find-the-last-three-digits-of-4-2-42-Mohammed-Alwan-

Question Number 191846 by malwan last updated on 01/May/23

f i n d t h e l a s t t h r e e d i g i t s

o f 4^{2^{42}}

M o h a m m e d A l w a n

Answered by AST last updated on 01/May/23

4^{2^{42}} = 2^{2^{43}} \equiv x (m o d 1000) \Rightarrow 2^{2^{43} - 3} \equiv \frac{x}{8} (m o d 125)

ϕ (125) = 100

2^{43} - 3 \equiv y (m o d 100) \Rightarrow 2^{41} \equiv \frac{y + 3}{4} (m o d 25)

ϕ (25) = 20 \Rightarrow 2^{41} \equiv {(2^{20})}^{2} \times 2 \equiv 1 \times 2 \equiv \frac{y + 3}{4} (m o d 25)

\Rightarrow 8 \equiv y + 3 (m o d 100) \Rightarrow y \equiv 5 (m o d 100)

\Rightarrow 2^{43} \equiv 8 (m o d 100)

\Rightarrow 2^{2^{43} - 3} \equiv (2^{100 q}) 2^{5} \equiv \frac{x}{8} (m o d 125)

\Rightarrow 8 \times 32 \equiv x (m o d 1000) \Rightarrow x \equiv 256 (m o d 1000)

\Rightarrow L a s t t h r e e d i g i t s o f 4^{2^{42}} = 256

Commented by malwan last updated on 01/May/23

t h a n k y o u s o m u c h s i r

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