2003 AMC 10A Problems/Problem 14

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

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 $1+5+3+3=12 \Rightarrow A$

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2003 AMC 10A Problems/Problem 14

Problem

Solution

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