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April 11th, 2014, 01:32 AM   #1
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Reminder

Reminder of 3^50 divide by 7
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April 11th, 2014, 01:48 AM   #2
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Math Focus: Abstract algebra
Hint: $\displaystyle 3^6\equiv1\!\pmod7$ (Fermat’s little theorem).
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April 11th, 2014, 02:04 AM   #3
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I make the remainder 2
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April 11th, 2014, 03:04 AM   #4
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You can also use

$$3^3 \equiv -1 \ (mod \ 7) $$

$$(3^3)^{16} \equiv (-1)^{16} \ (mod \ 7)$$

$$3^{48} \equiv 1 \ (mod \ 7)$$

$$3^2 \cdot 3^{48} \equiv 3^2 \cdot 1 \ (mod \ 7)$$

$$3^{50} \equiv 9 \ (mod \ 7)$$

$$3^{50} \equiv 2 \ (mod \ 7)$$

Alternatively ,

$$3^{-1} \ (mod \ 7) \ = \ 5$$

Because

$$3 \cdot 5 \equiv 1 \ (mod \ 7)$$

$$3^3 \equiv -1 \ (mod \ 7)$$

$$3^{51} \equiv -1 \ (mod \ 7)$$

$$3^{-1} \cdot 3^{51} \equiv 3^{-1} \cdot (-1) \ (mod \ 7)$$

$$3^{50} \equiv 5 \cdot (-1) \ (mod \ 7)$$

$$3^{50} \equiv -5 \ (mod \ 7)$$

$$3^{50} \equiv 2 \ (mod \ 7)$$

I left out a few minor details which I'm sure you can work out yourself , if not , please feel free to ask for clarification.

The trick is to get congruent to $ \pm 1$ , then use

IF

$$a \equiv b \ (mod \ m)$$

THEN

$$a^n \equiv b^n \ (mod \ m) $$

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