AMC 10 · 2024 · #1

Grade 7 arithmeticalgebra

Archetype Arithmetic Expression Decomposition

multi-digit-arithmeticplace-valuepattern-recognition identify-subproblemsconvert-to-algebra ↑ Prerequisites: multi-digit-arithmeticorder-of-operations

📏 Short solution 💡 2 insights

Problem

What is the value of $9901\cdot101-99\cdot10101?$

Pick an answer.

(A)

(B)

~20

(C)

~200

(D)

~202

(E)

~2020

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Toolkit + CCSS Solution

Understand

Restated: Compute the value of $9901 \cdot 101 - 99 \cdot 10101$ and match it to one of the five answer choices.

Givens: Expression: $9901 \cdot 101 - 99 \cdot 10101$; Four key numbers: $9901$, $101$, $99$, $10101$; Answer choices: (A) $2$, (B) $20$, (C) $200$, (D) $202$, (E) $2020$

Unknowns: The numerical value of the expression

Understand

Restated: Compute the value of $9901 \cdot 101 - 99 \cdot 10101$ and match it to one of the five answer choices.

Givens: Expression: $9901 \cdot 101 - 99 \cdot 10101$; Four key numbers: $9901$, $101$, $99$, $10101$; Answer choices: (A) $2$, (B) $20$, (C) $200$, (D) $202$, (E) $2020$

Plan

Primary tool: #16 Use Symmetry / Transform

Secondary: #4 Introduce a Variable

Brute-force multiplication of $9901 \cdot 101$ and $99 \cdot 10101$ would give two five-digit numbers and an easy place for a careless slip. Instead, notice that $9901 = 99 \cdot 100 + 1$ and $10101 = 101 \cdot 100 + 1$ — both numbers split into a product of the other given factors plus $1$. That symmetry is exactly what Tool #16 looks for: rewrite the expression so a large piece cancels. Tool #4 (let $a = 99$, $b = 101$) makes the symmetry visible at a glance and turns the whole problem into one line of algebra.

Execute — Answer: A

#4 Introduce a Variable 6.EE.A.2 Step 1

Name the two small numbers.
Let $a = 99$ and $b = 101$.
Now look at the two big numbers and write each one in terms of $a$ and $b$.

$$a = 99,\; b = 101,\; 9901 = 99 \cdot 100 + 1 = 100a + 1,\; 10101 = 101 \cdot 100 + 1 = 100b + 1$$

💡 Giving the repeated numbers short names is the Grade 6 "letters stand for numbers" move; it makes the hidden symmetry pop out.

#16 Use Symmetry / Transform 6.EE.A.3 Step 2

Rewrite the original expression using $a$ and $b$.
Both factors of $100$ now appear in a parallel way.

$$9901 \cdot 101 - 99 \cdot 10101 = (100a + 1) \cdot b - a \cdot (100b + 1)$$

💡 Same expression, new clothing — the substitution doesn't change the value, but it lines up matching pieces.

#16 Use Symmetry / Transform 7.EE.A.1 Step 3

Use the distributive property to expand both products.
The $100ab$ term shows up in each part with opposite signs.

$$(100a + 1) \cdot b - a \cdot (100b + 1) = 100ab + b - 100ab - a$$

💡 Distribute carefully, especially the minus sign across the second parenthesis.

#16 Use Symmetry / Transform 7.EE.A.1 Step 4

Cancel the matching $100ab$ terms and plug $a$ and $b$ back in.
The huge calculation collapses to a single subtraction.

$$100ab + b - 100ab - a = b - a = 101 - 99 = 2 \;\Rightarrow\; \textbf{(A)}$$

💡 Once the heavy term cancels, only the two tiny extras $+b$ and $-a$ survive — and their difference is just $2$.

[1] #4 6.EE.A.2 Name the two small numbers. Let $a = 99$ and $b = 101$. Now look at the two big

[2] #16 6.EE.A.3 Rewrite the original expression using $a$ and $b$. Both factors of $100$ now app

[3] #16 7.EE.A.1 Use the distributive property to expand both products. The $100ab$ term shows up

[4] #16 7.EE.A.1 Cancel the matching $100ab$ terms and plug $a$ and $b$ back in. The huge calcula

Review

Reasonableness: The answer choices jump from $2$ to $20$ to $200$ to $202$ to $2020$, so they're testing whether the student noticed the cancellation. A direct check confirms it: $9901 \cdot 101 = 1{,}000{,}001$ and $99 \cdot 10101 = 999{,}999$, and $1{,}000{,}001 - 999{,}999 = 2$. The two giant products are within $2$ of each other, exactly matching $b - a = 101 - 99$, so $\textbf{(A)}$ is the only consistent choice.

Alternative: Tool #6 (Guess and Check by direct computation): multiply $9901 \cdot 101 = 1{,}000{,}001$ and $99 \cdot 10101 = 999{,}999$, then subtract to get $2$. It works but burns time and invites arithmetic mistakes — the algebraic transform is much safer under contest pressure.

CCSS standards used (min grade 7)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Letting $a = 99$ and $b = 101$ so the four numerical inputs can be written as $a$, $b$, $100a + 1$, and $100b + 1$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Substituting the rewritten forms of $9901$ and $10101$ to turn the original expression into the equivalent expression $(100a + 1)b - a(100b + 1)$.)
7.EE.A.1 Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients (Expanding $(100a + 1)b - a(100b + 1)$ with the distributive property, then cancelling the $100ab$ terms to reduce the expression to $b - a$.)

⭐ Big-looking numbers often hide a small structure. Naming $99$ and $101$ as $a$ and $b$ exposes the matching $100ab$ piece that cancels — leaving the AMC 10 opener as a Grade 7 distributive-property exercise!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 7)

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