Question
The number $${\text{2}}{{\text{5}}^{64}} \times {64^{25}}$$ is the square of a natural number n. The sum of the digits of n is = ?
Answer: Option B
$$\eqalign{
& \Leftrightarrow {{\text{n}}^2} = {\left( {{\text{25}}} \right)^{64}} \times {\left( {64} \right)^{25}} \cr
& \Leftrightarrow {{\text{n}}^2} = {\left( {{{\text{5}}^2}} \right)^{64}} \times {\left( {{2^6}} \right)^{25}} \cr
& \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{150}} \cr
& \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{128}} \times {2^{22}} \cr
& \Leftrightarrow n = {5^{64}} \times {2^{64}} \times {2^{11}} \cr
& \Leftrightarrow n = {\left( {5 \times 2} \right)^{64}} \times {2^{11}} \cr
& \Leftrightarrow n = {10^{64}} \times 2048 \cr} $$
∴ Sum of digits of n
= 2 + 0 + 4 + 8
= 14
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$$\eqalign{
& \Leftrightarrow {{\text{n}}^2} = {\left( {{\text{25}}} \right)^{64}} \times {\left( {64} \right)^{25}} \cr
& \Leftrightarrow {{\text{n}}^2} = {\left( {{{\text{5}}^2}} \right)^{64}} \times {\left( {{2^6}} \right)^{25}} \cr
& \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{150}} \cr
& \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{128}} \times {2^{22}} \cr
& \Leftrightarrow n = {5^{64}} \times {2^{64}} \times {2^{11}} \cr
& \Leftrightarrow n = {\left( {5 \times 2} \right)^{64}} \times {2^{11}} \cr
& \Leftrightarrow n = {10^{64}} \times 2048 \cr} $$
∴ Sum of digits of n
= 2 + 0 + 4 + 8
= 14
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