Find the number of digits in 2222{ 2 }^{ 2^{ 22 } }

Note by Swapnil Das
4 years, 2 months ago

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log10(2222)+1\left\lfloor log_{ 10 }(2^{ 2^{ 22 } }) \right\rfloor +1= 1262612

Raghav Vaidyanathan - 4 years, 2 months ago

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The answer is log2222 \log {2^{2^{22}}}

Rajdeep Dhingra - 4 years, 2 months ago

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No, it is not, it is log10(2222)+1\left\lfloor log_{ 10 }(2^{ 2^{ 22 } }) \right\rfloor +1

Raghav Vaidyanathan - 4 years, 2 months ago

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I forgot to put it in Round function

Rajdeep Dhingra - 4 years, 2 months ago

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@Rajdeep Dhingra You forgot to add 11. Floor function is understood.

Raghav Vaidyanathan - 4 years, 2 months ago

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@Raghav Vaidyanathan I meant Round off function

Rajdeep Dhingra - 4 years, 2 months ago

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@Rajdeep Dhingra I get the round off function bit. The thing that I meant is that the number of digits nn of a natural number xx is given by:

n=log10(x)+1n=\left\lfloor log_{ 10 }(x) \right\rfloor +1

nlog10(x)n\ne \left\lfloor log_{ 10 }(x) \right\rfloor

Raghav Vaidyanathan - 4 years, 2 months ago

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@Raghav Vaidyanathan Is it always true ?

Rajdeep Dhingra - 4 years, 2 months ago

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@Rajdeep Dhingra Yes, this is due to the fact that log10(x)log_{10}(x) gives us the number of zeroes in the largest power of 1010 which is less than xx. For example, if x=12345x=12345, log10(x)=4log_{10}(x)=4(approx.) as 1000010000 is the largest power of ten below 1234512345. But we need to add one to this value to account for the leading number as we have now counted only the zeroes and left out the one in 1000010000. Therefore, the right answer is 4+1=54+1=5.

In general: 10k10^k has k+1\left\lfloor k \right\rfloor+1 digits.

Raghav Vaidyanathan - 4 years, 2 months ago

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22log2+1 = 7 (log2 =.3010)

Nithin Nithu - 4 years, 2 months ago

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Is there any way of calculation without using logarithms?

Swapnil Das - 4 years, 2 months ago

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yes, we could try writing it as 10 to a certain power(changing base/power.)

David Holcer - 4 years, 2 months ago

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14

Joe Doe - 3 years, 6 months ago

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2

Joe Doe - 3 years, 6 months ago

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