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Difference between revisions of "Simon's Favorite Factoring Trick"

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==The General Statement==
 
==The General Statement==
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Simon's Favorite Factoring Trick (SFFT) (made by AoPS user [https://artofproblemsolving.com/community/user/1233 Complex Zeta]) is often used in a Diophantine equation where factoring is needed. The most common form it appears is when there is a constant on one side of the equation and a product of variables with each of those variables in a linear term on the other side. An example would be: <cmath>xy+66x-88y=23333</cmath>where <math>23333</math> =1<math>. (2021 CEMC Galois #4b)
+
Simon's Favorite Factoring Trick (SFFT) (made by AoPS user [https://artofproblemsolving.com/community/user/1233 Complex Zeta]) is often used in a Diophantine equation where factoring is needed. The most common form it appears is when there is a constant on one side of the equation and a product of variables with each of those variables in a linear term on the other side. An example would be: <cmath>xy+66x-88y=23333</cmath>where <math>23333=1</math>. (2021 CEMC Galois #4b)
  
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Let's remove the denominators: </math>4y+5x=xy<math>. Then </math>xy-5x-4y=0<math>. Take out the </math>x<math>: </math>x(y-5)-4(y-5)=0+20<math> (notice how I artificially grouped up the </math>y<math> terms by adding </math>4*5=20<math>).  
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Let's remove the denominators: <math>4y+5x=xy</math>. Then <math>xy-5x-4y=0</math>. Take out the <math>x</math>: <math>x(y-5)-4(y-5)=0+20</math> (notice how I artificially grouped up the <math>y</math> terms by adding <math>4*5=20</math>).  
  
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Now, </math>(x-4)(y-5)=20<math> (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs to solve this problem.
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Now, <math>(x-4)(y-5)=20</math> (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs to solve this problem.
  
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Look at all factor pairs of 20: </math>1*20, 2*10, 4*5, 5*4, 10*2, 20*1<math>. The first factor is for </math>x<math>, the second is for </math>y<math>. Solving for each of the equations, we have the solutions as </math>\boxed{(5, 25), (24, 6), (6, 15), (14, 7), (8, 10), (9, 9)}$.
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Look at all factor pairs of 20: <math>1\times 20, 2\times 10, 4\times 5, 5\times 4, 10\times 2, 20\times 1</math>. The first factor is for <math>x</math>, the second is for <math>y</math>. Solving for each of the equations, we have the solutions as <math>\boxed{(5, 25), (24, 6), (6, 15), (14, 7), (8, 10), (9, 9)}</math>.
  
 
For more info on the solution: https://www.cemc.uwaterloo.ca/contests/past_contests/2021/2021GaloisSolution.pdf
 
For more info on the solution: https://www.cemc.uwaterloo.ca/contests/past_contests/2021/2021GaloisSolution.pdf

Revision as of 20:16, 28 August 2024

The General Statement

Simon's Favorite Factoring Trick (SFFT) (made by AoPS user Complex Zeta) is often used in a Diophantine equation where factoring is needed. The most common form it appears is when there is a constant on one side of the equation and a product of variables with each of those variables in a linear term on the other side. An example would be: \[xy+66x-88y=23333\]where $23333=1$. (2021 CEMC Galois #4b)

Let's remove the denominators: $4y+5x=xy$. Then $xy-5x-4y=0$. Take out the $x$: $x(y-5)-4(y-5)=0+20$ (notice how I artificially grouped up the $y$ terms by adding $4*5=20$).

Now, $(x-4)(y-5)=20$ (you can just do SFFT directly, but I am guiding you through the thinking behind SFFT). Now we use factor pairs to solve this problem.

Look at all factor pairs of 20: $1\times 20, 2\times 10, 4\times 5, 5\times 4, 10\times 2, 20\times 1$. The first factor is for $x$, the second is for $y$. Solving for each of the equations, we have the solutions as $\boxed{(5, 25), (24, 6), (6, 15), (14, 7), (8, 10), (9, 9)}$.

For more info on the solution: https://www.cemc.uwaterloo.ca/contests/past_contests/2021/2021GaloisSolution.pdf

Applications

This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually $x$ and $y$ are variables and $j,k$ are known constants. Sometimes, you have to notice that the variables are not in the form $x$ and $y.$ Additionally, you almost always have to subtract or add the $x, y,$ and $xy$ terms to one side so you can isolate the constant and make the equation factorable. It can be used to solve more than algebra problems, sometimes going into other topics such as number theory.

Fun Practice Problems

Introductory

  • Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

$\mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }$

(Source)

Intermediate

Problem 1

  • If $kn+54k+2n+108$ has a remainder of $4$ when divided by $5$, and $k$ has a remainder of $1$ when divided by $5$, find the value of the remainder of when $n$ is divided by $5$.

$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 0 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 }$

- icecreamrolls8

Solution

We have solution $\boxed{(C)}$. Note that $kn+54k+2n+108$ can be factored into \[(k+2)(n+54)\] using Simon's Favorite Factoring Trick. Now, look at n. Then, since the problem tells us that $k$ has a remainder of $1$ when divided by 5, we see that the $(k+2)$ factor in the $(k+2)(n+54)$ expression has a remainder of $3$ when divided by 5. Now, the $(n+54)$ must have a remainder of $3$ when divided by $5$ as well (because then the main expression has a remainder of $4$ when divided by $5$). Therefore, since 54 has a remainder of $4$ when divided by $5$, $n$ must have a remainder of $4$, so that the entire factor has a remainder of $3$ when divided by $5$.

- icecreamrolls8

Problem 2

  • $m, n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.

(Source)

Solution

\[m^2 + 3m^2n^2 = 30n^2 + 517\] \[(m^2-10)(3n^2+1)=507\] \[507=13*39\] \[(3n^2+1)=13,(m^2-10)=39\] \[3m^2n^2=588\]

Problem 3

(Source) A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers with $b > a$. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair $(a,b)$?

Solution: $A_{outer}=ab$

$A_{inner}=(a-2)(b-2)$

$A_{outer}=2A_{inner}$

$ab=2(a-2)(b-2)=2ab-4a-4b+8$

$0=ab-4a-4b+8$

By Simon's Favorite Factoring Trick:

$8=ab-4a-4b+16=(a-4)(b-4)$

Since $8=1\times8$ and $8=2\times4$ are the only positive factorings of $8$.

$(a,b)=(5,12)$ or $(a,b)=(6,8)$ yielding $\Rightarrow\textbf{(B)}$ $2$ solutions. Notice that because $b>a$, the reversed pairs are invalid.

Olympiad

  • The integer $N$ is positive. There are exactly $2005$ ordered pairs $(x, y)$ of positive integers satisfying:

\[\frac 1x +\frac 1y = \frac 1N\]

Prove that $N$ is a perfect square.

Solution: https://socratic.org/questions/given-the-integer-n-0-there-are-exactly-2005-ordered-pairs-x-y-of-positive-integ

Source: British Mathematical Olympiad Round 2 #1 https://bmos.ukmt.org.uk/home/bmo2-2005.pdf

See More

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