2003 AMC 10A Problem 14

Question

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?

12

15

18

21

24

Answer: A

Concepts:prime digits systematic listing

Difficulty rating: 1500

Small hint:

The single-digit primes are

2,

3,

5,

7,

and

10d + e

must also be prime

Big hint:

Test candidate pairs in descending order of

d,

discarding any pair for which

10d+e

is composite

Solution:

Both

d

and

e

are distinct members of

\{2,3,5,7\},

and

10d+e

must also be prime.

Start with the largest possible tens digit. For

d=7,

the choices

e=5

and

e=2

give the composite numbers

75

and

72,

while

e=3

gives the prime

73.

This produces

n=7\cdot3\cdot73=1533.

For

d=5,

the choices

e=7

and

e=2

give the composite numbers

57

and

52,

while

e=3

gives only

5\cdot3\cdot53=795.

Every case with

d\le3

is at most

3\cdot7\cdot37=777.

Hence

1533

is the largest valid value.

The sum of its digits is

1 + 5 + 3 + 3 = 12.

Thus, the correct answer is A.