MCQ
Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and $b+c+d$ is square of an integer. The least possible value of the number of digits of $c$ is
- A$2$
- ✓$3$
- C$4$
- D$5$
We have,
$a, b, c, d, e$ are natural number and in AP.
Let $D$ is common difference of $AP$.
$\therefore$ Let $c=C$
$a =C-2 D$
$b =C-D$
$d =C+D$
$e =C+2 D$
$a+b+c+d+e=5 C$
and $b+c+d=3 C$
Given, $a+b+c+d+e$ is a cube of number
$\therefore 5 C=\lambda^3$
and $b+c+d$ is a square of number
$\therefore 3 C=u^2$
From Eqs.$(i)$ and $(ii)$, we get
$\frac{\lambda^3}{5}=\frac{u^2}{3}$
$\lambda^3$ and $u^2$ is a multiple of 15
$\therefore$ Smallest possible value of $\lambda=15$ and $u=45$
$\therefore \quad c=\frac{u^2}{3}=\frac{(45)^2}{3}=675$
$\therefore$ Number of digits $=3$
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$I.$ If $n$ is a composite number, then $n$ divides $(n-1) ! .$
$II$. There are infinitely many natural numbers $n$ such that $n^3+2 n^2+n$ divides $n !$