128th Problem 2016

Algebra Level 3

If ( 100 20 ) 5 \left( { 100 }^{ 20 } \right) ^{ 5 } is expanded . How many digits will you have?

Check out the set: 2016 Problems
102 200 100 201

Angela Fajardo by Angela Fajardo , 19, Philippines

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2 solutions

The number of digits n n of a number N N is given by:

n = log ⁡ 10 N + 1 = log ⁡ 10 ( 10 0 20 ) 5 + 1 = log ⁡ 10 ( 1 0 40 ) 5 + 1 = log ⁡ 10 1 0 200 + 1 = 200 + 1 = 201 \begin{aligned} n & = \log_{10} N + 1 \\ & = \log_{10} \left(100^{20}\right)^5 + 1 \\ & = \log_{10} \left(10^{40}\right)^5 + 1 \\ & = \log_{10} 10^{200} + 1 \\ & = 200 + 1 \\ & = \boxed{201} \end{aligned} .

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This is the (2 mod 9, 8 mod 12, and 7 mod 11)th problem!!

Razzi Masroor - 4 years, 7 months ago

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What do you mean?

Chew-Seong Cheong - 4 years, 7 months ago

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the numbers that satisfy this is 128

Razzi Masroor - 4 years, 7 months ago
Ayush G Rai
Nov 4, 2016

Related Wiki : Logarithms
( 10 0 20 ) 5 = 10 0 100 . {(100^{20})}^5=100^{100}. Using formula based in logarithms, the number of digits = 100 × l o g 10 100 + 1 = 100 × 2 + 1 = 201 . =100\times log_{10}100+1=100\times 2+1=\boxed{201}.

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