Difference between revisions of "2023 AMC 8 Problems/Problem 4"
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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> | 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 | + | ~MathFun1000, MRENTHUSIASM, SophiaKim |
==Solution 2== | ==Solution 2== | ||
Revision as of 11:28, 3 March 2023
Contents
Problem
The numbers from 
to 
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 
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]](http://latex.artofproblemsolving.com/6/e/8/6e88180afd53a16ea3b9e0389e44933a31876e70.png)
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]](http://latex.artofproblemsolving.com/6/e/c/6ec6820e4734959c2e1ea32455e6e37e6ee38967.png)
From the four numbers that appear in the shaded squares, 
of them are prime: 
and 
~MathFun1000, MRENTHUSIASM, SophiaKim
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]](http://latex.artofproblemsolving.com/2/2/5/22579266b483266ee0dff48378bb0dd384c538a4.png)
From the four numbers that appear in the shaded squares, of them are prime: and
~TheMathGuyd
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/1bA7fD7Lg54?t=176
(Creative Thinking) Video Solution
~Education the Study of everything
Video Solution by WhyMath
~savannahsolver
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. 
