Difference between revisions of "2021 AMC 10B Problems/Problem 8"
MRENTHUSIASM (talk | contribs) |
|||
(66 intermediate revisions by 10 users not shown) | |||
Line 59: | Line 59: | ||
<math>\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380</math> | <math>\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380</math> | ||
− | |||
− | |||
− | |||
− | |||
− | {{AMC10 box|year=2021|ab=B| | + | ==Solution 1 (Observations and Patterns: Considers Only 5 Squares)== |
+ | 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 <math>D</math> and <math>E,</math> respectively. Note that the numbers in the yellow cells are consecutive odd perfect squares, as we can prove by induction. | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | 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); | ||
+ | |||
+ | label("$A$",(14.5,14.5)); | ||
+ | label("$B$",(13.5,13.5)); | ||
+ | label("$C$",(0.5,14.5)); | ||
+ | label("$E$",(1.5,13.5)); | ||
+ | label("$D$",(0.5,13.5)); | ||
+ | |||
+ | add(grid(15,15,linewidth(1.25))); | ||
+ | |||
+ | draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); | ||
+ | </asy> | ||
+ | By observations, we proceed as follows: | ||
+ | <cmath>\begin{alignat*}{6} | ||
+ | A=15^2=225, \ B=13^2=169 | ||
+ | \quad &\implies \quad &C &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ | ||
+ | \quad &\implies \quad &D &= &&C-1 &= 210& \\ | ||
+ | \quad &\implies \quad &E &= &&B-12 &= 157&. | ||
+ | \end{alignat*}</cmath> | ||
+ | Therefore, the answer is <math>D+E=\boxed{\textbf{(A)} ~367}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Observations and Patterns: Considers Only 7 Squares)== | ||
+ | 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 <math>C</math> and <math>G,</math> respectively. | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(11.5cm); | ||
+ | |||
+ | fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); | ||
+ | fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); | ||
+ | |||
+ | label("$A$",(14.5,14.5)); | ||
+ | label("$B$",(0.5,14.5)); | ||
+ | label("$C$",(0.5,13.5)); | ||
+ | label("$D$",(0.5,0.5)); | ||
+ | label("$E$",(14.5,0.5)); | ||
+ | label("$F$",(14.5,13.5)); | ||
+ | label("$G$",(1.5,13.5)); | ||
+ | |||
+ | add(grid(15,15,linewidth(1.25))); | ||
+ | |||
+ | draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); | ||
+ | </asy> | ||
+ | By observations, we proceed as follows: | ||
+ | <cmath>\begin{alignat*}{6} | ||
+ | A=15^2=225 | ||
+ | \quad &\implies \quad &B &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ | ||
+ | \quad &\implies \quad &C &= &&B-1 &= 210& \\ | ||
+ | \quad &\implies \quad &D &= &&C-13 &= 197& \\ | ||
+ | \quad &\implies \quad &E &= &&D-14 &= 183& \\ | ||
+ | \quad &\implies \quad &F &= &&E-13 &= 170& \\ | ||
+ | \quad &\implies \quad &G &= &&F-13 &= 157&. | ||
+ | \end{alignat*}</cmath> | ||
+ | Therefore, the answer is <math>C+G=\boxed{\textbf{(A)} ~367}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM ~Dynosol | ||
+ | |||
+ | ==Solution 3 (Brute Force: Draws All 225 Squares Out)== | ||
+ | From the full diagram below, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math> | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(11.5cm); | ||
+ | |||
+ | 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> | ||
+ | <b>This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.</b> | ||
+ | |||
+ | ~MRENTHUSIASM ~Taco12 | ||
+ | |||
+ | == Video Solution by OmegaLearn (Using Pattern Finding) == | ||
+ | https://youtu.be/bb4HB7pwO3Q | ||
+ | |||
+ | ~ 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== | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:06, 23 August 2024
Contents
- 1 Problem
- 2 Solution 1 (Observations and Patterns: Considers Only 5 Squares)
- 3 Solution 2 (Observations and Patterns: Considers Only 7 Squares)
- 4 Solution 3 (Brute Force: Draws All 225 Squares Out)
- 5 Video Solution by OmegaLearn (Using Pattern Finding)
- 6 Video Solution by TheBeautyofMath
- 7 Video Solution by Interstigation
- 8 See Also
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: Considers Only 5 Squares)
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: Considers Only 7 Squares)
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.
By observations, we proceed as follows:
Therefore, the answer is
~MRENTHUSIASM ~Dynosol
Solution 3 (Brute Force: Draws All 225 Squares Out)
From the full diagram below, the answer is
This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.
~MRENTHUSIASM ~Taco12
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.