anonymous
  • anonymous
If 7^(1983) is divided by 100, what is the remainder?
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
43
anonymous
  • anonymous
but how? could you explain it a bit more?
anonymous
  • anonymous
There are many ways we could solve this, I have used Binomial + Euler-Fermat's theorem.

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anonymous
  • anonymous
I am only a year 8 and I can understand the binomial formula but it would take too long and I don't really understand the Euler Fermat's theorem.
anonymous
  • anonymous
and this is from your text book?
anonymous
  • anonymous
from tutoring haha
anonymous
  • anonymous
well they wouldn't have put it in the test if it was too hard but thanks anyway (:
anonymous
  • anonymous
You don't need Fermat or Euler or Binomial Theorem! Here : \[7^{3} = 43 \mod 100 \implies (7^{3})^{661} = 43^{661} \mod 100\] Observe that 1983 = 3 * 661.Therefore it suffices to find the remainder when 43^661 is divided by 100! \[43^3 = 7 \mod 100 \implies 43^6 = 49 \mod 100\] \[\implies 43^{12} = 49^2 = 2401 = 1 \mod 100\] \[\implies (43^{12}) ^{55} = 1 \mod 100 \implies 43^{660} = 1 \mod 100\]
anonymous
  • anonymous
Now just multiply both sides by 43.You have your answer! NOTE : The" = " stands for congruence.
nikvist
  • nikvist
\[7^4\equiv 1\enspace(mod\enspace 100)\]\[7^{1980}=(7^4)^{495}\equiv 1\enspace(mod\enspace 100)\]\[7^{3}\equiv 43\enspace(mod\enspace 100)\]\[7^{1983}=7^{1980}\cdot 7^{3}\equiv 43\enspace(mod\enspace 100)\]
anonymous
  • anonymous
very fancy math. here is how a bone head does it, in case it is not clear. \[7^0=1,7^1=7,7^2=49,7^3=343,7^4=2401\] and once you see the "one" at the end of 2401, you know the pattern will repeat so that when you divide by 100 you only have 3 choices of remainders, 1,7,49,43 and they repeat in that order. to figure out which one you have, divide 1883 by 4 and take the integer remainder, which is fairly clearly 3 because 4 divides 1880 evenly, leaving 3 as the remainder. so your choice is \[7^3=43\]
anonymous
  • anonymous
*4 choices of remainder obviously, sorry
anonymous
  • anonymous
Don't complicate it satellite!

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