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

Latest revision as of 17:13, 14 March 2023

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:

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

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

1956 AHSC (Problems • Answer Key • Resources)
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

@@ Line 6: / Line 6: @@
 ==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.
+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==
-{{AHSME box|year=1956|num-b=4|num-a=6}}
+{{AHSME 50p box|year=1956|num-b=4|num-a=6}}
 [[Category:Introductory Algebra Problems]]
 {{MAA Notice}}