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

Latest revision as of 00:17, 19 May 2024

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

1 Problem
2 Solution 1
3 Solution 2 (Bash)(Only if you don't know how to do the rest of the problems and have about 20 minutes left)
4 Video Solution by Math-X (First understand the problem!!!)
5 Video Solution by CosineMethod [🔥Fast and Easy🔥]
6 Video Solution
7 Video Solution (⚡ Under 2 min ⚡)
8 Video Solution (easy to digest) by Power Solve
9 See Also

Problem

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

$\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad$

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 $2$ more digits. $16+2=\text{\boxed{\textbf{(E) }18}}$

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

~zhenghua

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

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

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

https://youtu.be/cMgngeSmFCY?si=b1khjbMn1i5rApCe&t=903

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

Video Solution (easy to digest) by Power Solve

https://youtu.be/Od1Spf3TDBs

2023 AMC 10A (Problems • Answer Key • Resources)
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 (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 AMC 12 Problems and Solutions

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

@@ 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==
-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 15^5</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\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad</math>
 ==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>2</math> more digits. <math>16+2=\text{\boxed{\textbf{(E) }18}}</math>
+<math>2.43*10^{17}</math> has <math>18</math> digits
 ~zhenghua
-==Solution 2==
+==Solution 2 (Bash)(Only if you don't know how to do the rest of the problems and have about 20 minutes left)==
-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
+==Video Solution by Math-X (First understand the problem!!!)==
+https://youtu.be/cMgngeSmFCY?si=b1khjbMn1i5rApCe&t=903
+== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
+https://www.youtube.com/watch?v=laHiorWO1zo
-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>.
+==Video Solution==
-Combining these two, we get <math>a<26-a\Rightarrow a<13</math>.
+https://youtu.be/MFPSxqtguQo
-The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
-~not_slay
+==Video Solution (⚡ Under 2 min ⚡)==
+https://youtu.be/Xy8vyymlPBg
-== Solution 3 (Fast) ==
+~Education, the Study of Everything
-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://artofproblemsolving.com/wiki/index.php/User:South South]
+==Video Solution (easy to digest) by Power Solve==
+https://youtu.be/Od1Spf3TDBs
-== 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}}
 {{MAA Notice}}

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