Difference between revisions of "1956 AHSME Problems/Problem 5"

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==Solution==
 
==Solution==
Arrange the nickels in a hexagonal fashion. One can see that one can only place <math>\boxed{\textbf{(C)} \quad 6}</math> nickels around the central nickel.
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Arranging the nickels in a hexagonal fashion, we see that only <math>\boxed{\textbf{(C) }6}</math> nickels can be placed around the central nickel.
  
 
==See Also==
 
==See Also==

Revision as of 16:12, 14 March 2023

Problem #5

A nickel is placed on a table. The number of nickels which can be placed around it, each tangent to it and to two others is:

$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

Solution

Arranging the nickels in a hexagonal fashion, we see that only $\boxed{\textbf{(C) }6}$ nickels can be placed around the central nickel.

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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