Find Remainder of $40^{110}$ and $3^{1000}$ Divided by 37 and 26

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The discussion focuses on calculating the remainders of $40^{110}$ when divided by 37 and $3^{1000}$ when divided by 26. Using modular arithmetic, it is established that the remainder of $40^{110} \mod 37$ is 3, while the remainder of $3^{1000} \mod 26$ is 1. These calculations utilize properties of congruences and the Chinese Remainder Theorem for efficient computation.

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(1)dividing $40^{110} \,\, by \,\, 37$

(2)dividing $3^{1000} \,\, by \,\, 26$


 
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Albert said:
(1)dividing $40^{110} \,\, by \,\, 37$

(2)dividing $3^{1000} \,\, by \,\, 26$


(1)
we have
40 = 3 mod 37
so we need
$3^{110}$ mod 37
as 37 is prime so we have as per flt
$3^{36} = 1$ mod 37
so
$3^{110} = 3^{3*36+2} = (3^{36})^{3} * 3^2$ mod 37 = 9 mod 37 = 9 that is the ans

so $40^{110}$ devided by 37 remainder is 9

(2)
$3^{3} = 27 = 1$ mod 26
hence
$3^{999} = (3^3)^{333} = 1$ mod 26
so $3^{1000}$ mod 26 = 3 so remainder divided by 26 is 3
 

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