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

m (→‎Diagram: Deleted commas in the diagram.)
(→‎Solution 3 (Full Version of Solution 2): Provided illustrations for Solutions 1 and 2.)
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~Taco12
 
~Taco12
  
−
==Solution 3 (Full Version of Solution 2)==
+
==Solution 3==
−
===Diagram===
+
In both solutions below, note that the numbers along the yellow cells are consecutive odd perfect squares. We can show this by induction. Two pictorial solutions follow from here.
 +
 
 +
===Solution 3.1 (Illustration of Solution 1--Only Considers 5 Squares)===
 +
<asy>
 +
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("$A$",(14.5,14.5));
 +
label("$B$",(13.5,13.5));
 +
label("$C$",(0.5,14.5));
 +
 
 +
fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green);
 +
label("$E$",(1.5,13.5));
 +
fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);
 +
label("$D$",(0.5,13.5));
 +
 
 +
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);
 +
 
 +
add(grid(15,15,linewidth(1.25)));
 +
</asy>
 +
 
 +
In the full diagram above, the red arrows indicate the progression of numbers. By observations, we find out the numbers in the key squares procedurally:
 +
<cmath>\begin{array}{llll}
 +
A=225, \ B=169 &\Longrightarrow C &= A-14 &= 211 \\
 +
&\Longrightarrow D &= C-1 &= 210 \\
 +
&\Longrightarrow E &= B-12 &= 157
 +
\end{array}</cmath>
 +
Therefore, the answer is <math>D+E=\boxed{\textbf{(A)} ~367}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
===Solution 3.2 (Illustration of Solution 2--Draws All 225 Squares Out)===
 
<asy>
 
<asy>
 
size(11.5cm);
 
size(11.5cm);
Line 116: Line 152:
 
</asy>
 
</asy>
  
−
~MRENTHUSIASM (by Asymptote)
 
−
 
−
===Answer===
 
 
From the full diagram above, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math>
 
From the full diagram above, the answer is <math>210+157=\boxed{\textbf{(A)} ~367}.</math>
−
 
−
Note that the numbers along the yellow cells are consecutive odd perfect squares. We can show this by induction.
 
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 23:03, 6 March 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

By observing that the right-top corner of a square will always be a square, we know that the top right corner of the $15$x$15$ 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 2nd 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 2nd from the left and 2nd from the top, which is $157$. We know that the least number in the 2nd 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 2: Draw It Out

Drawing out the diagram, we get $\boxed{\mathbf{(A)}\ 367}$. Note that this should mainly be used just to check your answer.

~Taco12

Solution 3

In both solutions below, note that the numbers along the yellow cells are consecutive odd perfect squares. We can show this by induction. Two pictorial solutions follow from here.

Solution 3.1 (Illustration of Solution 1--Only Considers 5 Squares)

[asy] 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("$A$",(14.5,14.5)); label("$B$",(13.5,13.5)); label("$C$",(0.5,14.5)); fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); label("$E$",(1.5,13.5)); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); label("$D$",(0.5,13.5)); 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); add(grid(15,15,linewidth(1.25))); [/asy]

In the full diagram above, the red arrows indicate the progression of numbers. By observations, we find out the numbers in the key squares procedurally: \[\begin{array}{llll} A=225, \ B=169 &\Longrightarrow C &= A-14 &= 211 \\ &\Longrightarrow D &= C-1 &= 210 \\ &\Longrightarrow E &= B-12 &= 157 \end{array}\] Therefore, the answer is $D+E=\boxed{\textbf{(A)} ~367}.$

~MRENTHUSIASM

Solution 3.2 (Illustration of Solution 2--Draws All 225 Squares Out)

[asy] 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]

From the full diagram above, the answer is $210+157=\boxed{\textbf{(A)} ~367}.$

~MRENTHUSIASM

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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