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The last digi...

Question

The last digit of $3^{3^{4 n}} + 1$ is ............

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Solution

$3^{4 n} = (3^{4})^{n} = 81^{n} = (1 + 80)^{n} = 1 +^{n} C_{1} (80) +^{n} C_{2} (80)^{2} + . . . . . +^{n} C_{n} (80)^{n} = 1 + 80 k$
where $k =^{n} C_{1} +^{n} C_{2} (80) + . . . . . +^{n} C_{n} (80)^{n - 1} =$ integer
Now $3^{3^{4 n}} = 3^{1 + 80 k} = 3 (3^{80 k}) = 3 (3)^{2 (16 k)} = 3 (9)^{16 k} = 3 (10 - 1)^{16 k}$
$= 3 [^{16 k} C_{0} (10)^{16 k} -^{16 k} C_{1} (10)^{6 k - 1} + . . . . . . . . . . . +^{16 k} C_{16 k} (10) 0 (- 1)^{16 k}]$
$= 3 [^{16 k} C_{0} (10)^{16 k} -^{16 k} C_{1} (10)^{6 k - 1} + . . . . . . . . . . . + 1] = 3 (10 λ + 1)$
where $λ =^{16 k} C_{0} (10)^{16 k - 1} -^{16 k} C_{1} (10)^{6 k - 2} + . . . . . . . . . . -^{16 k} C_{16 k - 1} (10)$
Thus $3^{3^{4 n}} + 1 = 30 λ + 4$
Hence last digit is $4$

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