Question

If $\alpha =888,.....8$ (a number of $n$ digits); \beta=666..6(a number of $n$ digits), and $\gamma =444,.....4$ (a number of $2n$ digits) then $\gamma -\alpha =$

• A. $\beta$
• B. ${\beta }^2$
• C. $-\beta$
• D. ${-\beta }^2$

Answer 1

The correct option is B ${\beta }^2$

$\alpha =8888$…. n digits

$\beta =6666$…. n digits

$\gamma =4444$…. $2n$ digits

$\gamma =4+40+400+...\dots +4\times {10}^{2n}$

$=4(1+10+100+...\dots +{10}^{2n})$

$=\frac{4({10}^{2n}-1)}{10-1}$

$\alpha =8+80+...\dots +8\times {10}^n$

$=8(1+10+100+....{.10}^n)$

$=\frac{8({10}^n-1)}{10-1}$

$\gamma -\alpha$

$=\frac{4({10}^{2n}-1)-8({10}^n-1)}{10-1}$

$=\frac{({10}^n-1)(4\times {10}^n+4-8)}{10-1}$

$=\frac{4({10}^n-1)({10}^n-1)}{(10-1)}\cdot \frac{9}{9}$

$={\left(\frac{6({10}^n-1)}{10-1}\right)}^2={\beta }^2$.