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GO Classes Weekly Quiz 1 | General Aptitude | Question: 3

in Quantitative Aptitude
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Compute $2^{32} \; \mod \; 37$
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$2^{-4} \ 2^{36}  \mod 37 = 2^{-4} \mod 37$  (By Fermat’s Little Theorem)

Since, $2 \times 19 \equiv 1 \mod 37. \ $ So, $2^{-1} \mod 37 = 19$

Hence, $2^{-4} \mod 37 = 19^4 \mod 37 = 7$
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$2^{-4} \mod 37  = 16^{-1} \mod 37$

Using Euclidean Algorithm

$37 = 2*16 + 5$ and $16 = 3*5 + 1 \Rightarrow 1 = 16 \ – \ 3*(37 \ – \ 2*16) \Rightarrow 1 = 7*16 \ – \ 3*37$

Hence, $16^{-1} \mod 37 = 7$
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Hello sir @ankitgupta.1729
where can i learn the above approaches?

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Fermat’s Little Theorem 

Extended Euclidean Algorithm

These are quite famous, so, if you don’t get it from here then you can search some text over web or videos on youtube also. 

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2 Answers

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$2^2\equiv 4\; \mod \; 37\\
2^4 \equiv 16 \; \mod \; 37\\
2^8 \equiv 256 \equiv 34 \; \mod \; 37\\
2^{16} \equiv (-3)^2 \equiv 9 \; \mod \; 37\\
2^{32} \equiv 81 \; \mod \; 37\\
2^{32} \equiv 7 \; \mod \; 37$
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