AMC 10 · 2013 · #5
Grade 7 algebraPositive integers a and b are each less than 6. What is the smallest possible value for 2⋅a−a⋅b?
Pick an answer.
AMC 10 2013 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.
Toolkit + CCSS Solution
Understand
Restated: Two positive integers $a$ and $b$ are each less than $6$, so each is one of $1,2,3,4,5$. Find the smallest value the expression $2\cdot a-a\cdot b$ can take.
Givens: $a$ and $b$ are positive integers; Each is less than $6$, so $a,b\in\{1,2,3,4,5\}$; The expression to minimize is $2\cdot a-a\cdot b$; Answer choices: (A) $-20$, (B) $-15$, (C) $-10$, (D) $0$, (E) $2$
Unknowns: The smallest possible value of $2a-ab$ over all allowed $a$ and $b$
Understand
Restated: Two positive integers $a$ and $b$ are each less than $6$, so each is one of $1,2,3,4,5$. Find the smallest value the expression $2\cdot a-a\cdot b$ can take.
Givens: $a$ and $b$ are positive integers; Each is less than $6$, so $a,b\in\{1,2,3,4,5\}$; The expression to minimize is $2\cdot a-a\cdot b$; Answer choices: (A) $-20$, (B) $-15$, (C) $-10$, (D) $0$, (E) $2$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The question asks for a minimum over a small, bounded set, which is the signature of Tool #14 (Extreme Principle): the smallest value lives at a boundary, so test the edge values rather than all $25$ pairs. First rewrite the expression by factoring (Tool #4, treating the two letters as one product) so the sign structure is visible. Finally use Tool #3 (Eliminate Possibilities) to confirm the winning value against the answer choices and to check that the tempting $-20$ is actually out of reach.
Execute — Answer: B
6.EE.A.3 Step 1 Factor the expression
- Both terms of $2a-ab$ share the common factor $a$.
- Pull it out to rewrite the expression as $a(2-b)$.
- This is the same expression, just organized so the value is a single product of two pieces: $a$ and $(2-b)$.
💡 Factoring out $a$ turns an add-subtract expression into one product, so the sign is easy to read.
7.NS.A.2 Step 2 Decide the sign
- A product is smallest (most negative) when one factor is positive and the other is a large negative.
- Here $a$ is always positive since $a\ge1$.
- So the whole thing is negative exactly when $(2-b)$ is negative, which needs $b>2$.
- In that case $a(2-b)=-a(b-2)$, so making the answer very negative means making the positive product $a(b-2)$ very large.
💡 Positive times negative is negative, so a bigger positive partner makes the product dive lower.
7.NS.A.1 Step 3 Push both factors to the edge
- To make $a(b-2)$ as large as possible, take the largest allowed values: $a=5$ and $b=5$.
- Then $b-2=3$ and $a(b-2)=5\cdot3=15$, so the expression equals $-15$.
- Checking directly: $2\cdot5-5\cdot5=10-25=-15$.
💡 The extreme value sits at the corner of the allowed range, where both letters are as big as possible.
6.NS.C.7 Step 4 Confirm against the choices
- Could the value drop below $-15$?
- Only if $a(b-2)$ could beat $15$, but the largest $a$ is $5$ and the largest $b-2$ is $3$, so $15$ is the ceiling of that product and $-15$ is the floor of the expression.
- That rules out $-20$ in (A), which would need $a(b-2)=20$.
- Among $-15,-10,0,2$ the smallest is $-15$, so the answer is (B).
💡 Since the positive product cannot exceed $15$, nothing beats $-15$ from below.
6.EE.A.3 Both terms of $2a-ab$ share the common factor $a$. Pull it out to rewrite the ex 7.NS.A.2 A product is smallest (most negative) when one factor is positive and the other 7.NS.A.1 To make $a(b-2)$ as large as possible, take the largest allowed values: $a=5$ an 6.NS.C.7 Could the value drop below $-15$? Only if $a(b-2)$ could beat $15$, but the larg Review
Reasonableness: The result $-15$ comes from the corner $a=b=5$, which matches the intuition that a minimum lives at a boundary. It clears the built-in trap: $-20$ looks reachable only if you forget that $a$ and $b$ both cap at $5$, so $a(b-2)$ cannot exceed $5\cdot3=15$. A quick scan of nearby corners agrees — $a=4,b=5$ gives $-12$ and $a=5,b=4$ gives $-10$, both larger than $-15$ — so $-15$ really is the lowest.
Alternative: Skip the algebra and reason about the shape $2a-ab=a(2-b)$ directly: to go as low as possible you want a big positive $a$ multiplied by a big negative $(2-b)$. The most negative $(2-b)$ is at $b=5$ (giving $-3$) and the biggest $a$ is $5$, so the product is $5\cdot(-3)=-15$ without testing any other pairs.
CCSS standards used (min grade 7)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring the common $a$ to rewrite $2a-ab$ as the single product $a(2-b)$.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Reasoning that a positive $a$ times a negative $(2-b)$ is negative, and grows more negative as the positive partner grows.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Computing $2\cdot5-5\cdot5=10-25=-15$ at the extreme values.)6.NS.C.7Understand ordering and absolute value of rational numbers (Comparing the negative candidates to confirm $-15$ is the smallest and that $-20$ is unreachable.)
⭐ For a smallest-value question over a small range, factor the expression and push each letter to the edge of what it's allowed to be.
⭐ For a smallest-value question over a small range, factor the expression and push each letter to the edge of what it's allowed to be.
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