Difference between revisions of "Simon's Favorite Factoring Trick"
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== Applications == | == 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. | + | 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. |
== Fun Practice Problems == | == Fun Practice Problems == | ||
Line 22: | Line 12: | ||
===Intermediate=== | ===Intermediate=== | ||
+ | |||
==Problem 1== | ==Problem 1== | ||
− | *If <math>kn+54k+2n+ | + | *If <math>kn+54k+2n+108</math> has a remainder of <math>3</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 when <math>n</math> is divided by <math>5</math>. |
− | <math> \mathrm{(A) \ | + | <math> \mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 0 } \qquad \mathrm{(C) \ 4 } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ 3 } </math> |
- icecreamrolls8 | - icecreamrolls8 | ||
==Solution== | ==Solution== | ||
− | We have solution <math> | + | We have solution <math>\boxed{(D)}</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>1</math> when divided by <math>5</math> as well [3(mod5) x 1(mod5) = 3(mod5)] matching the main expression which has a remainder of <math>3</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>2</math>, so that the entire factor has a remainder of <math>1</math> when divided by <math>5</math>. |
- icecreamrolls8 | - icecreamrolls8 | ||
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([[1987 AIME Problems/Problem 5|Source]]) | ([[1987 AIME Problems/Problem 5|Source]]) | ||
+ | |||
+ | ==Solution== | ||
+ | <cmath>m^2 + 3m^2n^2 = 30n^2 + 517</cmath> | ||
+ | <cmath>(m^2-10)(3n^2+1)=507</cmath> | ||
+ | <cmath>507=13*39</cmath> | ||
+ | <cmath>(3n^2+1)=13,(m^2-10)=39</cmath> | ||
+ | <cmath>3m^2n^2=588</cmath> | ||
+ | |||
+ | ==Problem 3== | ||
+ | |||
+ | ([[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>? | ||
+ | |||
+ | Solution: <math>A_{outer}=ab</math> | ||
+ | |||
+ | <math>A_{inner}=(a-2)(b-2)</math> | ||
+ | |||
+ | <math>A_{outer}=2A_{inner}</math> | ||
+ | |||
+ | <math>ab=2(a-2)(b-2)=2ab-4a-4b+8</math> | ||
+ | |||
+ | <math>0=ab-4a-4b+8</math> | ||
+ | |||
+ | By Simon's Favorite Factoring Trick: | ||
+ | |||
+ | <math>8=ab-4a-4b+16=(a-4)(b-4)</math> | ||
+ | |||
+ | Since <math>8=1\times8</math> and <math>8=2\times4</math> are the only positive factorings of <math>8</math>. | ||
+ | |||
+ | <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. | ||
===Olympiad=== | ===Olympiad=== | ||
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Prove that <math>N</math> is a perfect square. | Prove that <math>N</math> is a perfect square. | ||
− | Source: | + | 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== | == See More== |
Latest revision as of 12:13, 8 December 2024
Contents
[hide]Applications
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually and are variables and are known constants. Sometimes, you have to notice that the variables are not in the form and Additionally, you almost always have to subtract or add the and 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 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 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 [3(mod5) x 1(mod5) = 3(mod5)] matching the main expression which has a remainder of when divided by . Therefore, since 54 has a remainder of when divided by , must have a remainder of , so that the entire factor has a remainder of when divided by .
- icecreamrolls8
Problem 2
- are integers such that . Find .
(Source)
Solution
Problem 3
(Source) A rectangular floor measures by feet, where and are positive integers with . 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 foot around the painted rectangle and occupies half of the area of the entire floor. How many possibilities are there for the ordered pair ?
Solution:
By Simon's Favorite Factoring Trick:
Since and are the only positive factorings of .
or yielding solutions. Notice that because , the reversed pairs are invalid.
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 2 #1 https://bmos.ukmt.org.uk/home/bmo2-2005.pdf