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Difference between revisions of "2023 AMC 8 Problems/Problem 4"
Line 90: Line 90:
 
     curRight = len^2 + 3*len + 4;
 
     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>
 +
Note that given time constraint, its better to only count from perfect squares (in pink).
 +
<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("$\mathbf{39}$",(0.5,4.5));
 +
ds((1.5,3.5));label("$\mathbf{19}$",(1.5,3.5));
 +
ds((3.5,1.5));label("$\mathbf{23}$",(3.5,1.5));
 +
ds((4.5,0.5));label("$\mathbf{47}$",(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));
 
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));
Line 95: Line 147:
 
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, TheMathGuyd
 +
 
 +
 
  
 
==Video Solution by Magic Square==
 
==Video Solution by Magic Square==

Revision as of 11:45, 27 January 2023

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

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] Note that given time constraint, its better to only count from perfect squares (in pink). [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("$\mathbf{39}$",(0.5,4.5)); ds((1.5,3.5));label("$\mathbf{19}$",(1.5,3.5)); ds((3.5,1.5));label("$\mathbf{23}$",(3.5,1.5)); ds((4.5,0.5));label("$\mathbf{47}$",(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 $47.$

~MathFun1000, MRENTHUSIASM, TheMathGuyd


Video Solution by Magic Square

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

Video Solution by SpreadTheMathLove

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

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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