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

Revision as of 01:19, 21 June 2021

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: $xy+66x-88y=23333$ 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, 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: $(x+k)(y+j)=a+jk$ 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

Problem 1

If $kn+54k+2n+106$ 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 $5$ .

$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 }$

- icecreamrolls8

Solution

We have solution $3$ . Note that $kn+54k+2n+106$ 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 $3$ , our answer, 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)

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

@@ Line 30: / Line 30: @@
 ==Solution==
-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 <math>Simon's Favorite Factoring Trick</math>. 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>.
+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

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