Find-the-three-last-digits-of-5-9999-

Question Number 86018 by Maclaurin Stickker last updated on 26/Mar/20

F i n d t h e t h r e e l a s t d i g i t s o f 5^{9999} .

Answered by MJS last updated on 26/Mar/20

5^{1} 005

5^{2} 025

5^{3} 125

5^{4} 625

5^{5} 125

5^{6} 625

\dots

5^{2 n + 1} 125

5^{2 n} 625

9999 = 2 \times 4999 + 1 \Rightarrow answer is 125

Answered by Serlea last updated on 26/Mar/20

5 = 5 (\mod 1000)

5^{2} = 25 (\mod 1000)

5^{4} = 625 (\mod 1000)

5^{6} = 625 (\mod 1000)

So for all even number^{″} x^{'}

5^{x} = 625 (\mod 1000) if x ⩾ 4

5^{3} = 125 (\mod 1000)

5^{5} = 125 (\mod 1000)

So for all odd number^{″} Y^{″}

5^{y} = 125 (mld 1000) if Y ⩾ 3

With that

5^{9999} = 5^{9998 + 1} = (5^{9998}) (5^{1})

= 625 \times 5 (\mod 1000)

= 5^{4} (\mod 1000)

= 125

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