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**Deveno** modular arithmetic is the perfect tool for investigating things concerning divisibilty (and thus about primes, for example).

the nice thing about it is that:

a (mod n) + b (mod n) = a+b (mod n), and:

(a mod n)(b mod n) = ab (mod n)

so it doesn't matter if you reduce mod n before or after adding (or multiplying).

so we have:

4 = 1 (mod 3)

7 = 1 (mod 3)

10 = 1 (mod 3)

therefore:

4^{n} + 7^{2n+1} + 10^{3n-1}

≡ (1)^{n} + (1)^{2n+1} + (1)^{3n-1} (mod 3)

≡ 1 + 1 + 1 (mod 3)

≡ 0 (mod 3), thus we have 4^{n} + 7^{2n+1} + 10^{3n-1} is divisible by 3.

one can also attempt an induction proof, that 4^{n} + 7^{2n+1} + 10^{3n-1} is divisible by 3 for all natural numbers n.

base case: n = 1:

4 + 7^{3} + 10^{2} = 4 + 343 + 100 = 447 = 3*149.

assume that:

4^{k} + 7^{2k+1} + 10^{3k-1} = 3x, for some positive integer x.

then:

4^{k+1} + 7^{2k+3} + 10^{3k+2} =

(4^{k+1} - 4^{k}) + (7^{2k+3} - 7^{2k+1}) + (10^{3k+2} - 10^{3k-1}) + 4^{k} + 7^{2k+1} + 10^{3k-1}

= 4^{k}(4 - 1) + 7^{2k+1}(7^{2} - 1) + 10^{3k-1}(10^{3} - 1) + 3x <---by our induction hypothesis

= 4^{k}(3) + 7^{2k+1}(48) + 10^{3k-1}(999) + 3x

= 3[4^{k} + (7^{2k+1})(16) + (10^{3k-1})(333)] + 3x

= 3y + 3x = 3(y + x), for y = 4^{k} + (7^{2k+1})(16) + (10^{3k-1})(333)

and is therefore true for all n.