AMC 10 · 2023 · #5

Grade 8 arithmetic

exponentsprime-factorizationplace-valuemulti-digit-arithmetic identify-subproblemspattern-recognition ↑ Prerequisites: exponentsprime-factorization

📏 Medium solution 💡 2 insights

Problem

How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2023 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.

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Toolkit + CCSS Solution

Understand

Restated: How many digits does the integer $8^5 \cdot 5^{10} \cdot 15^5$ have when written in base ten?

Givens: The integer $N = 8^5 \cdot 5^{10} \cdot 15^5$; Base ten — i.e. count the decimal digits of $N$; Answer choices: (A) $14$, (B) $15$, (C) $16$, (D) $17$, (E) $18$

Unknowns: The number of base-ten digits of $N$

Understand

Restated: How many digits does the integer $8^5 \cdot 5^{10} \cdot 15^5$ have when written in base ten?

Givens: The integer $N = 8^5 \cdot 5^{10} \cdot 15^5$; Base ten — i.e. count the decimal digits of $N$; Answer choices: (A) $14$, (B) $15$, (C) $16$, (D) $17$, (E) $18$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #5 Look for a Pattern

Tool #7 (Identify Subproblems) splits the work cleanly: (A) rewrite every base in prime form using exponent laws, then (B) pair up $2$s and $5$s into copies of $10$, leaving a small leftover, then (C) count the digits of (leftover) $\times 10^{15}$. Tool #5 (Look for a Pattern) supports step (C) — multiplying any integer by $10^{15}$ adds exactly $15$ zeros to the end, so the digit count is (digits of the leftover) $+ 15$. Algebra (#13) is the wrong frame here; this is a pure exponent / place-value problem.

Execute — Answer: E

#7 Identify Subproblems 8.EE.A.1 Step 1

Subproblem A: rewrite each base in prime form.
$8 = 2^3$ and $15 = 3 \cdot 5$; the base $5$ is already prime.
Use $(a^m)^n = a^{mn}$ and $(ab)^n = a^n b^n$ to flatten the expression.

$$8^5 \cdot 5^{10} \cdot 15^5 = (2^3)^5 \cdot 5^{10} \cdot (3 \cdot 5)^5 = 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5$$

💡 Pushing exponents through products and powers is the core Grade 8 "properties of integer exponents" move — the expression turns into a clean prime factorization.

#7 Identify Subproblems 8.EE.A.1 Step 2

Subproblem B: collect like bases.
Combine the two powers of $5$ using $a^m \cdot a^n = a^{m+n}$, leaving $2^{15} \cdot 5^{15} \cdot 3^5$.

$$5^{10} \cdot 5^5 = 5^{15} \;\Longrightarrow\; 2^{15} \cdot 5^{15} \cdot 3^5$$

💡 $a^m \cdot a^n = a^{m+n}$ is the second Grade 8 exponent law — same base, add exponents.

#5 Look for a Pattern 8.EE.A.1 Step 3

Pair the $2$s and $5$s into copies of $10$ using $a^n b^n = (ab)^n$.
Every $2$ now has a $5$ partner, producing exactly $15$ tens.
The leftover factor is the $3^5$.

$$2^{15} \cdot 5^{15} \cdot 3^5 = (2 \cdot 5)^{15} \cdot 3^5 = 10^{15} \cdot 3^5$$

💡 Spotting that $2 \cdot 5 = 10$ and pairing exponents is the "trailing-zero" pattern — the heart of digit-counting for products with $2$s and $5$s.

#7 Identify Subproblems 6.EE.A.1 Step 4

Compute the leftover $3^5$ by repeated multiplication.

$$3^5 = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 9 \cdot 9 \cdot 3 = 81 \cdot 3 = 243$$

💡 Evaluating $3^5$ by chaining multiplications is a Grade 6 "whole-number exponent" calculation.

#5 Look for a Pattern 5.NBT.A.2 Step 5

Multiplying $243$ by $10^{15}$ appends $15$ zeros to the end of $243$.
So the decimal numeral is the three digits of $243$ followed by fifteen zeros — $3 + 15 = 18$ digits total.

$$N = 243 \cdot 10^{15} = 243\underbrace{000\ldots0}_{15 \text{ zeros}} \;\Rightarrow\; 3 + 15 = 18 \;\Rightarrow\; \textbf{(E)}$$

💡 Multiplying by a power of $10$ just shifts the digits to the left and pads with zeros — the Grade 5 "powers of $10$ and decimal point" pattern.

[1] #7 8.EE.A.1 Subproblem A: rewrite each base in prime form. $8 = 2^3$ and $15 = 3 \cdot 5$; t

[2] #7 8.EE.A.1 Subproblem B: collect like bases. Combine the two powers of $5$ using $a^m \cdot

[3] #5 8.EE.A.1 Pair the $2$s and $5$s into copies of $10$ using $a^n b^n = (ab)^n$. Every $2$ n

[4] #7 6.EE.A.1 Compute the leftover $3^5$ by repeated multiplication.

[5] #5 5.NBT.A.2 Multiplying $243$ by $10^{15}$ appends $15$ zeros to the end of $243$. So the de

Review

Reasonableness: Magnitude check. $N = 243 \cdot 10^{15}$ is between $10^{17}$ and $10^{18}$ because $100 \leq 243 < 1000$. An integer in $[10^{17}, 10^{18})$ has exactly $18$ digits, matching choice (E). Also $\log_{10}(243 \cdot 10^{15}) = 15 + \log_{10} 243 \approx 15 + 2.385 = 17.385$, so $\lfloor 17.385 \rfloor + 1 = 18$ — same answer by the log-based digit formula.

Alternative: Tool #8 (Analyze the Units): track "digits" as the unit. Each factor of $10$ contributes one trailing zero (one digit slot), and the leftover non-$10$ part determines the leading-digit count. We get $15$ digit-slots from $10^{15}$ and $3$ digit-slots from $243$, totaling $18$. Same answer (E).

CCSS standards used (min grade 8)

5.NBT.A.2 Explain patterns in the number of zeros and placement of decimal point when multiplying by powers of 10 (Recognizing that multiplying $243$ by $10^{15}$ appends $15$ zeros, so the total digit count is $3 + 15 = 18$.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Evaluating $3^5 = 243$ by chained multiplication.)
8.EE.A.1 Know and apply the properties of integer exponents (Applying $(a^m)^n = a^{mn}$, $(ab)^n = a^n b^n$, and $a^m \cdot a^n = a^{m+n}$ to rewrite $8^5 \cdot 5^{10} \cdot 15^5$ as $10^{15} \cdot 3^5$.)

⭐ This AMC 10 problem only needs Grade 8 "integer exponent rules" — pair every $2$ with a $5$ to make tens, then the leftover times $10^{15}$ tells you the digit count.