# Find the remainder when $21^3$ +$23^3$+$25^3$+$27^3$ is divided by $96$?

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.

• @lhf Thanku It was a typo mistake. – Jack Nov 2 '15 at 11:44

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.

• How a=$24$ and d=1? – Jack Nov 2 '15 at 23:21
• @Jack, $$24\pm1=?, 24\pm3=?$$ – lab bhattacharjee Nov 3 '15 at 1:25

By exploiting $a^3+b^3=(a+b)(a^2-ab+b^2)$ we have: $$21^3+23^3+25^3+27^3 = 48\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$$ hence $21^3+23^3+25^3+27^3$ is a multiple of $96$, since $\left(27^2-21\cdot 27+21^2+23^3-23\cdot 25+25^2\right)$ is an even number.

• On solving $21^3$+$23^3$+$25^3$+$27^3$=$44$. $551$ +$52$ . $679$.Which on divided by $96$ gives Remainder not equal to 0? – Jack Nov 2 '15 at 23:39

This is essentially a simpler version of lab's answer:

For any $a,b$, $(a-b)^3+(a+b)^3 = 2a^3 + 6ab^2 \equiv 0 \mod a$ for all $a,b$. In you case, you have two such pairs with $a=24$ (and $b=1,3$ respectively). So the answer is $0$.

• Draegere How a=24 and d=1,3 ? – Jack Nov 2 '15 at 23:24
• @Jack Because $23,25 = 24\pm 1$ and $21,27 = 24\pm 3$. – Klaus Draeger Nov 3 '15 at 10:11