Question

If 'a', 'b', 'c' and 'd' are consecutive natural numbers and $a^3=b^3+c^3+d^3$, what is the least value of 'a'?

• A. $6$
• B. $9$
• C. $3$
• D. $12$

Answer 1

The correct option is A $6$

$a^3=b^3+c^3+d^3$ is possible for consecutive natural numbers only if $a>b>c>d$
Therefore, $b=(a-1),c=(a-2),d=(a-3)$

$\Rightarrow a^3=(a-1{)}^3+(a-2{)}^3+(a-3{)}^3$

$a$ cannot be $3$ as this will make $d=0$ and it is given in the question that all $a,b,c,d$ are natural numbers
Substituting for $a$ from the answer option, starting with the least number

If $a=6$, then $LHS=a^3=3^3=216$, and

$\text{RHS}=$ $(a-1{)}^3+(a-2{)}^3+(a-3{)}^3$ $=5^3+4^3+3^3=125+64+27=216$

We can see $\text{LHS}=\text{RHS}$

We can stop here as we want the least value for $a$.

Hence, $6$ is the correct answer.