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Revision as of 13:55, 29 August 2021
Contents
[hide]Problem
Mr. Zhou places all the integers from to into a by grid. He places in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?
Solution 1 (Observations and Patterns)
In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are and respectively. Note that the numbers in the yellow cells are consecutive odd perfect squares, as we can prove by induction. By observations, we proceed as follows: Therefore, the answer is
~MRENTHUSIASM
Solution 2 (Observations and Patterns)
By observing that the right-top corner of a square will always be a square, we know that the top right corner of the grid is . We can subtract to get the value of the top-left corner; . We can then find the value of the bottom left and right corners similarly. From there, we can find the value of the box on the far right in the nd row from the top by subtracting , since the length of the side will be one box shorter. Similarly, we find the value for the box nd from the left and nd from the top, which is . We know that the least number in the nd row will be , and the greatest will be the number to its left, which is less than . We then sum and to get .
~Dynosol
Solution 3 (Draws All 225 Squares Out)
From the full diagram below, the answer is ~MRENTHUSIASM (Solution)
~Taco12 (Proposal)
Video Solution by OmegaLearn (Using Pattern Finding)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/GYpAm8v1h-U?t=412
~IceMatrix
Video Solution by Interstigation
https://youtu.be/DvpN56Ob6Zw?t=667
~Interstigation
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.