A number is divisible by 3 if the sum of the digits is divisible by 3. How about a number of the type (2^{2n} – 1), where n is a positive integer? Can we show that it is always divisible by 3?

**Answer**

We can use the following factoring rule:

a^{n} – 1 = (a – 1)(a^{n-1} + a^{n-2} + … + a + 1).

We can rewrite 2^{2n} – 1 as:

(2^{2})^{n} – 1 = (2^{2} – 1)[ (2^{2})^{n-1} + (2^{2})^{n-2} + … + 2^{2} + 1]

= 3(4^{n-1} + 4^{n-2} + … + 4 + 1), which is divisible by 3!

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