AMC 10 · 2002 · #15
Grade 4 number-theoryUsing the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
Pick an answer.
AMC 10 2002 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: Use each of the digits $1,2,3,4,5,6,7,9$ exactly once to build four two-digit prime numbers, then find the sum of those four primes.
Givens: The eight digits available are $1,2,3,4,5,6,7,9$ (note there is no $8$); Each digit is used exactly once, and the eight digits fill four two-digit numbers; All four two-digit numbers must be prime; Answer choices: (A) $150$, (B) $160$, (C) $170$, (D) $180$, (E) $190$
Unknowns: The sum of the four two-digit prime numbers
Understand
Restated: Use each of the digits $1,2,3,4,5,6,7,9$ exactly once to build four two-digit prime numbers, then find the sum of those four primes.
Givens: The eight digits available are $1,2,3,4,5,6,7,9$ (note there is no $8$); Each digit is used exactly once, and the eight digits fill four two-digit numbers; All four two-digit numbers must be prime; Answer choices: (A) $150$, (B) $160$, (C) $170$, (D) $180$, (E) $190$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #3 Eliminate Possibilities, #7 Identify Subproblems
The tempting plan is to hunt for four actual primes, but that is slow and unnecessary. Tool #16 (Change Focus) says look at what the question really asks: only the sum, and the sum of four two-digit numbers is $10\times(\text{sum of tens digits}) + (\text{sum of units digits})$. So the sum is fixed the moment we know which four digits land in the units place. To find those, Tool #3 (Eliminate Possibilities) rules out any digit that cannot end a prime: a two-digit prime is never even and never ends in $5$. That forces the split of the eight digits into a tens group and a units group. Tool #7 (Identify Subproblems) then finishes with two easy sub-sums.
Execute — Answer: E
4.OA.B.4 Step 1 Which digits can end a prime
- A two-digit prime is bigger than $10$, so it has no factor of $2$ and no factor of $5$.
- A number is divisible by $2$ when its units digit is even, and divisible by $5$ when its units digit is $0$ or $5$.
- So a prime's units digit must be odd and not $5$.
- Among $1,2,3,4,5,6,7,9$, that rules out $2,4,6$ (even) and $5$.
- The only digits that can sit in a units place are $1,3,7,9$.
💡 Any two-digit number ending in an even digit or in $5$ already has $2$ or $5$ as a factor, so it can never be prime.
1.NBT.B.2 Step 2 Force the tens and units groups
- Four primes need four units digits, and only $1,3,7,9$ are allowed there — exactly four digits, so all four are used as units digits.
- The four leftover digits $2,4,5,6$ have nowhere else to go, so they must all be tens digits.
- Now use place value: every two-digit number equals $10\times(\text{its tens digit}) + (\text{its units digit})$.
- Adding the four primes, the tens digits each get multiplied by $10$ and the units digits are added as-is, no matter how the pairing works out.
💡 With exactly four legal units digits for four units slots, the other four digits are squeezed into the tens slots, and place value fixes the sum before the primes are even built.
4.NBT.B.5 Step 3 Add the two place-value groups
- Split the sum into a tens part and a units part.
- The tens digits add to $2+4+5+6=17$, and each counts for $10$, giving $10\times 17 = 170$.
- The units digits add to $1+3+7+9=20$.
- Total sum $= 170 + 20 = 190$.
- That matches choice (E).
💡 Grouping all the tens together and all the units together turns four two-digit sums into one multiply-by-ten plus one small addition.
4.OA.B.4 A two-digit prime is bigger than $10$, so it has no factor of $2$ and no factor 1.NBT.B.2 Four primes need four units digits, and only $1,3,7,9$ are allowed there — exact 4.NBT.B.5 Split the sum into a tens part and a units part. The tens digits add to $2+4+5+6 Review
Reasonableness: The argument never built a single prime, so check that a valid set actually exists: $23,\ 41,\ 59,\ 67$ are all prime and use the digits $2,3,4,1,5,9,6,7$ — each of the eight digits once. Their sum is $23+41+59+67 = 190$, matching (E). A size check also fits: four two-digit numbers between $10$ and $99$ average about $47.5$, so a sum near $4\times 47.5 = 190$ is exactly the right ballpark. The trap answers $150$–$180$ come from mis-sorting a digit (for example letting a $5$ or a $6$ slip into a units place), which the primality rule forbids.
Alternative: Brute construction (Tool #6, Guess and Check): place $1,3,7,9$ as units digits and try to pair the tens digits $2,4,5,6$ so all four numbers are prime, e.g. $23, 41, 59, 67$. Then add directly: $23+41=64$, $64+59=123$, $123+67=190$. Same answer (E), but it does far more work than the place-value shortcut because it insists on finding the actual primes.
CCSS standards used (min grade 4)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Ruling out any digit that makes a two-digit number even or a multiple of $5$, so only $1,3,7,9$ can be units digits of a prime.)1.NBT.B.2Understand that the two digits of a two-digit number represent tens and ones (Writing each prime as $10\times(\text{tens digit}) + (\text{units digit})$ so the sum depends only on which digits are tens and which are units.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing $10\times 17 = 170$ for the tens part and adding the units part $20$ to get $190$.)
⭐ You never have to find the primes: even digits and $5$ can't end a prime, so $1,3,7,9$ are the units and $2,4,5,6$ are the tens, and the sum is $10\times 17 + 20 = 190$.
⭐ You never have to find the primes: even digits and $5$ can't end a prime, so $1,3,7,9$ are the units and $2,4,5,6$ are the tens, and the sum is $10\times 17 + 20 = 190$.
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