AMC 10 · 2003 · #14
Grade 4 number-theoryLet 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?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Choose single-digit numbers $d$ and $e$ so that $d$, $e$, and the two-digit number $10d+e$ are all prime and all different. Among every such choice, take the one that makes the product $n = d \cdot e \cdot (10d+e)$ as large as possible, then add up the digits of that $n$.
Givens: $d$ and $e$ are single digits.; $d$, $e$, and $10d+e$ must each be prime.; The three primes $d$, $e$, and $10d+e$ must be distinct.; $n = d \cdot e \cdot (10d+e)$, and $n$ is to be the largest value possible.; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $24$.
Unknowns: The digits $d$ and $e$ that maximize $n$.; The sum of the digits of the largest possible $n$.
Understand
Restated: Choose single-digit numbers $d$ and $e$ so that $d$, $e$, and the two-digit number $10d+e$ are all prime and all different. Among every such choice, take the one that makes the product $n = d \cdot e \cdot (10d+e)$ as large as possible, then add up the digits of that $n$.
Givens: $d$ and $e$ are single digits.; $d$, $e$, and $10d+e$ must each be prime.; The three primes $d$, $e$, and $10d+e$ must be distinct.; $n = d \cdot e \cdot (10d+e)$, and $n$ is to be the largest value possible.; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $24$.
Plan
Primary tool: #14 Extreme Principle
Secondary: #3 Eliminate Possibilities, #2 Make a Systematic List
The problem asks for the largest product, so Tool #14 (Extreme Principle) drives the search: push the biggest factor to its limit. The product $n = d \cdot e \cdot (10d+e)$ is dominated by the two-digit factor $10d+e$, and that factor grows fastest by making the tens digit $d$ as large as possible. Because $d$ and $e$ can only be single-digit primes, Tool #3 (Eliminate Possibilities) shrinks the search to the four numbers $2,3,5,7$, and Tool #2 (Make a Systematic List) checks each candidate for the second digit until $10d+e$ comes out prime. That turns an open-ended "largest" question into checking just a handful of cases.
Execute — Answer: A
4.OA.B.4 Step 1 List the single-digit primes
- Both $d$ and $e$ must be prime and a single digit.
- The single-digit primes are exactly $2$, $3$, $5$, and $7$ — no other single digit is prime ($1$ is not prime, and $4,6,8,9$ are composite).
- So every valid $d$ and $e$ is drawn from $\{2,3,5,7\}$, and they must be different from each other.
💡 Only four single digits are prime, so the whole problem lives inside a set of just four numbers.
4.OA.B.4 Step 2 Make the tens digit as large as possible
- The product $n = d \cdot e \cdot (10d+e)$ is largest when its biggest factor $10d+e$ is largest, and $10d+e$ grows fastest by making the tens digit $d$ as big as it can be.
- The largest single-digit prime is $7$, so try $d = 7$ first.
- This aims straight at the maximum instead of testing every pair blindly.
💡 A bigger tens digit lifts the two-digit factor by tens at a time, so start the search at the top.
4.OA.B.4 Step 3 Find e so that 70+e is prime
- With $d = 7$, the units digit $e$ must be a single-digit prime different from $7$, so $e \in \{2,3,5\}$, and $70+e$ must be prime.
- Check them: $70+2 = 72 = 8 \cdot 9$ is composite; $70+5 = 75 = 3 \cdot 25$ is composite; $70+3 = 73$ is prime.
- So the only working choice at $d=7$ is $e = 3$, giving the three distinct primes $7$, $3$, and $73$.
💡 Only one of the three possible units digits turns $7\_$ into a prime, and it is $3$.
4.NBT.B.5 Step 4 Multiply to get n, then add its digits
- Since $d=7$ already gives the largest possible tens digit, no smaller $d$ can beat it, so $n$ is largest at $d=7$, $e=3$.
- Multiply the three primes: $n = 7 \cdot 3 \cdot 73 = 21 \cdot 73 = 1533$.
- The sum of its digits is $1+5+3+3 = 12$, which is choice (A).
💡 Pin down the biggest factor first, multiply, then just read off and add the four digits.
4.OA.B.4 Both $d$ and $e$ must be prime and a single digit. The single-digit primes are e 4.OA.B.4 The product $n = d \cdot e \cdot (10d+e)$ is largest when its biggest factor $10 4.OA.B.4 With $d = 7$, the units digit $e$ must be a single-digit prime different from $7 4.NBT.B.5 Since $d=7$ already gives the largest possible tens digit, no smaller $d$ can be Review
Reasonableness: The choice $d=7,\,e=3$ uses the largest single-digit prime as the tens digit, so its two-digit factor $73$ is the largest achievable of the required form; any other start ($d=5$ gives at most $53$, $d=3$ gives at most $37$, $d=2$ gives at most $23$) produces a smaller top factor and hence a smaller product. So $n=1533$ is genuinely the maximum. Its digit sum $12$ is one of the listed choices, matching (A), and the smaller-$d$ products would give different digit sums, so the answer is not ambiguous.
Alternative: List every valid triple directly (Tool #2, Make a Systematic List): the two-digit primes $10d+e$ with both digits prime and distinct are $23\,(2\cdot 3),\,37\,(3\cdot 7),\,53\,(5\cdot 3),\,73\,(7\cdot 3)$, giving products $138,\,777,\,795,\,1533$. The largest is $1533$, whose digits sum to $12$ — the same answer (A).
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Identifying the single-digit primes $\{2,3,5,7\}$ and testing whether each candidate two-digit number $10d+e$ (such as $72,75,73$) is prime or composite.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing the product $n = 7 \cdot 3 \cdot 73 = 1533$ before summing its digits.)
⭐ To make the product biggest, make the two-digit prime biggest: start the tens digit at the largest single-digit prime $7$, find the one units digit that keeps it prime ($73$), and $7 \cdot 3 \cdot 73 = 1533$ has digit sum $12$.
⭐ To make the product biggest, make the two-digit prime biggest: start the tens digit at the largest single-digit prime $7$, find the one units digit that keeps it prime ($73$), and $7 \cdot 3 \cdot 73 = 1533$ has digit sum $12$.
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