Difference between revisions of "2007 AMC 12B Problems/Problem 24"
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− | + | Also refer to the [https://artofproblemsolving.com/wiki/index.php/2007_AMC_10B_Problems/Problem_25#Solution| 2007 AMC 10B #25] (same problem) | |
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− | |||
− | |||
+ | == Problem == | ||
+ | How many pairs of positive integers <math>(a,b)</math> are there such that <math>\text{gcd}(a,b)=1</math> and <math>\frac{a}{b} + \frac{14b}{9a}</math> is an integer? | ||
+ | |||
+ | <math>\mathrm {(A)}\ 4\quad\mathrm {(B)}\ 6\quad\mathrm {(C)}\ 9\quad\mathrm {(D)}\ 12\quad\mathrm {(E)}\ \text{infinitely many}</math> | ||
+ | |||
+ | == Solution 1 == | ||
Combining the fraction, <math>\frac{9a^2 + 14b^2}{9ab}</math> must be an integer. | Combining the fraction, <math>\frac{9a^2 + 14b^2}{9ab}</math> must be an integer. | ||
Since the denominator contains a factor of <math>9</math>, <math>9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b</math> | Since the denominator contains a factor of <math>9</math>, <math>9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b</math> | ||
− | + | Since <math>b = 3n</math> for some positive integer <math>n</math>, we can rewrite the fraction(divide by <math>9</math> on both top and bottom) as <math>\frac{a^2 + 14n^2}{3an}</math> | |
Since the denominator now contains a factor of <math>n</math>, we get <math>n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2</math>. | Since the denominator now contains a factor of <math>n</math>, we get <math>n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2</math>. | ||
− | But since <math>1=gcd(a,b)=gcd(a,3n)=gcd(a,n)</math>, we must have <math>n=1</math>, and thus <math>b=3</math>. | + | But since <math>1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)</math>, we must have <math>n=1</math>, and thus <math>b=3</math>. |
For <math>b=3</math> the original fraction simplifies to <math>\frac{a^2 + 14}{3a}</math>. | For <math>b=3</math> the original fraction simplifies to <math>\frac{a^2 + 14}{3a}</math>. | ||
− | For that to be an integer, <math>a</math> must | + | For that to be an integer, <math>a</math> must be a factor of <math>14</math>, and therefore we must have <math>a\in\{1,2,7,14\}</math>. Each of these values does indeed yield an integer. |
− | |||
− | |||
− | |||
+ | Thus there are four solutions: <math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math> and the answer is <math>\mathrm{(A)}</math> | ||
− | + | == Solution 2 == | |
− | + | Let's assume that <math>\frac{a}{b} + \frac{14b}{9a} = m</math> We get | |
− | ==Solution 2== | ||
− | Let's assume that <math>\frac{a}{b} + \frac{14b}{9a} = m | ||
<math>9a^2 - 9mab + 14b^2 = 0</math> | <math>9a^2 - 9mab + 14b^2 = 0</math> | ||
Line 65: | Line 64: | ||
For example, in the case | For example, in the case | ||
− | <math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not equal by <math>-9m</math> for | + | <math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not equal by <math>-9m</math> for any integer <math>m</math>. |
+ | |||
+ | Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total <math>\mathrm {(A)}</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | Let <math>u = \frac{a}{b}</math>. Then the given equation becomes <math>u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}</math>. | ||
+ | |||
+ | Let's set this equal to some value, <math>k \Rightarrow \frac{9u^2 + 14}{9u} = k</math>. | ||
+ | |||
+ | Clearing the denominator and simplifying, we get a quadratic in terms of <math>u</math>: | ||
+ | |||
+ | <math>9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}</math> | ||
+ | |||
+ | Since <math>a</math> and <math>b</math> are integers, <math>u</math> is a rational number. This means that <math>\sqrt{(9k)^2 - 504}</math> is an integer. | ||
+ | |||
+ | Let <math>\sqrt{(9k)^2 - 504} = x</math>. Squaring and rearranging yields: | ||
+ | |||
+ | <math>(9k)^2 - x^2 = 504</math> | ||
+ | |||
+ | <math>(9k+x)(9k-x) = 504</math>. | ||
+ | |||
+ | In order for both <math>x</math> and <math>a</math> to be an integer, <math>9k + x</math> and <math>9k - x</math> must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let <math>9k + x = 2m</math> and <math>9k - x = 2n</math>. | ||
+ | |||
+ | Then: | ||
+ | |||
+ | <math>2m \cdot 2n = 504</math> | ||
+ | |||
+ | <math>mn = 126</math>. | ||
+ | |||
+ | Factoring 126, we get <math>6</math> pairs of numbers: <math>(1,126), (2,63), (3,42), (6,21), (7,18),</math> and <math>(9,14)</math>. | ||
+ | |||
+ | Looking back at our equations for <math>m</math> and <math>n</math>, we can solve for <math>k = \frac{2m + 2n}{18} = \frac{m+n}{9}</math>. Since <math>k</math> is an integer, there are only <math>2</math> pairs of <math>(m,n)</math> that work: <math>(3,42)</math> and <math>(6,21)</math>. This means that there are <math>2</math> values of <math>k</math> such that <math>u</math> is an integer. But looking back at <math>u</math> in terms of <math>k</math>, we have <math>\pm</math>, meaning that there are <math>2</math> values of <math>u</math> for every <math>k</math>. Thus, the answer is <math>2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | Rewriting the expression over a common denominator yields <math>\frac{9a^2 + 14b^2}{9ab}</math>. This expression must be equal to some integer <math>m</math>. | ||
+ | |||
+ | Thus, <math>\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm</math>. Taking this <math>\pmod{a}</math> yields <math>14b^2 \equiv 0\pmod{a}</math>. Since <math>\gcd(a,b)=1</math>, <math>14 \equiv 0\pmod{a}</math>. This implies that <math>a|14</math> so <math>a = 1, 2, 7, 14</math>. | ||
+ | |||
+ | We can then take <math>9a^2 + 14b^2 = 9abm \pmod{b}</math> to get that <math>9 \equiv 0 \pmod{b}</math>. Thus <math>b = 1, 3, 9</math>. | ||
+ | |||
+ | However, taking <math>9a^2 + 14b^2 = 9abm \pmod{3}</math>, <math>b^2 \equiv 0\pmod{3}</math> so <math>b</math> cannot equal 1. | ||
+ | |||
+ | Also, note that if <math>b = 9</math>, <math>\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}</math>. Since <math>a|14</math>, <math>\frac{14}{a}</math> will be an integer, but <math>\frac{a}{9}</math> will not be an integer since none of the possible values of <math>a</math> are multiples of 9. Thus, <math>b</math> cannot equal 9. | ||
+ | |||
+ | Thus, the only possible values of <math>b</math> is 3, and <math>a</math> can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is <math>\mathrm{(A)}</math>. | ||
+ | |||
+ | == Solution 5 (Similar to Solution 1) == | ||
+ | Rewriting <math>\frac{a}{b} + \frac{14b}{9a}</math> over a common denominator gives <math>\frac{9a^2 + 14b^2}{9ab}.</math> | ||
+ | |||
+ | Thus, we have <math>9 \mid 9a^2 + 14b^2 \Rightarrow 3 \mid b.</math> | ||
+ | |||
+ | Next, we have <math>ab \mid 9a^2+14b^2 \Rightarrow ab \mid 14b^2 \Rightarrow a \mid 14b.</math> | ||
+ | |||
+ | Thus, <math>a \in (1,2,7,4).</math> | ||
+ | |||
+ | Next, we have <math>b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.</math> | ||
+ | |||
+ | Thus, <math>b \in (1,3,9).</math> | ||
+ | |||
+ | Now, we simply do casework on <math>b.</math> | ||
+ | |||
+ | Plugging in <math>b = 1,3</math> and <math>9</math> gives that there are <math>4</math> total solutions for <math>(a,b).</math> | ||
+ | |||
+ | ~coolmath2017 | ||
+ | == Solution 6 (Similar to solution 3) == | ||
+ | Let <math>\frac{a}{b} = r.</math> So <math>r + \frac{14}{9r}=I</math>, where I is an integer. Algebraic manipulations yield: <math>r^2-Ir+\frac{14}{9}=0</math>. The discriminant of this must be the square of a rational number, call this R. So <math>I^2-\frac{56}{9}=R^2 \longrightarrow I^2-R^2=(I-R)(I+R)=\frac{56}{9}</math>. I is <math>\frac{1}{2}</math> the sum of <math>I-R</math> and <math>I+R</math>. To have an integer sum, <math>I-R</math> and <math>I+R</math> must have the same denominator, namely 3. We proceed with casework. | ||
+ | |||
+ | Case 1. | ||
+ | <math>I+R=56/3</math>, <math>I-R=1/3</math>. This yields <math>I=19/2</math>, which is not an integer. This case produces 0 solutions. | ||
+ | |||
+ | Case 2. | ||
+ | <math>I+R=28/3</math>, <math>I-R=2/3</math>. This yields <math>I=5</math>. Substituting into our original equation yields: <math>r^2-5r+\frac{14}{9}=0</math>. Factoring gives: <math>r=\frac{1}{3}</math>, <math>r=\frac{14}{3}</math>. This case produces 2 solutions, namely (1,3) and (14,3). | ||
+ | |||
+ | Case 3. | ||
+ | <math>I+R=14/3</math>, <math>I-R=4/3</math>. This yields <math>I=3</math>. Substituting into our original equation yields: <math>r^2-3r+\frac{14}{9}=0</math>. Factoring gives: <math>r=\frac{2}{3}</math>, <math>r=\frac{7}{3}</math>. This case produces 2 solutions, namely (2,3) and (7,3). | ||
+ | |||
+ | Case 4. | ||
+ | <math>I+R=8/3</math>, <math>I-R=7/3</math>. This yields <math>I=5/2</math>, which is not an integer. This case produces 0 solutions. | ||
+ | |||
+ | Altogether, we have 4 solutions, so our answer is <math>\boxed{(A)}</math>. | ||
+ | |||
+ | ~Math4Life2020 | ||
− | + | == Solution 7 == | |
− | = | + | Rewrite the equation |
− | + | <math>\frac{a}{b}+\frac{14b}{9a}=k</math> | |
+ | in two different forms. First, multiply both sides by <math>b</math> and subtract <math>a</math> to obtain | ||
+ | <math>\frac{14b^2}{9a}=bk-a.</math> | ||
+ | Because <math>a</math>, <math>b</math>, and <math>k</math> are integers, <math>14b^2</math> must be a multiple of <math>a</math>, and because <math>a</math> and <math>b</math> have no common factors greater than 1, it follows that 14 is divisible by <math>a</math>. Next, multiply both sides of the original equation by <math>9a</math> and subtract <math>14b</math> to obtain | ||
+ | <math>\frac{9a^2}{b}=9ak-14b.</math> | ||
+ | This shows that <math>9a^2</math> is a multiple of <math>b</math>, so 9 must be divisible by <math>b</math>. Thus if <math>(a,b)</math> is a solution, then <math>b=1</math>, <math>3</math>, or <math>9</math>, and <math>a=1</math>, 2, 7, or 14. This gives a total of twelve possible solutions <math>(a,b)</math>, each of which can be checked quickly.<math> | ||
+ | \frac{a}{b}+\frac{14b}{9a}</math> | ||
+ | will only be an integer when <math>(a,b) \in \{(1,3),(2,3), (7,3), (14,3)\},</math> for a total of <math>\boxed{4}</math> pairs. | ||
− | ==See Also== | + | == See Also == |
{{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− | |||
− |
Latest revision as of 22:41, 13 September 2024
Also refer to the 2007 AMC 10B #25 (same problem)
Contents
Problem
How many pairs of positive integers are there such that and is an integer?
Solution 1
Combining the fraction, must be an integer.
Since the denominator contains a factor of ,
Since for some positive integer , we can rewrite the fraction(divide by on both top and bottom) as
Since the denominator now contains a factor of , we get .
But since , we must have , and thus .
For the original fraction simplifies to .
For that to be an integer, must be a factor of , and therefore we must have . Each of these values does indeed yield an integer.
Thus there are four solutions: , , , and the answer is
Solution 2
Let's assume that We get
Factoring this, we get equations-
(It's all negative, because if we had positive signs, would be the opposite sign of )
Now we look at these, and see that-
This gives us solutions, but we note that the middle term needs to give you back .
For example, in the case
, the middle term is , which is not equal by for any integer .
Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total
Solution 3
Let . Then the given equation becomes .
Let's set this equal to some value, .
Clearing the denominator and simplifying, we get a quadratic in terms of :
Since and are integers, is a rational number. This means that is an integer.
Let . Squaring and rearranging yields:
.
In order for both and to be an integer, and must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let and .
Then:
.
Factoring 126, we get pairs of numbers: and .
Looking back at our equations for and , we can solve for . Since is an integer, there are only pairs of that work: and . This means that there are values of such that is an integer. But looking back at in terms of , we have , meaning that there are values of for every . Thus, the answer is .
Solution 4
Rewriting the expression over a common denominator yields . This expression must be equal to some integer .
Thus, . Taking this yields . Since , . This implies that so .
We can then take to get that . Thus .
However, taking , so cannot equal 1.
Also, note that if , . Since , will be an integer, but will not be an integer since none of the possible values of are multiples of 9. Thus, cannot equal 9.
Thus, the only possible values of is 3, and can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is .
Solution 5 (Similar to Solution 1)
Rewriting over a common denominator gives
Thus, we have
Next, we have
Thus,
Next, we have
Thus,
Now, we simply do casework on
Plugging in and gives that there are total solutions for
~coolmath2017
Solution 6 (Similar to solution 3)
Let So , where I is an integer. Algebraic manipulations yield: . The discriminant of this must be the square of a rational number, call this R. So . I is the sum of and . To have an integer sum, and must have the same denominator, namely 3. We proceed with casework.
Case 1. , . This yields , which is not an integer. This case produces 0 solutions.
Case 2. , . This yields . Substituting into our original equation yields: . Factoring gives: , . This case produces 2 solutions, namely (1,3) and (14,3).
Case 3. , . This yields . Substituting into our original equation yields: . Factoring gives: , . This case produces 2 solutions, namely (2,3) and (7,3).
Case 4. , . This yields , which is not an integer. This case produces 0 solutions.
Altogether, we have 4 solutions, so our answer is .
~Math4Life2020
Solution 7
Rewrite the equation in two different forms. First, multiply both sides by and subtract to obtain Because , , and are integers, must be a multiple of , and because and have no common factors greater than 1, it follows that 14 is divisible by . Next, multiply both sides of the original equation by and subtract to obtain This shows that is a multiple of , so 9 must be divisible by . Thus if is a solution, then , , or , and , 2, 7, or 14. This gives a total of twelve possible solutions , each of which can be checked quickly. will only be an integer when for a total of pairs.
See Also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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