Difference between revisions of "2006 AMC 8 Problems/Problem 22"
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smallest sum, place <math>1</math> in the center cell and <math>2</math> and <math>3</math> in the outer ones. The top | smallest sum, place <math>1</math> in the center cell and <math>2</math> and <math>3</math> in the outer ones. The top | ||
number will be <math>7</math>. For the largest sum, place <math>9</math> in the center cell and <math>7</math> and <math>8</math> in | number will be <math>7</math>. For the largest sum, place <math>9</math> in the center cell and <math>7</math> and <math>8</math> in | ||
− | the outer ones. This top number will be <math>33</math>. The difference is <math>33 - 7 = \boxed{\textbf{(D)} 26 | + | the outer ones. This top number will be <math>33</math>. The difference is <math>33 - 7 = \boxed{\textbf{(D)}\ 26 |
}</math>. | }</math>. | ||
Latest revision as of 14:57, 19 October 2020
Problem
Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?
Solution
If the lower cells contain and , then the second row will contain and , and the top cell will contain . To obtain the smallest sum, place in the center cell and and in the outer ones. The top number will be . For the largest sum, place in the center cell and and in the outer ones. This top number will be . The difference is .
See Also
2006 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.