AMC 8 · 2020 · #17

Grade 6 number-theorycounting

Archetype Small-Domain Constrained Search

prime-factorizationfactorsdivisor-countcomplementary-countingprime-numbersperfect-squares complementary-countingcasework ↑ Prerequisites: prime-factorizationfactors

📏 Medium solution 💡 4 insights

Problem

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 $12.$ )

$\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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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

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

#2 Make a Systematic List 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 $2020 = 2^2 \times 5 \times 101$.

$$2020 = 2^2 \times 5 \times 101$$

💡 Breaking a number into its prime building blocks is the Grade 4 "prime or composite / factor pairs" idea.

#2 Make a Systematic List 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 $2$, $1$, $1$.

$$\text{total divisors} = (2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12$$

💡 Working with whole-number exponents like $2^2$ inside an expression is exactly the Grade 6 exponents standard.

#16 Change Focus / Count the Complement 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 $1$, $p$, $p^2$).

$$\text{categories: } 1, \; p, \; p^2$$

💡 Classifying small divisor counts uses the Grade 4 idea that primes have exactly two factors.

#2 Make a Systematic List 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.

$$\underbrace{1}_{1} \;+\; \underbrace{2,\,5,\,101}_{3} \;+\; \underbrace{4}_{1} \;=\; 5$$

💡 Listing in a fixed order — by category, then by prime — guarantees no duplicates and no misses.

#16 Change Focus / Count the Complement 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.

$$12 - 5 = 7 \;\Rightarrow\; \textbf{(B)}$$

💡 The final "total minus bad cases" subtraction is a Grade 4 multi-step word-problem move.

[1] #2 4.OA.B.4 Find the prime factorization of $2020$. Pull out factors of $10$ first: $2020 =

[2] #2 6.EE.A.1 Count the total number of divisors of $2020$ using the divisor-counting rule: if

[3] #16 4.OA.B.4 Apply Tool #16 by listing the divisors with $1$, $2$, or $3$ divisors — the "bad

[4] #2 4.OA.B.4 Make the systematic list inside each category, keeping only entries that actuall

[5] #16 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.4 Find 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.1 Write 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.3 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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