AMC 10 · 2007 · #25
Grade 6 arithmeticHow many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and:
ba+9a14b
is an integer?
Pick an answer.
AMC 10 2007 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: Count the pairs of positive whole numbers $(a,b)$ that share no common factor bigger than $1$ and make the expression $\dfrac{a}{b}+\dfrac{14b}{9a}$ come out to a whole number.
Givens: $a$ and $b$ are positive integers; $a$ and $b$ have no common factor greater than $1$ (they are coprime); The value $\dfrac{a}{b}+\dfrac{14b}{9a}$ must be an integer; Answer choices: (A) $4$, (B) $6$, (C) $9$, (D) $12$, (E) infinitely many
Unknowns: How many coprime pairs $(a,b)$ make the expression a whole number
Understand
Restated: Count the pairs of positive whole numbers $(a,b)$ that share no common factor bigger than $1$ and make the expression $\dfrac{a}{b}+\dfrac{14b}{9a}$ come out to a whole number.
Givens: $a$ and $b$ are positive integers; $a$ and $b$ have no common factor greater than $1$ (they are coprime); The value $\dfrac{a}{b}+\dfrac{14b}{9a}$ must be an integer; Answer choices: (A) $4$, (B) $6$, (C) $9$, (D) $12$, (E) infinitely many
Plan
Primary tool: #7 Identify Subproblems
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
Two separate fractions are hard to judge, so first combine them into one fraction $\dfrac{9a^2+14b^2}{9ab}$; then "is an integer" simply means the bottom divides the top. Tool #7 (Identify Subproblems) splits that one divisibility question into smaller pieces: what must $a$ divide, what must $b$ divide, and how does the shared factor of $3$ behave. The coprime rule turns each piece into a short divisor list, so $a$ and $b$ each have only a handful of possible values. Tool #2 (Make a Systematic List) then tests those few candidates, and Tool #3 (Eliminate Possibilities) throws out the ones that fail, leaving the exact count.
Execute — Answer: A
6.EE.A.2 Step 1 Combine into one fraction
- Put both fractions over the common bottom $9ab$.
- The first becomes $\dfrac{9a^2}{9ab}$ and the second becomes $\dfrac{14b^2}{9ab}$, so together they make a single fraction.
- Now the whole question is: when is this one fraction a whole number?
- That happens exactly when the bottom $9ab$ divides the top $9a^2+14b^2$ evenly.
💡 One fraction is easy to test for wholeness: the bottom just has to divide the top.
6.NS.B.4 Step 2 What a must divide
- Look only at the factor $a$ sitting on the bottom.
- For the bottom to divide the top, $a$ must divide $9a^2+14b^2$.
- Since $a$ already divides $9a^2$, it must divide what is left over, $14b^2$.
- But $a$ and $b$ share no common factor bigger than $1$, so $a$ shares no factor with $b^2$ either.
- That forces $a$ to divide $14$, and the positive divisors of $14$ are $1, 2, 7, 14$.
💡 Coprime numbers share nothing, so a factor of the bottom cannot hide inside $b$ — it must live in the $14$.
6.NS.B.4 Step 3 What b must divide
- Do the same for the factor $b$ on the bottom.
- For the bottom to divide the top, $b$ must divide $9a^2+14b^2$.
- Since $b$ divides $14b^2$, it must divide the rest, $9a^2$.
- Again $b$ shares no factor with $a$, so it shares none with $a^2$, and $b$ must divide $9$.
- The positive divisors of $9$ are $1, 3, 9$.
💡 The same coprime trick pins $b$ to the divisors of the $9$ it sits under.
4.OA.B.4 Step 4 Narrow to a short candidate list
- So only $a \in \{1,2,7,14\}$ and $b \in \{1,3,9\}$ are even possible.
- Being a divisor is necessary but not yet enough: the full bottom $9ab$ must divide the top, and the $9$ and the $b$ can share the factor $3$.
- So the value of $b$ decides everything.
- Test the three choices of $b$ one at a time, remembering that none of $1,2,7,14$ is a multiple of $3$.
💡 The candidates are few, so a short organised check will settle it.
4.OA.B.4 Step 5 Test each value of b
- Case $b=1$: the top is $9a^2+14$ and the bottom is $9a$.
- Since $9a$ divides $9a^2$, it would have to divide the leftover $14$; but $9a$ is a multiple of $9$ and $14$ is not, so this never works.
- Case $b=9$: the fraction becomes $\dfrac{9a^2+1134}{81a}$, which is a whole number only if $9a$ divides $a^2+126$; but $126$ is a multiple of $9$ while $a^2$ leaves a remainder of $1$, $4$, or $7$ when divided by $9$ (because $a$ is not a multiple of $3$), so $a^2+126$ is never a multiple of $9$ — no solutions.
- Case $b=3$: the fraction becomes $\dfrac{9a^2+126}{27a}=\dfrac{a^2+14}{3a}$, and we need $3a$ to divide $a^2+14$.
- Here $a$ divides $a^2+14$ (since $a \mid 14$), and $3$ divides $a^2+14$ (since $a^2$ leaves remainder $1$, so $a^2+14$ acts like $15$, a multiple of $3$).
- So every $b=3$ case works.
💡 The shared factor of $3$ only lines up when $b$ is exactly $3$; too little ($b=1$) or too much ($b=9$) breaks it.
4.OA.B.4 Step 6 Count the surviving pairs
- Only $b=3$ survives, and every allowed $a$ works with it, giving the pairs $(1,3), (2,3), (7,3), (14,3)$ — each coprime, since none of $1,2,7,14$ shares the factor $3$.
- Checking the value in each case confirms a whole number: $\frac{1+14}{3}=5$, $\frac{4+14}{6}=3$, $\frac{49+14}{21}=3$, and $\frac{196+14}{42}=5$.
- That is $4$ pairs in total, so the answer is $\textbf{(A)}\ 4$.
💡 One good value of $b$, four good values of $a$: multiply to four honest pairs.
6.EE.A.2 Put both fractions over the common bottom $9ab$. The first becomes $\dfrac{9a^2} 6.NS.B.4 Look only at the factor $a$ sitting on the bottom. For the bottom to divide the 6.NS.B.4 Do the same for the factor $b$ on the bottom. For the bottom to divide the top, 4.OA.B.4 So only $a \in \{1,2,7,14\}$ and $b \in \{1,3,9\}$ are even possible. Being a di 4.OA.B.4 Case $b=1$: the top is $9a^2+14$ and the bottom is $9a$. Since $9a$ divides $9a^ 4.OA.B.4 Only $b=3$ survives, and every allowed $a$ works with it, giving the pairs $(1,3 Review
Reasonableness: The four pairs were each plugged back in and gave the whole numbers $5, 3, 3, 5$, so all four truly work. No pair was double-counted, and the case analysis on $b$ proved that $b=1$ and $b=9$ contribute nothing, so nothing was missed — this rules out (E) infinitely many and the larger counts (B) $6$, (C) $9$, (D) $12$. Exactly $4$ pairs remain, matching (A).
Alternative: Set the expression equal to an integer $k$ and clear denominators to get $9a^2 - 9k\,ab + 14b^2 = 0$, a quadratic in $a$. For $a$ to be an integer the discriminant $9b^2(9k^2-56)$ must be a perfect square, so $9k^2-56$ must be a perfect square $t^2$. Then $(3k-t)(3k+t)=56$, whose factorisations force $k=3$ or $k=5$. Feeding those back and keeping only coprime pairs yields exactly $(1,3),(2,3),(7,3),(14,3)$ — the same four, which independently confirms the count is finite and equal to $4$.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Combining the two variable fractions over the common bottom $9ab$ into the single expression $\dfrac{9a^2+14b^2}{9ab}$.)6.NS.B.4Find the greatest common factor of two whole numbers and reason about common factors (Using $\gcd(a,b)=1$ to pass from $a \mid 14b^2$ to $a \mid 14$, and from $b \mid 9a^2$ to $b \mid 9$, shrinking each variable to a short divisor list.)4.OA.B.4Find factor pairs and determine whether a whole number is a multiple of a given number (Listing the divisors of $14$ and $9$, and testing for each candidate $b$ whether the full bottom $9ab$ divides the top $9a^2+14b^2$.)
⭐ Glue the fractions into one, then let the no-shared-factor rule force each letter onto a tiny divisor list — after that you only have a handful of cases to check.
⭐ Glue the fractions into one, then let the no-shared-factor rule force each letter onto a tiny divisor list — after that you only have a handful of cases to check.
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