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

Latest revision as of 14:49, 19 August 2023

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 we want $n$ to be as large as possible, we would like $d$ in $10d+e$ to be as large as possible. So, $d=7,$ the greatest single-digit prime. Then, $e$ cannot be $5$ because $10(7)+5 = 75,$ which is not prime. So $e = 3$ . Therefore, $d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533$ . So, the sum of the digits of $n$ is $1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}$ ~ MathGenius_ (Edited by Sophia866)

Solution 2

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 $1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}$

Video Solution(s)

https://youtu.be/NbkwhS7k2oU

~savannahsolver

https://www.youtube.com/watch?v=yApq-Vny_A0

~David

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.

@@ Line 5: / Line 5: @@
 == Solution 1 ==
+Since we want <math>n</math> to be as large as possible, we would like <math>d</math> in <math>10d+e</math> to be as large as possible. So, <math>d=7,</math> the greatest single-digit prime. 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>. Therefore, <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_ (Edited by Sophia866)
+== Solution 2 ==
 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>.
@@ Line 27: / Line 32: @@
 So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math>
-== Solution 2 ==
+==Video Solution(s)==
-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 7s = 1533</math>.
+https://youtu.be/NbkwhS7k2oU
-So, the sum of the digits of <math>n</math> is <math>1+5+3+3=12 \Rightarrow \boxed{\mathrm{(A)}\ 12}</math>
+~savannahsolver
+https://www.youtube.com/watch?v=yApq-Vny_A0
+~David
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