2008 AMC 10B Problems/Problem 2

Revision as of 01:35, 25 April 2008 by I like pie (talk | contribs) (New page: ==Problem== 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, <m...)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

A $4\times 4$ block of calendar dates has the numbers $1$ through $4$ in the first row, $8$ though $11$ in the second, $15$ though $18$ in the third, and $22$ through $25$ 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?

$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 AMC 10 Problems and Solutions