My Approach
I assumed the prime number of the form $6k+1$ when the prime number is greater than $3$.
$$(6k+1)^3=\frac {216k^3 + 1+108k^2+18k}6.$$Assuming $\color{blue}{k=1}$ I get The answer as $\color{red}{1}$.
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There are the special primes $2$ and $3$. Their cubes have remainders $2$ and $3$ respectively. All the rest have shape $6k+1$ or $6k+5$. You should end up with $11$. – André Nicolas Oct 31 '15 at 17:08
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the answer is 6 since the possible remainders of an odd prime is 1 or 5. And once cubed and summed we get 6. – Adelafif Oct 31 '15 at 17:18
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@AndréNicolas How? $2$ and $3$ have a remainder sum of 5.Any other number will have a remainder of $1$.So,I get 6.How you get 11? – justin takro Oct 31 '15 at 17:23
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2Primes of the form $6k+1$ give remainder $1$. Primes of the form $6k+5$ give remainder $5$. Total is $2+3+1+5$. – André Nicolas Oct 31 '15 at 17:26
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By the Congruence Power Rule $\bmod 6!:\ (6k\pm1)^3\equiv (\pm1)^3\equiv \pm1 \equiv 1,5,,$ and $2^3\equiv 2,\ 3^3\equiv 3,,$ so the sum $= 1+5+2+3=11\ \ $ – Bill Dubuque May 01 '23 at 18:09
1 Answers
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In fact, any prime $>3,$ can be written as $6k\pm1$
and $(6k\pm1)^3\equiv\pm1\pmod6$ as $6k\pm1\equiv\pm1$
lab bhattacharjee
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Yes I assumed only 6k+1 form.But the Ans is not given in the options according to you. – justin takro Oct 31 '15 at 17:09