Question

Which odd number should be excluded to form a cube number?
$41+43+45+47+49+51+53+55$

• A. $43$
• B. $45$
• C. $47$
• D. $41$

Answer 1

The correct option is D $41$

We know,

$1=1=1^3$,
$3+5=8=2^3$,
$7+9+11=27=3^3$,

$13+15+17+19=64=4^3$,

$21+23+25+27+29=125=5^3$,

$31+33+35+37+39+41=216=6^3$,

$43+45+47+49+51+53+55=343=7^3$.

Therefore, $41$ must be taken out to form a cube number, $7^3=343$.

Hence, option $D$ is correct.