Difference between revisions of "2023 AMC 8 Problems/Problem 4"
MRENTHUSIASM (talk | contribs) (→Solution: Considering the perfect squares should be an alternate approach, but harder in my opinion.) |
MRENTHUSIASM (talk | contribs) (→Solution) |
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<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== | + | ==Solution 1== |
We fill out the grid, as shown below: | We fill out the grid, as shown below: |
Revision as of 18:17, 27 January 2023
Contents
[hide]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?
Solution 1
We fill out the grid, as shown below: From the four numbers that appear in the shaded squares, of them are prime: and
~MathFun1000, MRENTHUSIASM
Solution 2
Note that given time constraint, it's better to only count from perfect squares (in pink), as shown below: 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
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.