MHB How many aluminum atoms can line up across a page?

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The discussion revolves around calculating how many aluminum atoms can fit across a standard page of paper, given the radius of an aluminum atom is approximately 140 picometers. The width of the paper is 21.59 cm, which converts to 2.159 x 10^11 picometers. By dividing the width of the page in picometers by the diameter of an aluminum atom (280 picometers), the calculation yields approximately 7.71 x 10^8 aluminum atoms. The initial calculation is confirmed as correct by another participant. The conclusion affirms the accuracy of the result.
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The problem statement, all variables and given/known data
The radius of an aluminum atom is about 140 picometers. How many spherical aluminum atoms could line up across a page of paper?The paper's width is 21.59 cm. Converting to picometers, the width is 2.159 x 1011 pm

I used the ratio of width of the page in picometers divided by the diameter of the aluminum atom

(2.159 x 1011pm)/(280pm)

I got 7.71x108 aluminum atoms.
I am not sure if this is the answer.
 
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mathcoral said:
The problem statement, all variables and given/known data
The radius of an aluminum atom is about 140 picometers. How many spherical aluminum atoms could line up across a page of paper?The paper's width is 21.59 cm. Converting to picometers, the width is 2.159 x 1011 pm

I used the ratio of width of the page in picometers divided by the diameter of the aluminum atom

(2.159 x 1011pm)/(280pm)

I got 7.71x108 aluminum atoms.
I am not sure if this is the answer.
Looks good to me.

-Dan
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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