1988 AJHSME Problems/Problem 16

Revision as of 22:55, 4 July 2013 by Nathan wailes (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Placing no more than one $\text{X}$ in each small square, what is the greatest number of $\text{X}$'s that can be put on the grid shown without getting three $\text{X}$'s in a row vertically, horizontally, or diagonally?

$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

[asy] for(int a=0; a<4; ++a)  {   draw((a,0)--(a,3));  } for(int b=0; b<4; ++b)  {   draw((0,b)--(3,b));  } [/asy]

Solution

By the Pigeonhole Principle, if there are at least $7$ $\text{X}$'s, then there will be some row with $3$ $\text{X}$'s. We can put in $6$ by leaving out the three boxes in one of the main diagonals.

$\rightarrow \boxed{\text{E}}$

See Also

1988 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png