Question

How many digits are there in the expression -
${6}^{2} \times {2}^{98} \times {5}^{99}$
NO SPAMMERS ALLOWED !
♈5 points only♈​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{ 101 \:digits}}}$

$\rule{200}{4}$

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

$\bf\bold{6^2 \times2^{98} \times 5^{99} }$

Breaking $\bf\bold{5^{99} \longrightarrow 5^{98+1} }$

$\bf\bold{\Rightarrow 6^2 \times2^{98} \times 5^{98+1} }$

We got

$\bf\bold{\Rightarrow6^2 \times2^{98} \times 5^{98}\times 5^1}$

$\bf\bold{\Rightarrow6^2 \times(2 \times 5) ^{98}\times5}$

$\bf\bold{\Rightarrow 36\times 5 \times(2 \times 5) ^{98}}$

$\bf\bold{\Rightarrow 180 \times(10) ^{98}}$

3 digits from 180  & 98 digits from 10⁹⁸

Therefore, 3 + 98 = 101

Hence, 101 digits are in the above expression

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Concept: -

$\bf\bold{a^{(m+n) } = a^m \times a^n }$

Answer 2

Answer:

There are total 101 digits in -

${6}^{2}\times {2}^{98}\times {5}^{99}$

Step-by-step explanation :

Given :

${6}^{2}\times {2}^{98}\times {5}^{99}$

To Find :

Number of digits in the given expression.

Solution :

$\sf \\\implies 6^2\times2^{98}\times5^{99}\\ \\\implies 6^2\times2^{98}\times5^{98+1}\\ \\\implies 6^2\times2^{98}\times5^{98}\times5^1\\ \\\implies 6^2\times5^1\times(2\times5)^{98}\\ \\\implies 36\times5\times(10)^{98}\\ \\\implies 180\times(10)^{98}$

After Simplifying,

You will got the Total number of digits in the expression is 101.

$\rule{300}{1.5}$

★ Things to Remember -

In this Question you can also do -

${5^{99}\rightarrow 5^{98+1}}$

★ Laws of Exponents :

$\implies \sf x^m \times x^n = x^{m-n}\\ \\\implies \sf x^n \times y^n = (x\times y)^n\\ \\\implies \sf x^n \div y^n = \left(\dfrac{x}{y}\right)^n$