AMC 10 · 2002 · #17
Easy mode Grade 5A prime is a whole number greater than 1 whose only factors are 1 and itself.
Use each of the digits 1 through 9 exactly once to build a group of numbers that are all prime. You choose how many numbers to make and how many digits each one has; for example, {7,83,421,659} uses all nine digits once each.
Add up the numbers in the group. What is the smallest total you can reach?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Split the nine nonzero digits $1,2,\ldots,9$ into a set of prime numbers so that each digit is used exactly once — the example $\{7,83,421,659\}$ shows one way to do it. Find the smallest sum such a set can have.
Givens: Every member of the set is a prime number; The primes together use each of the digits $1,2,3,4,5,6,7,8,9$ exactly once; $\{7,83,421,659\}$ is one valid set; Answer choices: (A) $193$, (B) $207$, (C) $225$, (D) $252$, (E) $447$
Unknowns: The smallest possible sum of such a set of primes
Understand
Restated: Split the nine nonzero digits $1,2,\ldots,9$ into a set of prime numbers so that each digit is used exactly once — the example $\{7,83,421,659\}$ shows one way to do it. Find the smallest sum such a set can have.
Givens: Every member of the set is a prime number; The primes together use each of the digits $1,2,3,4,5,6,7,8,9$ exactly once; $\{7,83,421,659\}$ is one valid set; Answer choices: (A) $193$, (B) $207$, (C) $225$, (D) $252$, (E) $447$
Plan
Primary tool: #14 Extreme Principle
Secondary: #15 Organize Information in More Ways, #7 Identify Subproblems, #6 Guess and Check
"Smallest possible" is exactly the signal for Tool #14 (Extreme Principle), and a minimum question always splits into two halves via Tool #7 (Identify Subproblems): prove a floor that no set can go under, then exhibit one set that sits exactly on the floor. Tool #15 (Organize Information in More Ways) supplies the floor — stop reading the set as a list of numbers and read it as nine digits, each charged according to its place, so the question becomes which digits are forbidden from ending a prime. Tool #6 (Guess and Check) closes the argument by producing an actual set of primes that pays exactly that price; without it, a floor is only a bound, not an answer.
Execute — Answer: B
5.NBT.A.1 Step 1 Write the sum by place value
- However the digits are grouped into numbers, the total is the sum of the nine digits weighted by where each digit sits.
- A digit in the units place adds its face value, in the tens place ten times its value, in the hundreds place a hundred times.
- So making the sum small means pushing every digit as far right as possible — ideally leaving all nine as single-digit numbers in the units place.
💡 Sliding a digit one place to the left multiplies what it adds to the total by ten, so a cheap total keeps every digit as far right as the rules permit.
4.OA.B.4 Step 2 Three digits cannot end a prime
- A number whose last digit is $4$, $6$, or $8$ is even.
- The only even prime is $2$, and none of $4$, $6$, $8$ equals $2$, so no prime ever ends in one of them.
- The one-digit numbers $4$, $6$, $8$ are not prime either, so those three digits can never stand alone and can never sit in a units place.
- In any valid set each of them must therefore occupy the tens place or further left, costing at least $40$, $60$, and $80$.
💡 An even last digit makes the whole number even, and the only even prime is $2$ — so these three digits get pushed left no matter what.
4.NBT.B.4 Step 3 Add up the floor: 207
- The six remaining digits $1,2,3,5,7,9$ are each worth at least their face value, since no place value is smaller than $1$.
- Charging $4$, $6$, $8$ the tens-place minimum and everything else the units-place minimum gives $(40+60+80) + (1+2+3+5+7+9) = 180 + 27 = 207$.
- No valid set can total less than $207$, which already kills choice (A) $193$.
💡 Charging every digit the cheapest position it is legally allowed gives a total that nothing can slip under.
4.OA.B.4 Step 4 Land on the floor with an example
- A floor is only the answer if some set reaches it.
- Take $\{41, 67, 89, 2, 3, 5\}$.
- Each member is prime; the digits $4,1,6,7,8,9,2,3,5$ are the nine nonzero digits used once each; and the sum is $41+67+89+2+3+5 = 207$.
- Every digit sits precisely where the floor assumed — $4$, $6$, $8$ in tens places and the other six in units places.
- Note the example must also handle $1$ and $9$, which are not prime on their own; here they ride along as the units digits of $41$ and $89$.
- Since $207$ is reachable and nothing smaller is, the answer is (B).
💡 A lower bound turns into the true minimum the instant you exhibit one set that matches it exactly.
5.NBT.A.1 However the digits are grouped into numbers, the total is the sum of the nine di 4.OA.B.4 A number whose last digit is $4$, $6$, or $8$ is even. The only even prime is $2 4.NBT.B.4 The six remaining digits $1,2,3,5,7,9$ are each worth at least their face value, 4.OA.B.4 A floor is only the answer if some set reaches it. Take ${41, 67, 89, 2, 3, 5} Review
Reasonableness: The floor argument bites hard enough to be checkable: choice (A) $193$ is below $207$, so it is unreachable no matter how the digits are arranged — a genuine impossibility, not a near miss. On the other side, $207$ is not a fluke of one lucky arrangement: $\{43, 61, 89, 2, 5, 7\}$ and $\{47, 61, 89, 2, 3, 5\}$ both consist of primes, both use each nonzero digit once, and both also sum to $207$. The problem's own example $\{7,83,421,659\}$ totals $1170$, far above the floor, exactly as expected once a digit is pushed into the hundreds place.
Alternative: Search the minimal shape exhaustively instead of guessing. Any set that actually meets the floor must have $4$, $6$, $8$ in tens places and the other six digits alone in units places — anything else costs more. So the question becomes: which units digit partners each of $4$, $6$, $8$? The leftover single digits must all be prime, and $1$ and $9$ are not, so $1$ and $9$ must both be spent as units digits. Since $49$ and $69$ are composite, $9$ must pair with $8$, giving $89$. Then $1$ pairs with $4$ or $6$, and the third tens digit takes $3$ or $7$: the surviving sets are $\{41,67,89,2,3,5\}$, $\{61,43,89,2,5,7\}$ and $\{61,47,89,2,3,5\}$. All three sum to $207$, confirming both that the floor is attainable and that (B) is the answer.
CCSS standards used (min grade 5)
5.NBT.A.1Recognize that a digit in one place represents ten times as much as to its right (Writing the set's sum as digits weighted by place value, which is what makes "keep digits to the right" the cost-minimising move.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Showing that $4$, $6$, $8$ can never end a prime, and verifying that $41$, $67$, $89$, $2$, $3$, $5$ are all prime.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Totalling the minimum contributions to $180 + 27 = 207$ and checking $41+67+89+2+3+5 = 207$.)
⭐ To find a smallest, first prove a floor that nothing can go under, then build one example that lands exactly on it.
⭐ To find a smallest, first prove a floor that nothing can go under, then build one example that lands exactly on it.
More like this
Same archetype — closest grade level first.
