Question

$\text{If the number}{\mathrm{1.2.2}}^2{.2}^3{.2}^4..........{.2}^{20}.{3.3}^2{.3}^3{.3}^4.......{.3}^{10}$
$\text{is written in power of '6' then the highest}$
$\text{power of '6' is \_\_\_\_\_}.$

• A. 10
• B. 20
• C. 30
• D. 55

Answer 1

The correct option is D

55

2 and 3 are factors of 6.

In the given problem the highest power of 2 is 20, whereas the highest power of 3 is 10.

The highest power of 6 will depend on the lowest power of 2 and 3.

$(2\times 3)\times (2\times 3{)}^2\times (2\times 3{)}^3\times (2\times 3{)}^4\times (2\times 3{)}^5$

$\times (2\times 3{)}^6\times (2\times 3{)}^7\times (2\times 3{)}^8\times (2\times 3{)}^9\times (2\times 3{)}^{10}$

= $6\times 6^2\times 6^3...........{.6}^{10}$

Using the product law of indices, we get $(6{)}^{1+2+3+4+...10}$

Therefore the highest power of 6 = 1+2+3+4+5+6+7+8+9+10 = 55.