2008 AMC 10B Problem 2

Attempt Problem 2 of the 2008 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

A 4×44 \times 4 block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?
22
44
66
88
1010
Answer: B
Concepts:date and timeprocess simulation
Small hint:
After reversing rows 22 and 4,4, write out the new grid
Big hint:
The main diagonal reads 1,10,17,221,10,17,22 and the other reads 4,9,16,254,9,16,25
Solution:
After reversing the second row to 11,10,9,811,10,9,8 and the fourth row to 25,24,23,22,25,24,23,22, the two diagonals are 1,10,17,221,10,17,22 and 4,9,16,25.4,9,16,25.
Their sums are 1+10+17+22=501+10+17+22=50 and 4+9+16+25=54,4+9+16+25=54, so the positive difference is 54−50=4.54-50=4.
Thus, the correct answer is B.
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