Difference between revisions of "2023 AMC 12A Problems/Problem 3"

Revision as of 16:05, 9 November 2023

How many positive perfect squares less than $2023$ are divisible by $5$ ?

$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$

Note that $45^{2}=2025$ so the list is $5,10,15,20,25,30,35,40$ there are $8$ elements so the answer is $\text{\boxed{(A)}}$

~zhenghua

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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 12 Problems and Solutions

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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

@@ Line 1: / Line 1: @@
-Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
+==Problem==
+How many positive perfect squares less than <math>2023</math> are divisible by <math>5</math>?
--Jonathan Yu
+<cmath>\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12</cmath>
+==Solution 1==
+Note that <math>45^{2}=2025</math> so the list is <math>5,10,15,20,25,30,35,40</math> there are <math>8</math> elements so the answer is <math>\text{\boxed{(A)}}</math>
+~zhenghua
+==See Also==
+{{AMC12 box|year=2023|ab=A|num-b=2|num-a=4}}
+{{AMC10 box|year=2023|ab=A|num-b=2|num-a=4}}
+{{MAA Notice}}