UKPSC PCS 2012 · Question 22 of 78
Quant
n = 8k+2 gives n^3 = 8(...) + 8, exactly divisible by 8, so the remainder 0 is not among the options.
Source: UKPSC 2012 Prelims CSAT Paper-II (adda247/studyfry PDF), exam held 30 Nov 2014
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