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2021 AMC 10B Problems/Problem 8

Revision as of 15:02, 29 August 2021 by MRENTHUSIASM (talk | contribs) (Made all solutions have illustrations. One picture is worth 1000 words.)

Problem

Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ 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]

$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$

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 $D$ and $E,$ 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((7.5,7.5)--(8.5,7.5)--(8.5,6.5)--(6.5,6.5)--(6.5,8.5)--(9.5,8.5)--(9.5,5.5)--(5.5,5.5)--(5.5,9.5)--(9.5,9.5),red+linewidth(1.125),EndArrow); draw((12.5,12.5)--(13.5,12.5)--(13.5,1.5)--(1.5,1.5)--(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); dot((7.5,7.5),10+red); [/asy] By observations, we proceed as follows: \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*} Therefore, the answer is $D+E=\boxed{\textbf{(A)} ~367}.$

~MRENTHUSIASM

Solution 2 (Observations and Patterns: Considers Only 8 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 $C$ and $G,$ 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((7.5,7.5)--(8.5,7.5)--(8.5,6.5)--(6.5,6.5)--(6.5,8.5)--(9.5,8.5)--(9.5,5.5)--(5.5,5.5)--(5.5,9.5)--(9.5,9.5),red+linewidth(1.125),EndArrow); draw((12.5,12.5)--(13.5,12.5)--(13.5,1.5)--(1.5,1.5)--(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); dot((7.5,7.5),10+red); [/asy] By observations, we proceed as follows: \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*} Therefore, the answer is $C+G=\boxed{\textbf{(A)} ~367}.$

~Dynosol (Fundamental Logic)

~MRENTHUSIASM (Reconstruction)

Solution 3 (Draws All 225 Squares Out)

From the full diagram below, the answer is $210+157=\boxed{\textbf{(A)} ~367}.$ [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] ~MRENTHUSIASM (Solution)

~Taco12 (Proposal)

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

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

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