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

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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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