Find the last four digits of each of the numbers $3^{1000}$ and $3^{1997}$.
Problem
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Tags: number theory
13.07.2021 04:46
How is this a grade 4 problem?
13.07.2021 07:28
By Euler, $3^{\varphi(625)}\equiv 1\pmod{625}$ . $$\implies 3^{500}\equiv 1\pmod{625}$$$$\implies 3^{1000}\equiv 1\pmod{625} \ \ \ \ \ \ \ \ (1)$$Because $6561\equiv 3^8\equiv 1\pmod{16}$ . $$\implies 3^{125\cdot 8}\equiv 1^{125}\pmod {16}$$$$\implies 3^{1000}\equiv 1\pmod{16} \ \ \ \ \ \ \ \ (2)$$$(1), (2)$ implied that $$3^{1000}\equiv 1\pmod{10000} \ \ \Box$$Or just use $3^{\varphi(10000)}\equiv 3^{1000}\equiv 1\pmod{10000}$ .
13.07.2021 07:41
Because $3^{1000}\equiv 1\pmod{10000}$ . $$\implies 3^{2000}\equiv 1\equiv 80001\equiv 27\cdot 2963\pmod{10000}$$$$\implies 3^{1997}\equiv 2963\pmod{10000} \ \ \Box$$
13.07.2021 07:46
Pi_Eater_31415 wrote: How is this a grade 4 problem? its not for 4th graders, its like the 4th year of high school or smth like that