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

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=== Statement of the factorization ===
 
Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization. SFFT is: <math>{xy}+{xk}+{yj}+{jk}=(x+j)(y+k)</math>.
 
  
=== Credit ===
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==The General Statement==
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Simon's Favorite Factoring Trick (SFFT) is often used in a Diophantine(Positive) 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. A extortive example would be: <cmath>xy+66x-88y=23333</cmath>where <math>23333</math> is the constant term, <math>xy</math> is the product of the variables, <math>66x</math> and <math>-88y</math> are the variables in linear terms.
  
This factorization was first popularized by AoPS user ComplexZeta, whose name is Simon.
 
  
=== Applications ===
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Let's put it in general terms. We have an equation <math>xy+jx+ky=a</math>, where <math>j</math>, <math>k</math>, and <math>a</math> are integral constants. According to Simon's Favourite Factoring Trick, this equation can be transformed into: <cmath>(x+k)(y+j)=a+jk</cmath>
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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Using the previous example, <math>xy+66x-88y=23333</math> is the same as: <cmath>(x-88)(y+66)=(23333)+(-88)(66)</cmath>
  
=== Examples ===
 
([[AIME]] 1987/5) <math>m</math> and <math>n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.
 
  
'''Outline Solution:''' Rearrange to <math>m^2 + 3m^2n^2 -30n^2= 517</math>. The key step is changing the equation to <math>m^2 + 3m^2n^2 -30n^2-10= 507</math>, where the equation factors to <math>(3n^2 + 1)(m^2 - 10) = 507 = 3\cdot 13^2</math>, from which the problem is trivial to solve by applying some simple number theory.
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If this is confusing or you would like to know the thought process behind SFFT, see this eight-minute video by Richard Rusczyk from AoPS: https://www.youtube.com/watch?v=0nN3H7w2LnI. For the thought process, start from https://youtu.be/0nN3H7w2LnI?t=366
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== Applications ==
 +
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>jk</math> term to both sides to perform the factorization.
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== Fun Practice Problems ==
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===Introductory===
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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) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }  </math>
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([[2000 AMC 12/Problem 6|Source]])
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===Intermediate===
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*<math>m, n</math> are integers such that <math>m^2 + 3m^2n^2 = 30n^2 + 517</math>. Find <math>3m^2n^2</math>.
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([[1987 AIME Problems/Problem 5|Source]])
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===Olympiad===
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*The integer <math>N</math> is positive. There are exactly 2005 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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Source: (British Mathematical Olympiad Round 3, 2005)
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== See More==
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* [[Algebra]]
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* [[Factoring]]
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[[Category:Elementary algebra]]
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[[Category:Theorems]]

Revision as of 20:18, 12 January 2021

The General Statement

Simon's Favorite Factoring Trick (SFFT) is often used in a Diophantine(Positive) 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. A extortive example would be: \[xy+66x-88y=23333\]where $23333$ is the constant term, $xy$ is the product of the variables, $66x$ and $-88y$ are the variables in linear terms.


Let's put it in general terms. We have an equation $xy+jx+ky=a$, where $j$, $k$, and $a$ are integral constants. According to Simon's Favourite Factoring Trick, this equation can be transformed into: \[(x+k)(y+j)=a+jk\] Using the previous example, $xy+66x-88y=23333$ is the same as: \[(x-88)(y+66)=(23333)+(-88)(66)\]


If this is confusing or you would like to know the thought process behind SFFT, see this eight-minute video by Richard Rusczyk from AoPS: https://www.youtube.com/watch?v=0nN3H7w2LnI. For the thought process, start from https://youtu.be/0nN3H7w2LnI?t=366

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. Also, it is typically necessary to add the $jk$ term to both sides to perform the factorization.

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

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

(Source)

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.

Source: (British Mathematical Olympiad Round 3, 2005)

See More

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