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April 11th, 2018, 05:17 AM  #1 
Member Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0  How to know amount of digits by calculator?
How can you tell how many digits there are in this question?

April 11th, 2018, 05:41 AM  #2 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
We can provide much better help if you tell us what your thoughts are. This is true even if you are pretty sure that you are on the wrong track because then we would have an idea about how to improve your thought process in the future. Just giving you an answer does not really do that.

April 11th, 2018, 05:44 AM  #3  
Member Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0  Quote:
 
April 11th, 2018, 06:17 AM  #4 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
If we say $x \in \mathbb Z \text { and } x = a * 10^2 \text { where } 1 \le a < 10$ How many digits are in x? Is 2^1800 an integer? The reason that 2^300 was chosen when the calculator could go as high as 2^302 is that 6 * 300 = 1800. $\therefore (2^{300})^6 = 2^{1800}.$ So the approximate value of 2^1800 equals what? 
April 11th, 2018, 07:56 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
$\displaystyle log_{10}2^1800= 1800 log(2)= 1800 (0.3010)= 541.853$ That means that $\displaystyle 2^{1800}= 10^{541.853}= 10^{541} 10^{0.853}$. $\displaystyle 10^{0.853}$ is less than 10. What does $\displaystyle 10^{54}$ tell you about the number of digits? 
April 12th, 2018, 03:37 AM  #6  
Member Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0  Quote:
 
April 12th, 2018, 05:06 AM  #7 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 
I may well have misread your original post. Exponents are in a small font in a small jpg that my computer will not expand legibly. I am assuming that I misread 2^1200 as 2^1800. Sorry about that. On that assumption, I have rewritten my previous post below. If we say $x \in \mathbb Z \text { and } x = a * 10^2 \text { where } 1 \le a < 10$ How many digits are in x? Is 2^1200 an integer? The reason that 2^300 was chosen when the calculator could go as high as 2^302 is that 4 * 300 = 1200. $\therefore (2^{300})^4 = 2^{1200}.$ So the approximate value of 2^1200 equals what? 
April 12th, 2018, 05:11 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
$2 \times 10^{\color{red}1} = 20$ has $1+{\color{red}1}=2$ digits. $2.0 \times 10^{\color{red}2} = 20$ has $1+{\color{red}2}=3$ digits. $2.03 \times 10^{\color{red}3} = 20$ has $1+{\color{red}3}=4$ digits. So $2^{300}=2.03\ldots \times 10^{\color{red}{90}} = 20$ has $1+{\color{red}{90}}=91$ digits. $(ab)^c=a^c b^c$ so $$ 2^{1200}=(2^{300})^4=(2.03\ldots \times 10^{\color{red}{90}})^4 = (2.03\ldots)^4 \times (10^{\color{red}{90}})^4=16.0\ldots\times 10^{\color{red}{360}}=1.60\ldots\times 10^{\color{red}{361}}$$ 
April 12th, 2018, 07:09 AM  #9  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  Quote:
$2 \times 10^{\color{red}1} = 20$ has $1+{\color{red}1}=2$ digits. $2.0 \times 10^{\color{red}2} = 200$ has $1+{\color{red}2}=3$ digits. $2.03 \times 10^{\color{red}3} = 2000$ has $1+{\color{red}3}=4$ digits.  
April 12th, 2018, 07:44 AM  #10  
Member Joined: Apr 2018 From: On Earth Posts: 34 Thanks: 0  Quote:
 

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