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

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'''Simon's Favorite Factoring Trick''' (abbreviated SFFT) is a special factorization first popularized by [[AoPS]] user [[user:ComplexZeta | Simon Rubinstein-Salzedo]].  <url>viewtopic.php?highlight=factoring&t=8215 This</url> appears to be the thread where Simon's favorite factoring trick was first introduced.
 
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== Statement of the factorization ==
 
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The general statement of SFFT is: <math>{xy}+{xk}+{yj}+{jk}=(x+j)(y+k)</math>.  Two special cases appear most commonly: <math>xy + x + y + 1 = (x+1)(y+1)</math> and <math>xy - x - y +1 = (x-1)(y-1)</math>.
 
  
 
== Applications ==
 
== Applications ==
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This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>{x}</math> and <math>{y}</math> are variables and <math>j,k</math> are known constants. Also it is typically necessary to add the <math>{j}{k}</math> term to both sides to perform the factorization.
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This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually <math>x</math> and <math>y</math> are variables and <math>j,k</math> are known constants. Sometimes, you have to notice that the variables are not in the form <math>x</math> and <math>y.</math> Additionally, you almost always have to subtract or add the <math>x, y,</math> and <math>xy</math> 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.
  
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== Problems ==
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== Fun Practice Problems ==
 
===Introductory===
 
===Introductory===
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*
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*Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
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Two different [[prime number]]s between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
 
  
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<math> \mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }  </math>
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<math> \mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }  </math>
  
 
([[2000 AMC 12/Problem 6|Source]])
 
([[2000 AMC 12/Problem 6|Source]])
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===Intermediate===
 
===Intermediate===
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==Problem 1==
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*If <math>kn+54k+2n+108</math> has a remainder of <math>4</math> when divided by <math>5</math>, and <math>k</math> has a remainder of <math>1</math> when divided by <math>5</math>, find the value of the remainder of when <math>n</math> is divided by <math>5</math>.
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<math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 0 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 }  </math>
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- icecreamrolls8
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==Solution==
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We have solution <math>\boxed{(C)}</math>. Note that <math>kn+54k+2n+108</math> can be factored into <cmath>(k+2)(n+54)</cmath> using Simon's Favorite Factoring Trick. Now, look at n. Then, since the problem tells us that <math>k</math> has a remainder of <math>1</math> when divided by 5, we see that the <math>(k+2)</math> factor in the <math>(k+2)(n+54)</math> expression has a remainder of <math>3</math> when divided by 5. Now, the <math>(n+54)</math> must have a remainder of <math>3</math> when divided by <math>5</math> as well (because then the main expression has a remainder of <math>4</math> when divided by <math>5</math>). Therefore, since 54 has a remainder of <math>4</math> when divided by <math>5</math>, <math>n</math> must have a remainder of <math>4</math>, so that the entire factor has a remainder of <math>3</math> when divided by <math>5</math>.
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- icecreamrolls8
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==Problem 2==
 
*<math>m, n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.
 
*<math>m, n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.
  
 
([[1987 AIME Problems/Problem 5|Source]])
 
([[1987 AIME Problems/Problem 5|Source]])
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==Solution==
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<cmath>m^2 + 3m^2n^2 = 30n^2 + 517</cmath>
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<cmath>(m^2-10)(3n^2+1)=507</cmath>
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<cmath>507=13*39</cmath>
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<cmath>(3n^2+1)=13,(m^2-10)=39</cmath>
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<cmath>3m^2n^2=588</cmath>
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 +
==Problem 3==
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([[2008 AMC 12B Problems/Problem 16|Source]]) A rectangular floor measures <math>a</math> by <math>b</math> feet, where <math>a</math> and <math>b</math> are positive integers with <math>b > a</math>. 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 <math>1</math> foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair <math>(a,b)</math>?
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Solution: <math>A_{outer}=ab</math>
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<math>A_{inner}=(a-2)(b-2)</math>
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<math>A_{outer}=2A_{inner}</math>
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<math>ab=2(a-2)(b-2)=2ab-4a-4b+8</math>
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<math>0=ab-4a-4b+8</math>
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By Simon's Favorite Factoring Trick:
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<math>8=ab-4a-4b+16=(a-4)(b-4)</math>
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Since <math>8=1\times8</math> and <math>8=2\times4</math> are the only positive factorings of <math>8</math>.
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<math>(a,b)=(5,12)</math> or <math>(a,b)=(6,8)</math> yielding <math>\Rightarrow\textbf{(B)}</math> <math>2</math> solutions. Notice that because <math>b>a</math>, the reversed pairs are invalid.
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===Olympiad===
 
===Olympiad===
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{{problem}}
 
  
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== See Also ==
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*The integer <math>N</math> is positive. There are exactly <math>2005</math> ordered pairs <math>(x, y)</math> of positive integers satisfying:
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<cmath>\frac 1x +\frac 1y = \frac 1N</cmath>
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Prove that <math>N</math> is a perfect square.
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Solution: https://socratic.org/questions/given-the-integer-n-0-there-are-exactly-2005-ordered-pairs-x-y-of-positive-integ
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Source: British Mathematical Olympiad Round 2 #1 https://bmos.ukmt.org.uk/home/bmo2-2005.pdf
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== See More==
 
* [[Algebra]]
 
* [[Algebra]]
 
* [[Factoring]]
 
* [[Factoring]]
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* [http://www.artofproblemsolving.com/Movies/simon1.php AoPS's SFFT Video 1]
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*[http://www.artofproblemsolving.com/Movies/simon2.php AoPS's SFFT Video 2]
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[[Category:Number theory]]
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[[Category:Theorems]]

Latest revision as of 20:05, 16 October 2024

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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