Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 3 (Full Version of Solution 2))
 
(55 intermediate revisions by 3 users not shown)
Line 60: Line 60:
 
<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==
+
==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);
  
−
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>.
+
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);
  
−
-Dynosol
+
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));
  
−
==Solution 2: Draw It Out==
+
add(grid(15,15,linewidth(1.25)));
  
−
Drawing out the diagram, we get <math>\boxed{\mathbf{(A)}\ 367}</math>. Note that this should mainly be used just to check your answer.
+
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>
  
−
~Taco12
+
~MRENTHUSIASM
  
−
==Solution 3 (Full Version of Solution 2)==
+
==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 */
 
/* 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>
 
<asy>
 +
/* Made by MRENTHUSIASM */
 
size(11.5cm);
 
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);
 
−
}
 
−
 
−
label("$1$", (7.5,7.5));
 
 
fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green);
 
fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green);
 
fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);
 
fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);
 +
 
add(grid(15,15,linewidth(1.25)));
 
add(grid(15,15,linewidth(1.25)));
  
Line 92: Line 146:
 
int curUp = 6;
 
int curUp = 6;
 
int curRight = 8;
 
int curRight = 8;
 +
 +
label("$1$",(7.5,7.5));
  
 
for (int len = 3; len<=15; len+=2)
 
for (int len = 3; len<=15; len+=2)
Line 97: Line 153:
 
for (int i=1; i<=len-1; ++i)  
 
for (int i=1; i<=len-1; ++i)  
 
   {
 
   {
−
label("$"+string(curDown)+"$", (7.5+adj,7.5+adj-i));
+
label("$"+string(curDown)+"$",(7.5+adj,7.5+adj-i));
−
     label("$"+string(curLeft)+"$", (7.5+adj-i,7.5-adj));  
+
     label("$"+string(curLeft)+"$",(7.5+adj-i,7.5-adj));  
−
     label("$"+string(curUp)+"$", (7.5-adj,7.5-adj+i));
+
     label("$"+string(curUp)+"$",(7.5-adj,7.5-adj+i));
−
     label("$"+string(curRight)+"$", (7.5-adj+i,7.5+adj));
+
     label("$"+string(curRight)+"$",(7.5-adj+i,7.5+adj));
 
     ++curDown;
 
     ++curDown;
 
     ++curLeft;
 
     ++curLeft;
Line 113: Line 169:
 
}
 
}
 
</asy>
 
</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>
  
−
From the full diagram above, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math>
+
~MRENTHUSIASM ~Taco12
−
 
 
−
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) ==
 
== 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_8&oldid=225623"