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Difference between revisions of "2021 AMC 10B Problems/Problem 8"

m (→‎Solution 3 (Draws All 225 Squares Out))
 
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<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>
  
−
==Solution 1 (Observations and Patterns)==
+
==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.
 
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>  
 
<asy>  
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add(grid(15,15,linewidth(1.25)));
 
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((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);
−
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>
 
</asy>
 
By observations, we proceed as follows:
 
By observations, we proceed as follows:
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 2 (Observations and Patterns)==
+
==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));
  
−
By observing that the right-top corner of a square will always be a square, we know that the top right corner of the <math>15\times15</math> grid is <math>225</math>. We can subtract <math>14</math> to get the value of the top-left corner; <math>211</math>. 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 <math>2</math>nd row from the top by subtracting <math>13</math>, since the length of the side will be one box shorter. Similarly, we find the value for the box <math>2</math>nd from the left and <math>2</math>nd from the top, which is <math>157</math>. We know that the least number in the <math>2</math>nd row will be <math>157</math>, and the greatest will be the number to its left, which is <math>1</math> less than <math>211</math>. We then sum <math>157</math> and <math>210</math> to get <math>\boxed{\mathbf{(A)}\ 367}</math>.
+
add(grid(15,15,linewidth(1.25)));
  
−
~Dynosol
+
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>
  
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==Solution 3 (Draws All 225 Squares Out)==
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~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>
 
From the full diagram below, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math>
−
<asy> /* Made by MRENTHUSIASM */
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<asy>
 +
/* Made by MRENTHUSIASM */
 
size(11.5cm);
 
size(11.5cm);
  
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}
 
}
 
</asy>
 
</asy>
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~MRENTHUSIASM (Solution)
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<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>
  
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~Taco12 (Proposal)
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~MRENTHUSIASM ~Taco12
  
 
== Video Solution by OmegaLearn (Using Pattern Finding) ==
 
== Video Solution by OmegaLearn (Using Pattern Finding) ==

Latest revision as of 22:06, 23 August 2024

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((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: \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 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 $C$ 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((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: \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 $C+G=\boxed{\textbf{(A)} ~367}.$

~MRENTHUSIASM ~Dynosol

Solution 3 (Brute Force: 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] 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)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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