2003 AMC 10A Problems/Problem 14: Difference between revisions

Revision as of 09:51, 27 May 2021

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_

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 by WhyMath

https://youtu.be/Rssxmgt62A0

~savannahsolver

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 32: / 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>
+==Video Solution by WhyMath==
+https://youtu.be/Rssxmgt62A0
+~savannahsolver
 == See Also ==

Page

Toolbox

Search