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

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== Solution 1 ==
 
== Solution 1 ==
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Since <math>d</math> is a single digit prime number, the set of possible values of <math>d</math> is <math>\{2,3,5,7\}</math>.
 
  
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Since <math>e</math> is a single digit prime number and is the units digit of the prime number <math>10d+e</math>, the set of possible values of <math>e</math> is <math>\{3,7\}</math>.  
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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> <math>d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533</math>.
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So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math> ~ MathGenius_
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Using these values for <math>d</math> and <math>e</math>, the set of possible values of <math>10d+e</math> is <math>\{23,27,33,37,53,57,73,77\}</math>
 
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Out of this set, the prime values are <math>\{23,37,53,73\}</math>
 
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Therefore the possible values of <math>n</math> are:
 
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<math>2\cdot3\cdot23=138</math>
 
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<math>3\cdot7\cdot37=777</math>
 
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<math>5\cdot3\cdot53=795</math>
 
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<math>7\cdot3\cdot73=1533</math>
 
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The largest possible value of <math>n</math> is <math>1533</math>.  
 
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So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math>
 
  
 
== See Also ==
 
== See Also ==

Revision as of 11:43, 18 March 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 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

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