Question

If $a=\underset{55}{\underset{}{\mathrm{111.........1}}},b=1+10+{10}^2+{10}^3+{10}^4$ and $c=1+{10}^5+{10}^{10}+....+{10}^{50}$, then show following.
(i) 'a' is a not composite number
(ii) $aisnotequalbc$

Answer 1

If the number of 1 repeating in the number are multiple of 3, the number is composite.

Eg: $111=3\left(37\right)111111=3\left(37037\right)$ etc

Since the number contains 55 ones and 55 is not a multiple of 3

Hence $a$ is not a composite number

$b=1+10+{10}^2+{10}^3+{10}^4=\frac{\left(1\right)({10}^5-1)}{9}c=1+{10}^5+{10}^{10}+.....+{10}^{50}=\frac{\left(1\right)({10}^{11}-1)}{({10}^5-1)}a=1+10+{10}^2+....+{10}^{55}=\frac{({10}^{56}-1)}{9}\Rightarrow a\ne bc$