Difference between revisions of "2023 AMC 12A Problems/Problem 4"

(shift to 12A)
(Problem)
(39 intermediate revisions by 21 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2023 AMC 10A Problems/Problem 5|2023 AMC 10A #5]] and [[2023 AMC 12A Problems/Problem 4|2023 AMC 12A #4]]}}
 +
 
==Problem==
 
==Problem==
A quadrilateral has all integer sides lengths, a perimeter of <math>26</math>, and one side of length <math>4</math>. What is the greatest possible length of one side of this quadrilateral?
+
How many digits are in the base-ten representation of <math>8^5 \cdot 5^{10} \cdot 15^2</math>?
  
<math>\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13</math>
+
<math>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\texbf{(D)}~17\qquad\textbf{(E)}~18</math>
  
 
==Solution 1==
 
==Solution 1==
Let's use the triangle inequality. We know that for a triangle, the 2 shorter sides must always be longer than the longest side. Similarly for a convex quadrilateral, the shortest 3 sides must always be longer than the longest side. Thus, the answer is <math>\frac{26}{2}-1=13-1=\boxed {\textbf{(D) 12}}</math>
+
Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}</math>.
 +
 
 +
<math>10^{15}</math> has <math>16</math> digits and <math>243</math> = <math>2.43*10^{2}</math> gives us <math>3</math> more digits. <math>16+2=\text{\boxed{\textbf{(E) }18}}</math>
 +
 
 +
<math>2.43*10^{17}</math> has <math>18</math> digits
  
 
~zhenghua
 
~zhenghua
  
==Solution 2==
+
==Solution 2 (Only if you don't know how to do the rest of the problems and have about 20 minutes left, IT IS  recommended)==
Say the chosen side is <math>a</math> and the other sides are <math>b,c,d</math>.
+
Multiplying it out, we get that <math>8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000</math>. Counting, we have the answer is <math>\text{\boxed{\textbf{(E) }18}}</math>
 +
~andliu766
 +
 
 +
==Solution 3 (Similar to Solution 1)==
 +
All the exponents have a common factor of <math>5</math> which we can factor out. This leaves us with <math>(8 \cdot 5^2 \cdot 15)^5 = (3000)^5 = (3 \cdot 1000)^5</math>. We can then distribute the power leaving us with <math>3^5 \cdot 10^{3 \cdot 5} = 243 \cdot 10^{15}</math>. This would be <math>243</math> followed by <math>15</math> zeros resulting in our answer being <math>15+3=\text{\boxed{\textbf{(E)}18}}</math>
 +
 
 +
~leon_0iler
 +
 
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=mJjq7vPLptdSe0AJ&t=872
 +
~little-fermat
 +
 
 +
==Video Solution (easy to digest) by Power Solve==
 +
https://youtu.be/Od1Spf3TDBs
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
  
By the Generalised Polygon Inequality, <math>a<b+c+d</math>. We also have <math>a+b+c+d=26\Rightarrow b+c+d=26-a</math>.
+
https://www.youtube.com/watch?v=laHiorWO1zo
  
Combining these two, we get <math>a<26-a\Rightarrow a<13</math>.
+
==Video Solution==
  
The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
+
https://youtu.be/MFPSxqtguQo
  
~not_slay
+
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
== Solution 3 (Fast) ==
+
==Video Solution (⚡ Under 2 min ⚡)==
By Brahmagupta's Formula, the area of the rectangle is defined by <math>\sqrt{(s-a)(s-b)(s-c)(s-d)}</math> where <math>s</math> is the semi-perimeter. If the perimeter of the rectangle is <math>26</math>, then the semi-perimeter will be <math>13</math>. The area of the rectangle must be positive so the difference between the semi-perimeter and a side length must be greater than <math>0</math> as otherwise, the area will be <math>0</math> or negative. Therefore, the longest a side can possibly be in this rectangle is <math>\boxed {\textbf{(D) 12}}</math>
+
https://youtu.be/Xy8vyymlPBg
  
~[https://artofproblemsolving.com/wiki/index.php/User:South South]
+
~Education, the Study of Everything
  
== See Also ==
+
==See Also==
{{AMC10 box|year=2023|ab=A|num-b=3|num-a=5}}
+
{{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}}
 
{{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}}
 
{{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:50, 13 November 2024

The following problem is from both the 2023 AMC 10A #5 and 2023 AMC 12A #4, so both problems redirect to this page.

Problem

How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^2$?

$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\texbf{(D)}~17\qquad\textbf{(E)}~18$ (Error compiling LaTeX. Unknown error_msg)

Solution 1

Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$.

$10^{15}$ has $16$ digits and $243$ = $2.43*10^{2}$ gives us $3$ more digits. $16+2=\text{\boxed{\textbf{(E) }18}}$

$2.43*10^{17}$ has $18$ digits

~zhenghua

Solution 2 (Only if you don't know how to do the rest of the problems and have about 20 minutes left, IT IS recommended)

Multiplying it out, we get that $8^5 \cdot 5^{10} \cdot 15^5 = 243000000000000000$. Counting, we have the answer is $\text{\boxed{\textbf{(E) }18}}$ ~andliu766

Solution 3 (Similar to Solution 1)

All the exponents have a common factor of $5$ which we can factor out. This leaves us with $(8 \cdot 5^2 \cdot 15)^5 = (3000)^5 = (3 \cdot 1000)^5$. We can then distribute the power leaving us with $3^5 \cdot 10^{3 \cdot 5} = 243 \cdot 10^{15}$. This would be $243$ followed by $15$ zeros resulting in our answer being $15+3=\text{\boxed{\textbf{(E)}18}}$

~leon_0iler

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=mJjq7vPLptdSe0AJ&t=872 ~little-fermat

Video Solution (easy to digest) by Power Solve

https://youtu.be/Od1Spf3TDBs

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=laHiorWO1zo

Video Solution

https://youtu.be/MFPSxqtguQo

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

Video Solution (⚡ Under 2 min ⚡)

https://youtu.be/Xy8vyymlPBg

~Education, the Study of Everything

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 (ProblemsAnswer KeyResources)
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 AMC 12 Problems and Solutions

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