Proving Divisibility: Modular Arithmetic and the Pattern of 16^43 - 10^26 Mod 21

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The discussion focuses on proving that 16^43 - 10^26 is divisible by 21 using modular arithmetic. It demonstrates that 16 is congruent to 2 mod 7, revealing a repeating pattern every three powers. By expressing 43 as 3 * 14 + 1, it simplifies the calculation to 16^1 mod 7, which equals 2. The reasoning behind this expression is that 16^43 can be rewritten as (16^3)^14 * 16, where 16^3 is congruent to 1 mod 7. Thus, the proof shows that the expression is indeed divisible by 21.
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Hi I'm reading a text about modular arithmetic,

Prove that 16^43 - 10^26 actually is divisible by 21.
They separate it by showing it is divisible by 7 and 3

they showed 16 \equiv 2 \textrm{ mod 7} \\ <br /> 16^2 \equiv 2^2 \equiv 4 \textrm{ mod 7} \\<br /> 16 \equiv 2^3 \equiv 1 \textrm{ mod 7} \\
So there is a pattern of length 3.

They later made 43 = 3 * 14 +1 . so,
<br /> 16^{43} \equiv 16^1 \equiv 2 \textrm{ mod 7} \\

whats the reasoning with 43 = 3 * 14 + 1 ?
 
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The Subject said:
Hi I'm reading a text about modular arithmetic,

Prove that 16^43 - 10^26 actually is divisible by 21.
They separate it by showing it is divisible by 7 and 3

they showed 16 \equiv 2 \textrm{ mod 7} \\<br /> 16^2 \equiv 2^2 \equiv 4 \textrm{ mod 7} \\<br /> 16 ^3\equiv 2^3 \equiv 1 \textrm{ mod 7} \\
So there is a pattern of length 3.

They later made 43 = 3 * 14 +1 . so,
<br /> 16^{43} \equiv 16^1 \equiv 2 \textrm{ mod 7} \\

whats the reasoning with 43 = 3 * 14 + 1 ?
##16^{43}=16^{42}\cdot 16 = (16^3)^{14}\cdot16##. You know that 16^3= 1 mod 7, 16=2 mod 7.
 
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