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

Revision as of 23:14, 9 November 2023

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 $2^{15}\cdot3^{5}\cdot5^{15}$ Pairing $2^{15}$ and $5^{15}$ gives us a number with $15$ zeros giving us 15 digits. $3^5=243$ and this adds an extra 3 digits. $15+3=\text{\boxed{(E)18}}$

~zhenghua

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 12 Problems and Solutions

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

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

@@ Line 1: / Line 1: @@
 ==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>
+<cmath>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad</cmath>
 ==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}</math> Pairing <math>2^{15}</math> and <math>5^{15}</math> gives us a number with <math>15</math> zeros giving us 15 digits. <math>3^5=243</math> and this adds an extra 3 digits. <math>15+3=\text{\boxed{(E)18}}</math>
 ~zhenghua
-==Solution 2==
+==See Also==
-Say the chosen side is <math>a</math> and the other sides are <math>b,c,d</math>.
-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>.
-Combining these two, we get <math>a<26-a\Rightarrow a<13</math>.
-The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
-~not_slay
-== Solution 3 (Fast) ==
-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]
-== See Also ==
-{{AMC10 box|year=2023|ab=A|num-b=3|num-a=5}}
 {{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}}
+{{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}}
 {{MAA Notice}}

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

Revision as of 23:14, 9 November 2023

Problem

Solution 1

See Also