Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?
MyApproach
Since I cannot form pattern above
I simplified this as $21^3$ = $3^3$ $. $ $7^3$
Similarly I did $27^3$ = $9^3$ $.$ $3^3$
Taking both I get $3^3$($7^3$+$9^3$)=$9$ $.$ $1072$/ $2^5$.
From solving this I get Remainder as $1$
And Similarly on solving $25^3$+$23^3$ separately on dividing by $96$.I get remainder as $73$ and $71$
On adding and Finding remainder I get Remainder as $49$.
I am getting wrong Ans.
Please correct me how to approach towards the problem.
Best Answer
As the terms are in Arithmetic Progression, let us try some generalization
As the number of terms is even, the terms can be $a\pm d,a\pm3d,a\pm5d$ etc.
$$(a-3d)^3+(a-d)^3+(a+d)^3+(a+3d)^3 =4a(a^2+15d^2)$$
Here $a=24$
and $d=1$(though any integer value of $d$ will do)
Had the number of terms been odd, the terms could be $a,a\pm d,a\pm2d$ etc.