## anonymous 4 years ago If 7^(1983) is divided by 100, what is the remainder?

1. anonymous

43

2. anonymous

but how? could you explain it a bit more?

3. anonymous

There are many ways we could solve this, I have used Binomial + Euler-Fermat's theorem.

4. 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.

5. anonymous

and this is from your text book?

6. anonymous

from tutoring haha

7. anonymous

well they wouldn't have put it in the test if it was too hard but thanks anyway (:

8. 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! $43^3 = 7 \mod 100 \implies 43^6 = 49 \mod 100$ $\implies 43^{12} = 49^2 = 2401 = 1 \mod 100$ $\implies (43^{12}) ^{55} = 1 \mod 100 \implies 43^{660} = 1 \mod 100$

9. anonymous

Now just multiply both sides by 43.You have your answer! NOTE : The" = " stands for congruence.

10. 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)$

11. 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 $7^3=43$

12. anonymous

*4 choices of remainder obviously, sorry

13. anonymous

Don't complicate it satellite!