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

(→Intermediate) |
Etmetalakret (talk | contribs) |
||

(10 intermediate revisions by 5 users not shown) | |||

Line 1: | Line 1: | ||

− | ==The General Statement | + | ==The General Statement== |

− | Simon's Favorite Factoring Trick (SFFT) is often used in a Diophantine | + | Simon's Favorite Factoring Trick (SFFT) 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. 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. |

− | 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 | + | 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, and the coefficient of xy must be 1(If it is not 1, then divide the coefficient off of the equation.). According to Simon's Favorite Factoring Trick, this equation can be transformed into: <cmath>(x+k)(y+j)=a+jk</cmath> |

Using the previous example, <math>xy+66x-88y=23333</math> is the same as: <cmath>(x-88)(y+66)=(23333)+(-88)(66)</cmath> | Using the previous example, <math>xy+66x-88y=23333</math> is the same as: <cmath>(x-88)(y+66)=(23333)+(-88)(66)</cmath> | ||

Line 27: | Line 27: | ||

<math> \mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 } </math> | <math> \mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 } </math> | ||

+ | - icecreamrolls8 | ||

==Solution== | ==Solution== | ||

− | We have solution <math>3</math>. Note that <math>kn+ | + | We have solution <math>3</math>. Note that <math>kn+54k+2n+106</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>3</math>, our answer, so that the entire factor has a remainder of <math>3</math> when divided by <math>5</math>. |

+ | |||

+ | - icecreamrolls8 | ||

==Problem 2== | ==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]]) |

===Olympiad=== | ===Olympiad=== | ||

− | *The integer <math>N</math> is positive. There are exactly 2005 ordered pairs <math>(x, y)</math> of positive integers satisfying: | + | *The integer <math>N</math> is positive. There are exactly <math>2005</math> ordered pairs <math>(x, y)</math> of positive integers satisfying: |

<cmath>\frac 1x +\frac 1y = \frac 1N</cmath> | <cmath>\frac 1x +\frac 1y = \frac 1N</cmath> | ||

Line 50: | Line 53: | ||

* [[Factoring]] | * [[Factoring]] | ||

− | [[Category: | + | [[Category:Number theory]] |

[[Category:Theorems]] | [[Category:Theorems]] |

## Latest revision as of 13:42, 14 July 2021

## Contents

## The General Statement

Simon's Favorite Factoring Trick (SFFT) 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. A extortive example would be: where is the constant term, is the product of the variables, and are the variables in linear terms.

Let's put it in general terms. We have an equation , where , , and are integral constants, and the coefficient of xy must be 1(If it is not 1, then divide the coefficient off of the equation.). According to Simon's Favorite Factoring Trick, this equation can be transformed into:
Using the previous example, is the same as:

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 and are variables and are known constants. Also, it is typically necessary to add the term to both sides to perform the factorization.

## Fun Practice Problems

### Introductory

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

(Source)

### Intermediate

## Problem 1

- If has a remainder of when divided by , and has a remainder of when divided by , find the value of the remainder of when is divided by .

- icecreamrolls8

## Solution

We have solution . Note that can be factored into using Simon's Favorite Factoring Trick. Now, look at n. Then, since the problem tells us that has a remainder of when divided by 5, we see that the factor in the expression has a remainder of when divided by 5. Now, the must have a remainder of when divided by as well (because then the main expression has a remainder of when divided by ). Therefore, since 54 has a remainder of when divided by , must have a remainder of , our answer, so that the entire factor has a remainder of when divided by .

- icecreamrolls8

## Problem 2

- are integers such that . Find .

(Source)

### Olympiad

- The integer is positive. There are exactly ordered pairs of positive integers satisfying:

Prove that is a perfect square.

Source: (British Mathematical Olympiad Round 3, 2005)