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 one year ago
What'd the last digit of the expansion of \(2^{12n}  6^{4n}\) be? I know that it is \(0\) by trying out \(n = 0\), but a more formalized approach would be...?
 one year ago
What'd the last digit of the expansion of \(2^{12n}  6^{4n}\) be? I know that it is \(0\) by trying out \(n = 0\), but a more formalized approach would be...?

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[2^{12n}  2^{4n}3^{4n}\]Then\[2^{4n}(2^{8n}  3^{4n})\]What next?
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