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

Revision as of 21:31, 10 November 2023

The following problem is from both the 2023 AMC 10A #3 and 2023 AMC 12A #3, so both problems redirect to this page.

Problem

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

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

Solution 1

Note that $40^2=1600$ but $45^{2}=2025$ (which is over our limit of $2023$ ). Therefore, the list is $5^2,10^2,15^2,20^2,25^2,30^2,35^2,40^2$ . There are $8$ elements, so the answer is $\boxed{\textbf{(A) 8}}$ .

~zhenghua ~walmartbrian (Minor edits for clarity by Technodoggo)

Solution 2 (slightly refined)

Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$ , there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$ perfect squares less than 2023.

~not_slay

Solution 3 (even better)

Since $5$ is prime, each solution must be divisible by $5^2=25$ . We take $\left \lfloor{\frac{2023}{25}}\right \rfloor = 80$ and see that there are $\boxed{\textbf{(A) 8}}$ positive perfect squares no greater than $80$ .

~jwseph

Solution 4

We know the highest value would be at least $40$ but less than $50$ so we check $45$ , prime factorizing 45. We get $3^2 \cdot 5$ . We square this and get $81 \cdot 25$ . We know that $80 \cdot 25 = 2000$ , then we add 25 and get $2025$ , which does not satisfy our requirement of having the square less than $2023$ . The largest multiple of $5$ that satisfies this is $40$ and the smallest multiple of $5$ that works is $5$ so all multiples of $5$ from $5$ to $40$ satisfy the requirements. Now we divide each element of the set by $5$ and get $1-8$ so there are $\boxed{\textbf{(A) 8}}$ solutions.

~kyogrexu (minor edits by vadava_lx)

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=E0a8wvcNRoeg2A3X&t=422

Video Solution

https://youtu.be/w7RBPIatRNE

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 26: / Line 26: @@
 We know the highest value would be at least <math>40</math> but less than <math>50</math> so we check <math>45</math>, prime factorizing 45. We get <math>3^2 \cdot 5</math>. We square this and get <math>81 \cdot 25</math>. We know that <math>80 \cdot 25 = 2000</math>, then we add 25 and get <math>2025</math>, which does not satisfy our requirement of having the square less than <math>2023</math>. The largest multiple of <math>5</math> that satisfies this is <math>40</math> and the smallest multiple of <math>5</math> that works is <math>5</math> so all multiples of <math>5</math> from <math>5</math> to <math>40</math> satisfy the requirements. Now we divide each element of the set by <math>5</math> and get <math>1-8</math> so there are <math>\boxed{\textbf{(A) 8}}</math> solutions.
-~kyogrexu(minor edits by vadava_lx)
+~kyogrexu (minor edits by vadava_lx)
 ==Video Solution by Math-X (First understand the problem!!!)==

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