Difference between revisions of "2008 AMC 10B Problems/Problem 2"

Latest revision as of 11:46, 29 April 2008

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-==Problem==
+#REDIRECT [[2008 AMC 12B Problems/Problem 2]]
-A <math>4\times 4</math> block of calendar dates has the numbers <math>1</math> through <math>4</math> in the first row, <math>8</math> though <math>11</math> in the second, <math>15</math> though <math>18</math> in the third, and <math>22</math> through <math>25</math> in the fourth. The order of the numbers in the second and the fourth rows are reversed. The numbers on each diagonal are added. What will be the positive difference between the diagonal sums?
-<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10</math>
-==Solution==
-{{solution}}
-==See also==
-{{AMC10 box|year=2008|ab=B|num-b=1|num-a=3}}