AMC 8 · 2011 · #15

Easy mode Grade 6
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Problem

The symbol 454^5 means 4×4×4×4×44 \times 4 \times 4 \times 4 \times 4. The symbol 5105^{10} means 55 multiplied by itself 1010 times.

Now multiply those two numbers together to get 455104^5 \cdot 5^{10}.

If you wrote this product out as a regular whole number, how many digits would it have?

(A) 8(B) 9(C) 10(D) 11(E) 12\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12

Pick an answer.

(A)
8
(B)
9
(C)
10
(D)
11
(E)
12
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Toolkit + CCSS Solution

Understand

Restated: Find how many digits the integer $4^5 \cdot 5^{10}$ has when written out in base $10$.

Givens: Expression: $4^5 \cdot 5^{10}$; Bases $4$ and $5$ are different, so the exponents cannot be combined directly; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Unknowns: The number of digits in the product $4^5 \cdot 5^{10}$

Understand

Restated: Find how many digits the integer $4^5 \cdot 5^{10}$ has when written out in base $10$.

Givens: Expression: $4^5 \cdot 5^{10}$; Bases $4$ and $5$ are different, so the exponents cannot be combined directly; Answer choices: (A) $8$, (B) $9$, (C) $10$, (D) $11$, (E) $12$

Plan

Primary tool: #16 Change Focus / Change Your Point of View

Secondary: #9 Solve an Easier Related Problem

Computing $4^5 \cdot 5^{10}$ as a brute number is painful, but the bases $4$ and $5$ are hiding a $2$ inside $4$. Tool #16 (Change Focus) says: rewrite $4$ as $2^2$ so the bases become $2$ and $5$. Now the $2$s and $5$s pair up perfectly into $10$s, turning the product into a clean power of $10$. Tool #9 (Easier Related Problem) then takes over: counting digits in $10^{10}$ is far easier than in $4^5 \cdot 5^{10}$, and we already know the rule $10^n$ has $n+1$ digits.

Execute — Answer: D

#16 Change Focus / Change Your Point of View 6.EE.A.1 Step 1
  • Rewrite $4$ using the smaller base $2$.
  • Since $4 = 2^2$, we get $4^5 = (2^2)^5 = 2^{10}$ by the power-of-a-power rule.
$$4^5 = (2^2)^5 = 2^{2 \cdot 5} = 2^{10}$$

💡 Changing the base from $4$ to $2$ lets the exponents talk to the $5^{10}$ next door.

#16 Change Focus / Change Your Point of View 6.EE.A.1 Step 2
  • Substitute back into the original product.
  • Now the two factors share the same exponent $10$, so we can combine them under a single power using $a^n \cdot b^n = (a \cdot b)^n$.
$$4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$$

💡 Every $2$ finds a partner $5$ and together they form a $10$ — the whole product collapses into $10^{10}$.

#9 Solve an Easier Related Problem 5.NBT.A.2 Step 3
  • Replace the original digit-counting question with an easier one: how many digits does $10^{10}$ have?
  • Use the pattern $10^1 = 10$ ($2$ digits), $10^2 = 100$ ($3$ digits), $10^3 = 1000$ ($4$ digits).
  • The rule is $10^n$ has $n+1$ digits — one $1$ followed by $n$ zeros.
$$10^{10} = 1\underbrace{00\cdots 0}_{10 \text{ zeros}}$$

💡 Multiplying by $10$ tacks on a zero — a Grade 5 place-value pattern.

#9 Solve an Easier Related Problem 5.NBT.A.2 Step 4

Apply the rule with $n = 10$: $10^{10}$ has $10 + 1 = 11$ digits.

$$10 + 1 = 11 \;\Rightarrow\; \textbf{(D)}$$

💡 One leading $1$ plus ten trailing zeros equals eleven digits total.

[1] #16 6.EE.A.1 Rewrite $4$ using the smaller base $2$. Since $4 = 2^2$, we get $4^5 = (2^2)^5 =
[2] #16 6.EE.A.1 Substitute back into the original product. Now the two factors share the same ex
[3] #9 5.NBT.A.2 Replace the original digit-counting question with an easier one: how many digits
[4] #9 5.NBT.A.2 Apply the rule with $n = 10$: $10^{10}$ has $10 + 1 = 11$ digits.

Review

Reasonableness: Quick sanity check on size: $4^5 = 1024$ ($4$ digits) and $5^{10} = 9{,}765{,}625$ ($7$ digits). Multiplying a $4$-digit number by a $7$-digit number gives a result with either $4 + 7 - 1 = 10$ or $4 + 7 = 11$ digits. Our answer $11$ lands in that range, and since the leading digits $1.024 \times 9.765\ldots \approx 10$ push the product over $10^{10}$, the upper case ($11$ digits) is correct. Choices (A)$8$ and (B)$9$ are far too small; (E)$12$ would require the product to exceed $10^{11}$, which it does not.

Alternative: Tool #5 (Look for a Pattern) on digit counts: $10^1$ has $2$ digits, $10^2$ has $3$, ..., $10^n$ has $n+1$. Once we've rewritten the product as $10^{10}$, the pattern instantly delivers $10 + 1 = 11$. Or, without rewriting, multiply $4^5 \cdot 5^{10} = 1024 \cdot 9{,}765{,}625 = 10{,}000{,}000{,}000$ directly and count $11$ digits — but recognizing the power of $10$ first avoids the long multiplication entirely.

CCSS standards used (min grade 6)

  • 6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Rewriting $4^5 = (2^2)^5 = 2^{10}$ and combining $2^{10} \cdot 5^{10} = (2 \cdot 5)^{10} = 10^{10}$ using the exponent rules.)
  • 5.NBT.A.2 Explain patterns in the number of zeros of the product when multiplying by powers of 10 (Recognizing that $10^{10}$ is written as one $1$ followed by ten zeros, giving $10 + 1 = 11$ digits.)

⭐ Whenever you see $2$s and $5$s with matching exponents, pair them into $10$s — the answer just falls out as a power of $10$.

⭐ Whenever you see $2$s and $5$s with matching exponents, pair them into $10$s — the answer just falls out as a power of $10$.