Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 10A Problems/Problem 14"

(→‎Solution 2)
(→‎Solution 2)
Line 28: Line 28:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Since you want <math>n</math> to be the largest number possible, you will want <math>d</math> in <math>10d+e</math> to be as large as possible. So <math>d = 7</math>.Then, <math>e</math> cannot be <math>5</math> because <math>10(7)+5 = 75</math> which is not prime. So <math>e = 3</math>.<math>~</math>
+
Since you want <math>n</math> to be the largest number possible, you will want <math>d</math> in <math>10d+e</math> to be as large as possible. So <math>d = 7</math>.Then, <math>e</math> cannot be <math>5</math> because <math>10(7)+5 = 75</math> which is not prime. So <math>e = 3</math>.   <math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533</math>.
−
<math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533</math>.
 
 
So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> ~ MathGenius_
 
So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> ~ MathGenius_
  

Revision as of 09:26, 14 January 2020

Problem

Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$, $e$, and $10d+e$, where $d$ and $e$ are single digits. What is the sum of the digits of $n$?

$\mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24$

Solution 1

Since $d$ is a single digit prime number, the set of possible values of $d$ is $\{2,3,5,7\}$.

Since $e$ is a single digit prime number and is the units digit of the prime number $10d+e$, the set of possible values of $e$ is $\{3,7\}$.

Using these values for $d$ and $e$, the set of possible values of $10d+e$ is $\{23,27,33,37,53,57,73,77\}$

Out of this set, the prime values are $\{23,37,53,73\}$

Therefore the possible values of $n$ are:

$2\cdot3\cdot23=138$

$3\cdot7\cdot37=777$

$5\cdot3\cdot53=795$

$7\cdot3\cdot73=1533$

The largest possible value of $n$ is $1533$.

So, the sum of the digits of $n$ is $1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}$

Solution 2

Since you want $n$ to be the largest number possible, you will want $d$ in $10d+e$ to be as large as possible. So $d = 7$.Then, $e$ cannot be $5$ because $10(7)+5 = 75$ which is not prime. So $e = 3$. $d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533$. So, the sum of the digits of $n$ is $1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}$ ~ MathGenius_

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_14&oldid=114718"