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Find the last 2 digits of 18^(37)

Guest Jul 19, 2018
 #1
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+2

problems like these are just patterns

write out powers of 18 until a pattern forms

 

18^2--last 2 didgets are 24

18^3--last 2 didgets are 32

18^4--last 2 didgets are 76

18^5--last 2 didgets are 68

18^6--last 2 didgets are 24

 

we see that it starts repeating when last 2 didgets are 24 thus their are 4 unique last 2 didgets

18^(2-5)

18^(6-9)

18^(10-13)

18^(14-17)

18^(18-21)

18^(22-25)

18^(26-29)

18^(30-33)

18^(34-37)

 

as you can see 37 is the last of the set or the last 2 didgets are 68

Guest Jul 19, 2018
 #2
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+2

Find the last 2 digits of 18^(37)


\(\begin{array}{|rcll|} \hline && 18^{37} \pmod {100} \\ &\equiv& \Big(18^4 \Big)^{9}\cdot 18 \pmod {100} \quad & | \quad 18^4 \equiv 76 \pmod {100} \\ &\equiv& 76^{9}\cdot 18 \pmod {100} \\ &\equiv& 76^8\cdot 76 \cdot 18 \pmod {100} \quad & | \quad 76^{2n} \equiv 76 \pmod {100} \text{ or } 76^8 \equiv 76 \pmod {100} \\ &\equiv& 76\cdot 76 \cdot 18 \pmod {100} \\ &\equiv& 76^2 \cdot 18 \pmod {100} \quad & | \quad 76^2 \equiv 76 \pmod {100} \\ &\equiv& 76 \cdot 18 \pmod {100} \\ &\equiv& 1368 \pmod {100} \\ &\mathbf{ \equiv } & \mathbf{ 68 \pmod {100} } \\ \hline \end{array}\)

 

The last 2 digits are 68

 

laugh

heureka  Jul 20, 2018

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