Difference between revisions of "2021 AMC 10B Problems/Problem 8"
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==Solution 3 (Full Version of Solution 2)== | ==Solution 3 (Full Version of Solution 2)== | ||
| + | ===Diagram=== | ||
<asy> | <asy> | ||
size(11.5cm); | size(11.5cm); | ||
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</asy> | </asy> | ||
| + | ~MRENTHUSIASM (by Asymptote) | ||
| + | |||
| + | ===Answer=== | ||
From the full diagram above, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math> | From the full diagram above, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math> | ||
Revision as of 13:46, 6 March 2021
Contents
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?
![[asy] /* Made by samrocksnature */ add(grid(7,7)); label("$\dots$", (0.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (2.5,0.5)); label("$\dots$", (3.5,0.5)); label("$\dots$", (4.5,0.5)); label("$\dots$", (5.5,0.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (1.5,0.5)); label("$\dots$", (0.5,1.5)); label("$\dots$", (0.5,2.5)); label("$\dots$", (0.5,3.5)); label("$\dots$", (0.5,4.5)); label("$\dots$", (0.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (6.5,0.5)); label("$\dots$", (6.5,1.5)); label("$\dots$", (6.5,2.5)); label("$\dots$", (6.5,3.5)); label("$\dots$", (6.5,4.5)); label("$\dots$", (6.5,5.5)); label("$\dots$", (0.5,6.5)); label("$\dots$", (1.5,6.5)); label("$\dots$", (2.5,6.5)); label("$\dots$", (3.5,6.5)); label("$\dots$", (4.5,6.5)); label("$\dots$", (5.5,6.5)); label("$\dots$", (6.5,6.5)); label("$17$", (1.5,1.5)); label("$18$", (1.5,2.5)); label("$19$", (1.5,3.5)); label("$20$", (1.5,4.5)); label("$21$", (1.5,5.5)); label("$16$", (2.5,1.5)); label("$5$", (2.5,2.5)); label("$6$", (2.5,3.5)); label("$7$", (2.5,4.5)); label("$22$", (2.5,5.5)); label("$15$", (3.5,1.5)); label("$4$", (3.5,2.5)); label("$1$", (3.5,3.5)); label("$8$", (3.5,4.5)); label("$23$", (3.5,5.5)); label("$14$", (4.5,1.5)); label("$3$", (4.5,2.5)); label("$2$", (4.5,3.5)); label("$9$", (4.5,4.5)); label("$24$", (4.5,5.5)); label("$13$", (5.5,1.5)); label("$12$", (5.5,2.5)); label("$11$", (5.5,3.5)); label("$10$", (5.5,4.5)); label("$25$", (5.5,5.5)); [/asy]](http://latex.artofproblemsolving.com/2/d/2/2d2408a8772066f5c86886246eef174525a5b6a1.png)
Solution 1
By observing that the right-top corner of a square will always be a square, we know that the top right corner of the x 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 2nd 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 2nd from the left and 2nd from the top, which is 
. We know that the least number in the 2nd 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 2: Draw It Out
Drawing out the diagram, we get . Note that this should mainly be used just to check your answer.
~Taco12
Solution 3 (Full Version of Solution 2)
Diagram
![[asy] size(11.5cm); for (real i=7.5; i<=14.5; ++i) { fill((i+0.5,i+0.5)--(i-0.5,i+0.5)--(i-0.5,i-0.5)--(i+0.5,i-0.5)--cycle,yellow); } fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); add(grid(15,15,linewidth(1.25))); int adj = 1; int curDown = 2; int curLeft = 4; int curUp = 6; int curRight = 8; label("$1$", (7.5,7.5)); for (int len = 3; len<=15; len+=2) { for (int i=1; i<=len-1; ++i) { label("$"+string(curDown)+"$", (7.5+adj,7.5+adj-i)); label("$"+string(curLeft)+"$", (7.5+adj-i,7.5-adj)); label("$"+string(curUp)+"$", (7.5-adj,7.5-adj+i)); label("$"+string(curRight)+"$", (7.5-adj+i,7.5+adj)); ++curDown; ++curLeft; ++curUp; ++curRight; } ++adj; curDown = len^2 + 1; curLeft = len^2 + len + 2; curUp = len^2 + 2*len + 3; curRight = len^2 + 3*len + 4; } [/asy]](http://latex.artofproblemsolving.com/1/d/b/1db62a2086825c29df002f594cb285d42feb89c9.png)
~MRENTHUSIASM (by Asymptote)
Answer
From the full diagram above, the answer is 
Note that the numbers along the yellow cells are consecutive odd perfect squares. We can show this by induction.
~MRENTHUSIASM
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. 
