Question 45

Challenge question #45 (27/02/2017)

Junior Question

Is $1\,111\,111\,111$ prime or composite?

Answer

$1\,111\,111\,111$ is composite.

1\,111\,111\,111=

1\,111\,100\,000+

11\,111

=

11\,111\times

100\,000+

11\,111

=

11\,111

(100\,000+1)

=

1\,111\,100\,000\times

100\,001

Thus, $1\,111\,111\,111$ is composite.

Senior Question

Show that the product of $3$ consecutive positive integers can never be a perfect cube.

Hint

Let the integers be $n$ , $n+1$ , and $n+2$ . Compare the product $n(n+1)(n+2)$ with $n^{3}$ and $(n+1)^{3}$ .

Let the integers be $n$ , $n+1$ , and $n+2$ .

Notice that $n(n+1)(n+2)>$ $n^{3}$ .... (1)

Also notice that $n(n+2)<(n+1)^{2}$ $\{$ since $n(n+2)=n^{2}+2n$ and $(n+1)^{2}=n^{2}+2n+1\}$

$\therefore\quad n(n+1)(n+2)<$ $(n+1)^{3}$ .... (2)

From (1) and (2), $n(n+1)(n+2)$ always lies between two consecutive perfect cubes, and so it cannot be a perfect cube.