136 mod 35 = 31 That trick, taking the last digit, only works for 10 in base 10, 2 in base 2, etc. – yesennes May 22 '16 at 17:23
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One way to show that $6^{2006} \equiv 56 \pmod{100}$: $6^7 \equiv 6^2 \pmod{100}$, so the last two digits of $6^n$ repeat on a cycle of length $5$ for $n\geq2$; but $2006 \equiv 1 \pmod5$, so $6^{2006} \equiv 6^6 \equiv 56 \pmod{100}$. – David K May 22 '16 at 19:50
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1For this type of problem in general, see How do I compute $a^b,\bmod c$ by hand? – Martin Sleziak May 22 '16 at 23:53
3 Answers
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Not sure how you've figured "The last two digits of $6^{2006}$ are $36$", but here's my approach:
$$6^2\equiv36\equiv1\pmod{35}\implies6^{2006}\equiv(6^2)^{1003}\equiv1^{1003}\equiv1\pmod{35}$$
barak manos
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The last two digits of $6^{2006}$ are $56$, not $36$. http://www.wolframalpha.com/input/?i=6%5E(2006)+mod+100 – user236182 May 22 '16 at 13:57
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1@user236182: And even if they were $36$, it still wouldn't imply that the remainder of the number divided by $35$ is $1$. – barak manos May 22 '16 at 14:00
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Here is another approach. For solving problems$\pmod{N}$, we can try finding the prime factors of $N$, then solving the problem modulo each of those factors. Sometimes this is easier than directly solving $\pmod{N}$.
Then something called the Chinese Remainder Theorem lets us combine the results for each factor to produce the result for $N$.
In this case the prime factors of $35$ are $5$ and $7$. Since $6 \equiv 1 \pmod{5}$ and $6 \equiv -1 \pmod{7}$, we see straightaway that:
$6^{2006} \equiv 1\pmod 5$
$6^{2006} \equiv 1 \pmod 7$ , since $2006$ is even
The Chinese Remainder Theorem tells us that, given congruences $N \equiv a \pmod{p}$ and $N \equiv b \pmod {q}$ , then there is a unique $x$ such that $N \equiv x \pmod{pq}$, and it is the same $x$ for all $N$ (and this generalizes to multiple factors).
In the general case you may have to follow an algorithm or use brute force to find $x$. In this case it is right in front of us: letting $N = 1$ satisfies our two congruences. So $1 \pmod{35}$ is the answer we are looking for.
M.M
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