Difference between revisions of "1996 AJHSME Problems/Problem 3"
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− | Obviously 1 is in the top left corner, 8 is in the top right corner, and 64 is in the bottom right corner. To find the bottom left corner, | + | ==Problem== |
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+ | The 64 whole numbers from 1 through 64 are written, one per square, on a checkerboard (an 8 by 8 array of 64 squares). The first 8 numbers are written in order across the first row, the next 8 across the second row, and so on. After all 64 numbers are written, the sum of the numbers in the four corners will be | ||
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+ | <math>\text{(A)}\ 130 \qquad \text{(B)}\ 131 \qquad \text{(C)}\ 132 \qquad \text{(D)}\ 133 \qquad \text{(E)}\ 134</math> | ||
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+ | ==Solution== | ||
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+ | Obviously <math>1</math> is in the top left corner, <math>8</math> is in the top right corner, and <math>64</math> is in the bottom right corner. To find the bottom left corner, subtract <math>7</math> from <math>64</math> which is <math>57</math>. Adding the results gives <math>1+8+57+64=130</math> which is answer <math>\boxed{A}</math>. | ||
== See also == | == See also == |
Revision as of 19:10, 1 August 2011
Problem
The 64 whole numbers from 1 through 64 are written, one per square, on a checkerboard (an 8 by 8 array of 64 squares). The first 8 numbers are written in order across the first row, the next 8 across the second row, and so on. After all 64 numbers are written, the sum of the numbers in the four corners will be
Solution
Obviously is in the top left corner, is in the top right corner, and is in the bottom right corner. To find the bottom left corner, subtract from which is . Adding the results gives which is answer .
See also
1996 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |