AMC 10 · 2013 · #24
Grade 6 number-theoryA positive integer n is nice if there is a positive integer m with exactly four positive divisors (including 1 and m) such that the sum of the four divisors is equal to n. How many numbers in the set {2010,2011,2012,…,2019} are nice?
Pick an answer.
AMC 10 2013 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: Call a positive integer $n$ "nice" if some positive integer $m$ has exactly four positive divisors (including $1$ and $m$) whose sum equals $n$. Count how many of the ten integers $2010, 2011, 2012, \ldots, 2019$ are nice.
Givens: $m$ is a positive integer with exactly four positive divisors, including $1$ and $m$.; $n$ is the sum of those four divisors.; Candidate values: $n \in \{2010, 2011, 2012, \ldots, 2019\}$ (ten consecutive integers).; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$.
Unknowns: How many of the ten integers $2010$ through $2019$ are nice.
Understand
Restated: Call a positive integer $n$ "nice" if some positive integer $m$ has exactly four positive divisors (including $1$ and $m$) whose sum equals $n$. Count how many of the ten integers $2010, 2011, 2012, \ldots, 2019$ are nice.
Givens: $m$ is a positive integer with exactly four positive divisors, including $1$ and $m$.; $n$ is the sum of those four divisors.; Candidate values: $n \in \{2010, 2011, 2012, \ldots, 2019\}$ (ten consecutive integers).; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$.
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities, #2 Make a Systematic List
The word "nice" hides a divisor-sum condition, so the first move is to name the primes. Tool #4 (Introduce a Variable): write $m = pq$ and turn its divisor sum $1 + p + q + pq$ into the factored form $(1+p)(1+q)$ — this converts the murky question "is $n$ nice?" into the sharp test "can $n$ be split into two factors that are each one-more-than-a-prime?" Tool #7 (Identify Subproblems) first splits four-divisor numbers into the two shapes $p^{3}$ and $pq$. Tool #3 (Eliminate Possibilities) uses parity to knock out all five odd numbers at once. Tool #2 (Make a Systematic List) then tests each surviving even number through its factor pairs.
Execute — Answer: A
4.OA.B.4 Step 1 Two shapes of a four-divisor number
- A positive integer has exactly four divisors in only two ways.
- If it is $p^{3}$ for a prime $p$, its divisors are $1, p, p^{2}, p^{3}$.
- If it is $pq$ for two different primes $p$ and $q$, its divisors are $1, p, q, pq$.
- No other factorization shape yields exactly four divisors.
💡 The divisor count equals the product of (each exponent $+1$), and $4$ only comes from $4$ itself or $2 \times 2$ — that is, $p^{3}$ or $pq$.
6.EE.A.3 Step 2 Factor the divisor sum
- Take the $m = pq$ shape.
- Its four divisors add to $1 + p + q + pq$.
- Pull $(1+p)$ out of the first two terms and out of the last two: $1 + p + q + pq = (1+p) + q(1+p) = (1+p)(1+q)$.
- So through this shape $n$ is nice exactly when $n = (1+p)(1+q)$ for distinct primes $p, q$ — equivalently, when $n$ can be written as a product $a \cdot b$ with $a - 1$ and $b - 1$ both prime.
💡 Grouping the four-term sum into a single product is the whole trick — it turns "is $n$ nice?" into a factoring question.
6.EE.A.2 Step 3 The cube shape overshoots
- Check the $m = p^{3}$ shape, whose divisor sum is $1 + p + p^{2} + p^{3}$.
- For primes $p = 2, 3, 5, 7, 11, 13$ this equals $15, 40, 156, 400, 1464, 2380$.
- The values leap straight from $1464$ to $2380$, skipping the entire window $2010$–$2019$.
- So no cube shape can produce a nice number here, and only the $pq$ shape matters.
💡 Cubes grow so fast that consecutive divisor sums jump right over the ten-number target.
2.OA.C.3 Step 4 Odd numbers cannot be nice
- At most one of $p, q$ can be the even prime $2$, so at least one of them is an odd prime.
- Then at least one of the factors $1 + p$ or $1 + q$ is even, which makes the product $n = (1+p)(1+q)$ even.
- Therefore every nice number is even, and the five odd entries $2011, 2013, 2015, 2017, 2019$ are eliminated at once.
💡 An odd prime plus one is even, and any product containing an even factor is even.
4.OA.B.4 Step 5 Test the five even candidates
- For each even candidate, list its factor pairs $a \cdot b$ with $a, b \ge 3$ and check whether $a - 1$ and $b - 1$ are both prime.
- $2010$: pairs $6 \cdot 335$ and $30 \cdot 67$ give partners $334$ and $66$ — never both prime.
- $2012 = 4 \cdot 503$: here $503 - 1 = 502 = 2 \cdot 251$ is not prime.
- $2014 = 38 \cdot 53$: $37$ is prime but $52$ is not.
- $2018 = 2 \cdot 1009$ has no factor pair with both parts $\ge 3$.
- All four fail.
💡 Running each factor pair through the "one-more-than-a-prime" test quickly exhausts every option.
4.OA.B.4 Step 6 2016 is the one nice number
- Now $2016$.
- Split it as $4 \cdot 504$: then $a - 1 = 3$ and $b - 1 = 503$ are both prime and distinct.
- So $m = 3 \cdot 503 = 1509$ works — its divisors $1, 3, 503, 1509$ sum to exactly $2016$.
- Since $2016$ is nice and the other nine numbers are not, exactly one number in the set is nice, so the count is $1$, choice $\textbf{(A)}$.
💡 One valid split — $3$ and $503$ both prime — is enough to make $2016$ nice, and it is the only number in the window that admits such a split.
4.OA.B.4 A positive integer has exactly four divisors in only two ways. If it is $p^{3}$ 6.EE.A.3 Take the $m = pq$ shape. Its four divisors add to $1 + p + q + pq$. Pull $(1+p)$ 6.EE.A.2 Check the $m = p^{3}$ shape, whose divisor sum is $1 + p + p^{2} + p^{3}$. For p 2.OA.C.3 At most one of $p, q$ can be the even prime $2$, so at least one of them is an o 4.OA.B.4 For each even candidate, list its factor pairs $a \cdot b$ with $a, b \ge 3$ and 4.OA.B.4 Now $2016$. Split it as $4 \cdot 504$: then $a - 1 = 3$ and $b - 1 = 503$ are bo Review
Reasonableness: The count $1$ matches choice (A). Double-check $2016$: $m = 1509 = 3 \times 503$ has divisors $1, 3, 503, 1509$ summing to $2016$, and $503$ is prime (no prime up to $22$ divides it). The parity filter is airtight — every nice number must be even — and among the five evens only $2016$ survived, because it is the one that factors as (small even) $\times$ (one-more-than-a-large-prime) with both parts one-more-than-a-prime.
Alternative: Tool #2 (Systematic List) pushed to the end: for each small prime $p$, ask which $q$ makes $(1+p)(1+q)$ land in $[2010, 2019]$. $p = 2 \to 3(1+q)$ forces $q = 669$ or $671$, both composite. $p = 3 \to 4 \cdot 504$ gives $q = 503$, prime, hitting $2016$. Larger small primes also land on $2016$ ($p = 7 \to 8 \cdot 252$ with $q = 251$; $p = 11 \to 12 \cdot 168$ with $q = 167$) and nowhere else in the window. No other integer in $[2010, 2019]$ is ever produced, so the count is $1$.
CCSS standards used (min grade 6)
2.OA.C.3Determine whether a group of objects has an odd or even number (Arguing by parity that every nice number is even, which eliminates the five odd candidates.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Classifying four-divisor numbers as $p^{3}$ or $pq$, and testing each even candidate's factor pairs for the one-more-than-a-prime condition.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating the cube-shape divisor sum $1 + p + p^{2} + p^{3}$ for small primes to show it skips the target window.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the divisor sum $1 + p + q + pq$ into $(1+p)(1+q)$ — the key that reframes niceness as a factoring test.)
⭐ A number with exactly four divisors is $pq$, and its divisor sum factors as $(1+p)(1+q)$ — so "nice" just means you can split $n$ into two parts that are each one-more-than-a-prime; only $2016 = 4 \cdot 504 = (1+3)(1+503)$ works, so the answer is $1$, choice (A).
⭐ A number with exactly four divisors is $pq$, and its divisor sum factors as $(1+p)(1+q)$ — so "nice" just means you can split $n$ into two parts that are each one-more-than-a-prime; only $2016 = 4 \cdot 504 = (1+3)(1+503)$ works, so the answer is $1$, choice (A).
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