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

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The discussion focuses on finding the remainders of two calculations: $40^{110}$ divided by 37 and $3^{1000}$ divided by 26. For $40^{110}$ mod 37, the calculation involves simplifying $40$ to $3$ mod 37, leading to $3^{110}$, which can be computed using properties of modular arithmetic. For $3^{1000}$ mod 26, the approach includes recognizing that $3$ and $26$ are coprime, allowing the use of Euler's theorem to simplify the exponent. The results from both calculations provide the respective remainders for each division. The thread emphasizes the application of modular arithmetic techniques to solve these problems efficiently.
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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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