AMC 8 · 2020 · #17
Grade 6 number-theorycountingProblem
How many positive integer factors of 2020 have more than 3 factors? (As an example, 12 has 6 factors, namely 1,2,3,4,6, and 12.)
Pick an answer.
AMC 8 2020 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: Among all positive integer divisors of $2020$, how many of them have more than $3$ divisors themselves? For example, $12$ qualifies because it has $6$ divisors $(1, 2, 3, 4, 6, 12)$, while $4$ does not qualify because it has only $3$ divisors $(1, 2, 4)$.
Givens: The target number is $2020$; We must count divisors $d$ of $2020$ such that $d$ has more than $3$ positive divisors; A worked example is provided: $12$ has $6$ divisors; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: The count of divisors of $2020$ that themselves have $4$ or more divisors
Understand
Restated: Among all positive integer divisors of $2020$, how many of them have more than $3$ divisors themselves? For example, $12$ qualifies because it has $6$ divisors $(1, 2, 3, 4, 6, 12)$, while $4$ does not qualify because it has only $3$ divisors $(1, 2, 4)$.
Givens: The target number is $2020$; We must count divisors $d$ of $2020$ such that $d$ has more than $3$ positive divisors; A worked example is provided: $12$ has $6$ divisors; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #2 Make a Systematic List
Counting divisors with "more than $3$" divisors directly would force us to check $12$ separate divisors of $2020$ one by one. Tool #16 (Complement) flips the question: count the divisors with $3$ or fewer divisors and subtract from the total $12$. That smaller class is tiny because numbers with $\leq 3$ divisors have a clean structure — exactly $1$, the primes $p$, and the prime-squares $p^2$. Tool #2 (Systematic List) then enumerates the three categories cleanly so nothing is missed and nothing is double-counted.
Execute — Answer: B
4.OA.B.4 Step 1 - Find the prime factorization of $2020$.
- Pull out factors of $10$ first: $2020 = 10 \times 202 = 2 \times 5 \times 2 \times 101$.
- Group the primes to get $2020 = 2^2 \times 5 \times 101$.
💡 Breaking a number into its prime building blocks is the Grade 4 "prime or composite / factor pairs" idea.
6.EE.A.1 Step 2 - Count the total number of divisors of $2020$ using the divisor-counting rule: if $N = p_1^{a} \cdot p_2^{b} \cdot p_3^{c}$, then $N$ has $(a+1)(b+1)(c+1)$ divisors.
- The exponents here are $2$, $1$, $1$.
💡 Working with whole-number exponents like $2^2$ inside an expression is exactly the Grade 6 exponents standard.
4.OA.B.4 Step 3 - Apply Tool #16 by listing the divisors with $1$, $2$, or $3$ divisors — the "bad" cases we will subtract.
- A number has $1$ divisor only if it is $1$.
- A number has $2$ divisors only if it is prime.
- A number has $3$ divisors only if it is the square of a prime (its divisors are $1$, $p$, $p^2$).
💡 Classifying small divisor counts uses the Grade 4 idea that primes have exactly two factors.
4.OA.B.4 Step 4 - Make the systematic list inside each category, keeping only entries that actually divide $2020$.
- (i) $1$ always divides — $1$ number.
- (ii) Primes dividing $2020$: $2$, $5$, $101$ — $3$ numbers.
- (iii) Prime-squares dividing $2020$: check $2^2 = 4$ (yes, since $2020 / 4 = 505$), $5^2 = 25$ (no, because $5$ appears only to the first power in $2020$), $101^2 = 10201$ (no, larger than $2020$) — $1$ number.
💡 Listing in a fixed order — by category, then by prime — guarantees no duplicates and no misses.
4.OA.A.3 Step 5 - Subtract the complement count from the total to get the answer.
- Out of $12$ divisors of $2020$, exactly $5$ have $\leq 3$ divisors, so $12 - 5 = 7$ have more than $3$ divisors.
💡 The final "total minus bad cases" subtraction is a Grade 4 multi-step word-problem move.
4.OA.B.4 Find the prime factorization of $2020$. Pull out factors of $10$ first: $2020 = 6.EE.A.1 Count the total number of divisors of $2020$ using the divisor-counting rule: if 4.OA.B.4 Apply Tool #16 by listing the divisors with $1$, $2$, or $3$ divisors — the "bad 4.OA.B.4 Make the systematic list inside each category, keeping only entries that actuall 4.OA.A.3 Subtract the complement count from the total to get the answer. Out of $12$ divi Review
Reasonableness: Sanity check by listing all $12$ divisors of $2020$ and counting directly: $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020$. Mark divisors with $\leq 3$ divisors: $1$ (one), $2$ (two), $5$ (two), $101$ (two), $4$ (three) — that's $5$ of them. The remaining $12 - 5 = 7$ are $10, 20, 202, 404, 505, 1010, 2020$, each of which has $4$ or more divisors. Answer $7$ matches choice (B).
Alternative: Tool #2 (Systematic List) alone: write out all $12$ divisors of $2020$ and compute each one's divisor count from its prime factorization (e.g. $20 = 2^2 \cdot 5$ has $3 \cdot 2 = 6$ divisors). Then count the ones with divisor count $\geq 4$ directly. This works but does extra bookkeeping that Tool #16 (Complement) avoids.
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Prime-factoring $2020 = 2^2 \times 5 \times 101$, and using the facts that primes have $2$ divisors and prime-squares have $3$ divisors to enumerate the complement.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Working with exponents in $2020 = 2^2 \times 5^1 \times 101^1$ and applying the divisor-counting product $(2+1)(1+1)(1+1) = 12$.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Performing the complementary-count subtraction $12 - 5 = 7$ as the final step.)
⭐ This AMC 8 problem only needs Grade 6 exponent thinking — once you write $2020 = 2^2 \times 5 \times 101$, everything else is counting you already know!
⭐ This AMC 8 problem only needs Grade 6 exponent thinking — once you write $2020 = 2^2 \times 5 \times 101$, everything else is counting you already know!
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