AMC 8 · 2005 · #6

Grade 5 arithmetic

Archetype Small-Domain Constrained Search

place-valuedigit-constraintsinterval-arithmeticsystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: place-value

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Problem

Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$ ?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: $d$ is a single digit (one of $0,1,2,\dots,9$). For how many of those values does $2.00d5 > 2.005$?

Givens: $d$ is a digit, so $d \in \{0,1,2,3,4,5,6,7,8,9\}$; The four-place decimal $2.00d5$ has $d$ in the thousandths place and $5$ in the ten-thousandths place; Answer choices: (A) $0$, (B) $4$, (C) $5$, (D) $6$, (E) $10$

Unknowns: The count of digit values $d$ that make $2.00d5 > 2.005$ true

Understand

Restated: $d$ is a single digit (one of $0,1,2,\dots,9$). For how many of those values does $2.00d5 > 2.005$?

Plan

Primary tool: #13 Count Carefully

Secondary: #7 Break Into Subproblems

The question asks "for how many values," so the goal is to count digits that pass a test (Tool #13). The test itself is the decimal comparison $2.00d5 > 2.005$. To run the test cleanly, split the $10$ digits into three cases by how $d$ compares to $5$ in the thousandths place (Tool #7). The case $d=5$ needs an extra look at the ten-thousandths place, so isolating it as its own subproblem keeps the reasoning honest.

Execute — Answer: C

#7 Break Into Subproblems 5.NBT.A.3 Step 1

Line up the two decimals so the place values match.
Write $2.005$ as $2.0050$ to give it the same four decimal places as $2.00d5$.

$$2.00d5 \;\text{vs.}\; 2.0050$$

💡 Comparing decimals is a Grade 5 skill: same place values lined up, then compare digit by digit from the left.

#7 Break Into Subproblems 5.NBT.A.3 Step 2

Check the matching places.
Both numbers have $2$ in the ones, $0$ in the tenths, and $0$ in the hundredths.
So the comparison is decided at the thousandths place — that is the first place where the digits might differ.

ones $=$ ones, tenths $=$ tenths, hundredths $=$ hundredths $\;\Rightarrow\;$ decide at thousandths

💡 Whichever number wins the first place where they differ wins the whole comparison.

#7 Break Into Subproblems 5.NBT.A.3 Step 3

Split $d$ into three cases based on the thousandths digit on the right side, which is $5$.
Case A: $d > 5$.
Then $2.00d5$ already wins at the thousandths place, no matter what comes after.
The digits that work are $d = 6, 7, 8, 9$ — four values.

$$d \in \{6,7,8,9\} \;\Rightarrow\; 2.00d5 > 2.0050 \;\checkmark$$

💡 A bigger digit in the first differing place makes the whole decimal bigger.

#7 Break Into Subproblems 5.NBT.A.3 Step 4

Case B: $d < 5$.
Then $2.00d5$ loses at the thousandths place, so $d = 0, 1, 2, 3, 4$ all fail.
None of these five digits count.

$$d \in \{0,1,2,3,4\} \;\Rightarrow\; 2.00d5 < 2.0050 \;\times$$

💡 A smaller digit in the first differing place makes the whole decimal smaller.

#7 Break Into Subproblems 5.NBT.A.3 Step 5

Case C: $d = 5$.
The thousandths digits tie at $5$, so move one place right.
The left side is $2.0055$ and the right side is $2.0050$.
The ten-thousandths digits are $5$ vs.
$0$, and $5 > 0$, so the left side wins.
$d = 5$ counts.

$$d = 5 \;\Rightarrow\; 2.0055 > 2.0050 \;\checkmark$$

💡 If the deciding place ties, slide one more place to the right and compare there.

#13 Count Carefully 5.NBT.A.3 Step 6

Count the winning digits across all cases.
From Case A: $6, 7, 8, 9$.
From Case C: $5$.
Total: $4 + 1 = 5$ digits.
The answer is $(C)$.

$$|\{5, 6, 7, 8, 9\}| = 5 \;\Rightarrow\; \textbf{(C)}$$

💡 Tool #13 in action: sweep through all $10$ digits, count the ones that pass the test.

[1] #7 5.NBT.A.3 Line up the two decimals so the place values match. Write $2.005$ as $2.0050$ to

[2] #7 5.NBT.A.3 Check the matching places. Both numbers have $2$ in the ones, $0$ in the tenths,

[3] #7 5.NBT.A.3 Split $d$ into three cases based on the thousandths digit on the right side, whi

[4] #7 5.NBT.A.3 Case B: $d < 5$. Then $2.00d5$ loses at the thousandths place, so $d = 0, 1, 2,

[5] #7 5.NBT.A.3 Case C: $d = 5$. The thousandths digits tie at $5$, so move one place right. The

[6] #13 5.NBT.A.3 Count the winning digits across all cases. From Case A: $6, 7, 8, 9$. From Case

Review

Reasonableness: Spot-check with the boundary digit $d = 5$: $2.0055$ vs. $2.005 = 2.0050$. The extra $5$ in the ten-thousandths place clearly makes $2.0055$ bigger, confirming $d = 5$ counts. Spot-check $d = 4$: $2.0045$ vs. $2.0050$. At the thousandths place $4 < 5$, so $2.0045 < 2.0050$ — correctly excluded. The winning set $\{5,6,7,8,9\}$ has $5$ elements, matching answer (C). Also, intuitively, about half of the $10$ digits should be larger than the cutoff, so $5$ feels right.

Alternative: Tool #6 (Guess and Check): test all $10$ digits one by one. $d = 0$: $2.0005 < 2.005$. $d = 1$: $2.0015 < 2.005$. ... $d = 4$: $2.0045 < 2.005$. $d = 5$: $2.0055 > 2.005$. $d = 6$: $2.0065 > 2.005$. ... $d = 9$: $2.0095 > 2.005$. Five digits ($5, 6, 7, 8, 9$) pass, confirming $(C)$. The case-split approach is faster, but the brute-force check works because there are only $10$ digits.

CCSS standards used (min grade 5)

5.NBT.A.3 Read, write, and compare decimals to thousandths (Comparing $2.00d5$ and $2.0050$ place by place — the core decimal-comparison skill from Grade 5.)
5.NBT.A.1 Recognize that a digit in one place represents 10 times what it represents in the place to its right (Knowing that the thousandths place decides the comparison before the ten-thousandths place, because a difference one place to the left is ten times bigger.)

⭐ Stack the decimals, find the first place where they differ, and count the digits that come out on top — that turns a tricky-looking inequality into a quick Grade 5 check.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 5)

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