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

m (→‎Solution 1: Added title.)
(Since the original Sol 3 have two completely different ideas, I broke it up into two solutions. Then, I group similar solutions together. Also, LaTeX'ed all solutions.)
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==Solution 1 (Observations and Patterns)==
 
==Solution 1 (Observations and Patterns)==
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Note that the numbers along the yellow cells are consecutive odd perfect squares. We can show this by induction.
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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</math>x<math>15</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 2nd 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 2nd from the left and 2nd from the top, which is <math>157</math>. We know that the least number in the 2nd 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>.
 
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-Dynosol
 
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==Solution 2 (Draw It Out)==
 
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Drawing out the diagram, we get <math>\boxed{\mathbf{(A)}\ 367}</math>. Note that this should mainly be used just to check your answer.
 
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~Taco12
 
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==Solution 3 (Illustrations)==
 
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In both solutions below, note that the numbers along the yellow cells are consecutive odd perfect squares. We can show this by induction.  
 
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Two pictorial solutions follow from here:
 
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===Solution 3.1 (Illustration of Solution 1: Considers Only 5 Squares)===
 
 
<asy>  
 
<asy>  
 
/* Made by MRENTHUSIASM */
 
/* Made by MRENTHUSIASM */
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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===Solution 3.2 (Illustration of Solution 2: Draws All 225 Squares Out)===
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==Solution 2 (Observations and Patterns)==
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 +
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>.
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~Dynosol
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==Solution 3 (Draws All 225 Squares Out)==
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From the full diagram below, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math>
 
<asy> /* Made by MRENTHUSIASM */
 
<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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From the full diagram above, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math>
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~Taco12 (Proposal)
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~MRENTHUSIASM
 
  
 
== Video Solution by OmegaLearn (Using Pattern Finding) ==
 
== Video Solution by OmegaLearn (Using Pattern Finding) ==

Revision as of 22:28, 25 August 2021

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)

Note that the numbers along the yellow cells are consecutive odd perfect squares. We can show this 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]

In the diagram above, 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. 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)

By observing that the right-top corner of a square will always be a square, we know that the top right corner of the $15\times15$ grid is $225$. We can subtract $14$ to get the value of the top-left corner; $211$. 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 $2$nd row from the top by subtracting $13$, since the length of the side will be one box shorter. Similarly, we find the value for the box $2$nd from the left and $2$nd from the top, which is $157$. We know that the least number in the $2$nd row will be $157$, and the greatest will be the number to its left, which is $1$ less than $211$. We then sum $157$ and $210$ to get $\boxed{\mathbf{(A)}\ 367}$.

~Dynosol

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); 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] ~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

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

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