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Difference between revisions of "2023 AMC 8 Problems/Problem 4"

(→‎Solution 1)
(→‎Solution 2)
 
(38 intermediate revisions by 18 users not shown)
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==Problem==
 
==Problem==
  
−
The numbers from 1 to 49 are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number 7. How many of these four numbers are prime?
+
The numbers from <math>1</math> to <math>49</math> are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number <math>7.</math> How many of these four numbers are prime?
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(175);
  
 +
void ds(pair p) {
 +
filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey);
 +
}
 +
 +
ds((0.5,4.5));
 +
ds((1.5,3.5));
 +
ds((3.5,1.5));
 +
ds((4.5,0.5));
 +
 +
add(grid(7,7,grey+linewidth(1.25)));
 +
 +
int adj = 1;
 +
int curUp = 2;
 +
int curLeft = 4;
 +
int curDown = 6;
 +
 +
label("$1$",(3.5,3.5));
 +
 +
for (int len = 3; len<=3; len+=2)
 +
{
 +
for (int i=1; i<=len-1; ++i)
 +
  {
 +
label("$"+string(curUp)+"$",(3.5+adj,3.5-adj+i));
 +
    label("$"+string(curLeft)+"$",(3.5+adj-i,3.5+adj));
 +
    label("$"+string(curDown)+"$",(3.5-adj,3.5+adj-i));
 +
    ++curDown;
 +
    ++curLeft;
 +
    ++curUp;
 +
}
 +
++adj;
 +
    curUp = len^2 + 1;
 +
    curLeft = len^2 + len + 2;
 +
    curDown = len^2 + 2*len + 3;
 +
}
 +
 +
draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2));
 +
</asy>
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4</math>
  
 
==Solution 1==
 
==Solution 1==
  
−
First, we fill out the entire grid. We find that the <math>4</math> numbers are <math>39,19,23,47</math>. The numbers <math>19,23,</math> and <math>47</math> are prime, so there are <math>\boxed{\textbf{(D) }3}</math> prime numbers.
+
We fill out the grid, as shown below:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(175);
 +
 
 +
void ds(pair p) {
 +
filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey);
 +
}
 +
 
 +
ds((0.5,4.5));
 +
ds((1.5,3.5));
 +
ds((3.5,1.5));
 +
ds((4.5,0.5));
 +
 
 +
add(grid(7,7,grey+linewidth(1.25)));
 +
 
 +
int adj = 1;
 +
int curUp = 2;
 +
int curLeft = 4;
 +
int curDown = 6;
 +
int curRight = 8;
  
−
~MathFun1000 (minor edits apex304 and [[User:ILoveMath31415926535|ILoveMath31415926535]])
+
label("$1$",(3.5,3.5));
 +
 
 +
for (int len = 3; len<=7; len+=2)
 +
{
 +
for (int i=1; i<=len-1; ++i)
 +
  {
 +
label("$"+string(curUp)+"$",(3.5+adj,3.5-adj+i));
 +
    label("$"+string(curLeft)+"$",(3.5+adj-i,3.5+adj));
 +
    label("$"+string(curDown)+"$",(3.5-adj,3.5+adj-i));
 +
    label("$"+string(curRight)+"$",(3.5-adj+i,3.5-adj));
 +
    ++curDown;
 +
    ++curLeft;
 +
    ++curUp;
 +
    ++curRight;
 +
}
 +
++adj;
 +
    curUp = len^2 + 1;
 +
    curLeft = len^2 + len + 2;
 +
    curDown = len^2 + 2*len + 3;
 +
    curRight = len^2 + 3*len + 4;
 +
}
 +
 
 +
draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2));
 +
</asy>
 +
From the four numbers that appear in the shaded squares, <math>\boxed{\textbf{(D)}\ 3}</math> of them are prime: <math>19,23,</math> and <math>47.</math>
 +
 
 +
~MathFun1000, MRENTHUSIASM
  
 
==Solution 2==
 
==Solution 2==
 +
Note that given time constraint, it's better to only count from perfect squares (in pink), as shown below:
 +
<asy>
 +
/* Grid Made by MRENTHUSIASM */
 +
/* Squares pattern solution input by TheMathGuyd */
 +
size(175);
  
−
Fill out the entire grid to count that there are <math>\boxed{\text{(D)}3}</math> prime numbers
+
void ds(pair p) {
−
-apex304
+
filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey);
 +
}
 +
 
 +
void ps(pair p) {
 +
filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,pink+opacity(0.3));
 +
}
 +
real ts = 0.5;
 +
 
 +
ds((0.5,4.5));label("$39$",(0.5,4.5));
 +
ds((1.5,3.5));label("$19$",(1.5,3.5));
 +
ds((3.5,1.5));label("$23$",(3.5,1.5));
 +
ds((4.5,0.5));label("$47$",(4.5,0.5));
 +
 
 +
ps((3.5,3.5));label("$1$",(3.5,3.5));
 +
ps((4.5,2.5));label("$9$",(4.5,2.5));
 +
ps((5.5,1.5));label("$25$",(5.5,1.5));
 +
ps((6.5,0.5));label("$49$",(6.5,0.5));
 +
ps((3.5,4.5));label("$4$",(3.5,4.5));
 +
ps((2.5,5.5));label("$16$",(2.5,5.5));
 +
ps((1.5,6.5));label("$36$",(1.5,6.5));
 +
label(scale(ts)*"$\leftarrow$",(1,6),NE);
 +
label(scale(ts)*"$+1$",(1,6),NW);
 +
label(scale(ts)*"$\downarrow$",(1,6),SW);
 +
label(scale(ts)*"$+2$",(1,5),NW);
 +
label(scale(ts)*"$\downarrow$",(1,5),SW);
 +
label(scale(ts)*"$+3$",(1,4),NW);
 +
label(scale(ts)*"$+1$",(2,5),NW);
 +
label(scale(ts)*"$\downarrow$",(2,5),SW);
 +
label(scale(ts)*"$+2$",(2,4),NW);
 +
label(scale(ts)*"$\downarrow$",(2,4),SW);
 +
label(scale(ts)*"$+3$",(2,3),NW);
 +
 
 +
label(scale(ts)*"$\leftarrow$",(5,1),NE);
 +
label(scale(ts)*"$-1$",(5,1),NW);
 +
label(scale(ts)*"$\leftarrow$",(4,1),NE);
 +
label(scale(ts)*"$-2$",(4,1),NW);
 +
label(scale(ts)*"$\leftarrow$",(6,0),NE);
 +
label(scale(ts)*"$-1$",(6,0),NW);
 +
label(scale(ts)*"$\leftarrow$",(5,0),NE);
 +
label(scale(ts)*"$-2$",(5,0),NW);
 +
 
 +
add(grid(7,7,grey+linewidth(1.25))); //USES OLYMPIAD.ASY
 +
 
 +
draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2));
 +
</asy>
 +
From the four numbers that appear in the shaded squares, <math>\boxed{\textbf{(D)}\ 3}</math> of them are prime: <math>19,23,</math> and <math>47.</math>
 +
 
 +
~TheMathGuyd
 +
 
 +
==Solution 3==
 +
 
 +
Since the last shaded square is very close to the final number, which is 49, we can easily subtract two instead of having to count all the way there. Then, we can count (:P).
 +
 
 +
~AliceDubbleYou
 +
 
 +
==Simple, Intuitive Solution by MathTalks_Now==
 +
* *Different Solution not shown before!*
 +
 
 +
https://studio.youtube.com/video/PMOeiGLkDH0/edit
 +
 
 +
==Video Solution (How to Creatively THINK!!!)==
 +
https://youtu.be/7gwhzjySKl0
 +
 
 +
~Education the Study of everything
 +
 
 +
==Video Solution by Math-X (Smart and Simple)==
 +
https://youtu.be/Ku_c1YHnLt0?si=cc_Ii2j2pmT6wOuZ&t=412
 +
 
 +
~Math-X
  
 
==Video Solution by Magic Square==
 
==Video Solution by Magic Square==
 
https://youtu.be/-N46BeEKaCQ?t=5392
 
https://youtu.be/-N46BeEKaCQ?t=5392
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=EcrktBc8zrM
 +
==Video Solution by Interstigation==
 +
https://youtu.be/DBqko2xATxs&t=233
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/1qwfPJDNYGc
 +
 +
~savannahsolver
 +
 +
==Video Solution by harungurcan==
 +
https://www.youtube.com/watch?v=35BW7bsm_Cg&t=402s
 +
 +
~harungurcan
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/9ynYd5V0r88
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=3|num-a=5}}
 
{{AMC8 box|year=2023|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:09, 3 November 2024

Problem

The numbers from $1$ to $49$ are arranged in a spiral pattern on a square grid, beginning at the center. The first few numbers have been entered into the grid below. Consider the four numbers that will appear in the shaded squares, on the same diagonal as the number $7.$ How many of these four numbers are prime? [asy] /* Made by MRENTHUSIASM */ size(175); void ds(pair p) { filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey); } ds((0.5,4.5)); ds((1.5,3.5)); ds((3.5,1.5)); ds((4.5,0.5)); add(grid(7,7,grey+linewidth(1.25))); int adj = 1; int curUp = 2; int curLeft = 4; int curDown = 6; label("$1$",(3.5,3.5)); for (int len = 3; len<=3; len+=2) { for (int i=1; i<=len-1; ++i) { label("$"+string(curUp)+"$",(3.5+adj,3.5-adj+i)); label("$"+string(curLeft)+"$",(3.5+adj-i,3.5+adj)); label("$"+string(curDown)+"$",(3.5-adj,3.5+adj-i)); ++curDown; ++curLeft; ++curUp; } ++adj; curUp = len^2 + 1; curLeft = len^2 + len + 2; curDown = len^2 + 2*len + 3; } draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2)); [/asy] $\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution 1

We fill out the grid, as shown below: [asy] /* Made by MRENTHUSIASM */ size(175); void ds(pair p) { filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey); } ds((0.5,4.5)); ds((1.5,3.5)); ds((3.5,1.5)); ds((4.5,0.5)); add(grid(7,7,grey+linewidth(1.25))); int adj = 1; int curUp = 2; int curLeft = 4; int curDown = 6; int curRight = 8; label("$1$",(3.5,3.5)); for (int len = 3; len<=7; len+=2) { for (int i=1; i<=len-1; ++i) { label("$"+string(curUp)+"$",(3.5+adj,3.5-adj+i)); label("$"+string(curLeft)+"$",(3.5+adj-i,3.5+adj)); label("$"+string(curDown)+"$",(3.5-adj,3.5+adj-i)); label("$"+string(curRight)+"$",(3.5-adj+i,3.5-adj)); ++curDown; ++curLeft; ++curUp; ++curRight; } ++adj; curUp = len^2 + 1; curLeft = len^2 + len + 2; curDown = len^2 + 2*len + 3; curRight = len^2 + 3*len + 4; } draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2)); [/asy] From the four numbers that appear in the shaded squares, $\boxed{\textbf{(D)}\ 3}$ of them are prime: $19,23,$ and $47.$

~MathFun1000, MRENTHUSIASM

Solution 2

Note that given time constraint, it's better to only count from perfect squares (in pink), as shown below: [asy] /* Grid Made by MRENTHUSIASM */ /* Squares pattern solution input by TheMathGuyd */ size(175); void ds(pair p) { filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,mediumgrey); } void ps(pair p) { filldraw((0.5,0.5)+p--(-0.5,0.5)+p--(-0.5,-0.5)+p--(0.5,-0.5)+p--cycle,pink+opacity(0.3)); } real ts = 0.5; ds((0.5,4.5));label("$39$",(0.5,4.5)); ds((1.5,3.5));label("$19$",(1.5,3.5)); ds((3.5,1.5));label("$23$",(3.5,1.5)); ds((4.5,0.5));label("$47$",(4.5,0.5)); ps((3.5,3.5));label("$1$",(3.5,3.5)); ps((4.5,2.5));label("$9$",(4.5,2.5)); ps((5.5,1.5));label("$25$",(5.5,1.5)); ps((6.5,0.5));label("$49$",(6.5,0.5)); ps((3.5,4.5));label("$4$",(3.5,4.5)); ps((2.5,5.5));label("$16$",(2.5,5.5)); ps((1.5,6.5));label("$36$",(1.5,6.5)); label(scale(ts)*"$\leftarrow$",(1,6),NE); label(scale(ts)*"$+1$",(1,6),NW); label(scale(ts)*"$\downarrow$",(1,6),SW); label(scale(ts)*"$+2$",(1,5),NW); label(scale(ts)*"$\downarrow$",(1,5),SW); label(scale(ts)*"$+3$",(1,4),NW); label(scale(ts)*"$+1$",(2,5),NW); label(scale(ts)*"$\downarrow$",(2,5),SW); label(scale(ts)*"$+2$",(2,4),NW); label(scale(ts)*"$\downarrow$",(2,4),SW); label(scale(ts)*"$+3$",(2,3),NW); label(scale(ts)*"$\leftarrow$",(5,1),NE); label(scale(ts)*"$-1$",(5,1),NW); label(scale(ts)*"$\leftarrow$",(4,1),NE); label(scale(ts)*"$-2$",(4,1),NW); label(scale(ts)*"$\leftarrow$",(6,0),NE); label(scale(ts)*"$-1$",(6,0),NW); label(scale(ts)*"$\leftarrow$",(5,0),NE); label(scale(ts)*"$-2$",(5,0),NW); add(grid(7,7,grey+linewidth(1.25))); //USES OLYMPIAD.ASY draw((4,4)--(3,4)--(3,3)--(5,3)--(5,5)--(2,5)--(2,2)--(6,2)--(6,6)--(1,6)--(1,1)--(7,1)--(7,7)--(0,7)--(0,0)--(7,0),linewidth(2)); [/asy] From the four numbers that appear in the shaded squares, $\boxed{\textbf{(D)}\ 3}$ of them are prime: $19,23,$ and $47.$

~TheMathGuyd

Solution 3

Since the last shaded square is very close to the final number, which is 49, we can easily subtract two instead of having to count all the way there. Then, we can count (:P).

~AliceDubbleYou

Simple, Intuitive Solution by MathTalks_Now

  • *Different Solution not shown before!*

https://studio.youtube.com/video/PMOeiGLkDH0/edit

Video Solution (How to Creatively THINK!!!)

https://youtu.be/7gwhzjySKl0

~Education the Study of everything

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=cc_Ii2j2pmT6wOuZ&t=412

~Math-X

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=5392

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=EcrktBc8zrM

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=233

Video Solution by WhyMath

https://youtu.be/1qwfPJDNYGc

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=35BW7bsm_Cg&t=402s

~harungurcan

Video Solution by Dr. David

https://youtu.be/9ynYd5V0r88

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 AJHSME/AMC 8 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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