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

Revision as of 15:41, 20 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. 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

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 37: / Line 37: @@
 *<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]]) NERD
+([[1987 AIME Problems/Problem 5|Source]])
 ===Olympiad===

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