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Algebraic Manipulation

Factoring tricks (SFFT, difference of squares, sum of cubes), telescoping products, clever substitution, symmetric expressions

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Algebraic ManipulationAMC 10 Step by Step

Name: Date: October 28, 2026

all 42 problems, easiest first

What you need to know

Squares. $a^2-b^2=(a-b)(a+b)$ and $(a\pm b)^2=a^2\pm2ab+b^2$ (read each $\pm$ as $+$ throughout, or as $-$ throughout). Example: $2024^2-2023^2=(2024-2023)(2024+2023)=4047$ .
Cubes. $a^3+b^3=(a+b)(a^2-ab+b^2)$ , $a^3-b^3=(a-b)(a^2+ab+b^2)$ , and $(a+b)^3=a^3+b^3+3ab(a+b)$ . Example: if $a+b=5$ and $ab=6$ , then $a^3+b^3=5^3-3\cdot6\cdot5=35$ . (Check: $a$ and $b$ are $2$ and $3$ , and $8+27=35$ .)
Simon's Favorite Factoring Trick. $xy+ax+by+ab=(x+b)(y+a)$ . So for $xy+ax+by=c$ , add $ab$ to both sides to get $(x+b)(y+a)=c+ab$ , then list the factor pairs of $c+ab$ . Example: $xy+x+2y=10$ becomes $(x+2)(y+1)=12$ .
Three variables. $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$ , and $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ . Example: if $a+b+c=6$ and $ab+bc+ca=11$ , then $a^2+b^2+c^2=36-22=14$ (check with $1$ , $2$ , $3$ ).
Two rarer factorizations. The Sophie Germain Identity $a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$ , and $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ . Example: with $a=4$ , $b=1$ : $4^4+4=260=26\cdot10$ .
A number plus its reciprocal. Let $t=x+\frac1x$ . Then $x^2+\frac1{x^2}=t^2-2$ (because $t^2=x^2+2+\frac1{x^2}$ ) and $x^3+\frac1{x^3}=t^3-3t$ . Example: if $x+\frac1x=3$ , then $x^2+\frac1{x^2}=7$ and $x^3+\frac1{x^3}=27-9=18$ .
Telescoping products. Factor every term and neighbors cancel. Example: $1-\frac1{k^2}=\frac{(k-1)(k+1)}{k\cdot k}$ , so $\left(1-\frac14\right)\left(1-\frac19\right)\cdots\left(1-\frac1{100}\right)=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdots\frac{9\cdot11}{10\cdot10}=\frac12\cdot\frac{11}{10}=\frac{11}{20}$ .

How AMC 10 tests it

Problems 3–10: a number expression such as $\frac{2024^2-2023^2}{\dots}$ or $\sqrt{31\cdot30\cdot29\cdot28+1}$ . Brute force is possible, but an identity makes it instant.
Problems 8–15: "find all positive integers $x,y$ with $xy-3x+2y=20$ ". Use Simon's Favorite Factoring Trick, then count or add up the solutions from the factor pairs.
Problems 10–16: given $x+\frac1x=k$ , find $x^n+\frac1{x^n}$ , such as $x^4+\frac1{x^4}$ .
Problems 12–18: a long product like $\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdots$ or $\left(1+\frac12\right)\left(1+\frac13\right)\cdots$ that telescopes after factoring.
Problems 15–22: three unknowns with $a+b+c$ , $ab+bc+ca$ and $abc$ given (or findable), asking for $a^3+b^3+c^3$ or a similar sum of powers.

Standard approaches

Before computing, ask whether the expression is a difference of squares, a perfect square, or a sum or difference of cubes in disguise.
If an equation has an $xy$ term plus $x$ and $y$ terms, rearrange it for Simon's Favorite Factoring Trick. List all factor pairs of the constant, including negative ones.
For $x+\frac1x$ , call it $t$ and climb one power at a time: $x^{n+1}+\frac1{x^{n+1}}=t\left(x^n+\frac1{x^n}\right)-\left(x^{n-1}+\frac1{x^{n-1}}\right)$ . Example: with $t=3$ , $x^4+\frac1{x^4}=3\cdot18-7=47$ . (Given $x-\frac1x$ instead? Square it and add $2$ to get $x^2+\frac1{x^2}$ .)
For long products or sums, factor every term completely and write out three or four terms in a row to see what cancels.
If a number expression looks designed, replace the big number by $n$ and simplify with algebra. Example: $n(n+1)(n+2)(n+3)+1=(n^2+3n+1)^2$ , so with $n=28$ , $\sqrt{28\cdot29\cdot30\cdot31+1}=28^2+3\cdot28+1=869$ .

Worked example

How many ordered pairs of integers $(x,y)$ satisfy $xy+4x-3y=30$ ?

(A) $6$ (B) $8$ (C) $10$ (D) $12$ (E) $18$

The first two terms share an $x$ , so write the equation as $x(y+4)-3y=30$ . To get a second $(y+4)$ we need $-3(y+4)=-3y-12$ , so subtract $12$ from both sides:
$x(y+4)-3(y+4)=18\quad\Longrightarrow\quad (x-3)(y+4)=18 .$
Now $x-3$ and $y+4$ are integers that multiply to $18$ . Each way to write $18=u\cdot v$ with integers $u$ , $v$ gives exactly one solution, $x=u+3$ and $y=v-4$ . For example, $u=2$ , $v=9$ gives $(5,5)$ , and indeed $25+20-15=30$ .

Count the pairs $(u,v)$ . Since $18=2\cdot3^2$ , it has $(1+1)(2+1)=6$ positive divisors: $1,2,3,6,9,18$ . Each one can be $u$ , with $v=\frac{18}u$ , so there are $6$ ordered pairs with both factors positive. Making both factors negative gives $6$ more, such as $(-1,-18)$ . That is $12$ pairs in all.

The answer is $\boxed{\textbf{(D)}\ 12}$ .

Pitfalls

Forgetting the negative factor pairs in Simon's Favorite Factoring Trick, which cuts the count in half. Or adding the extra constant to only one side.
Sign mistakes between $a^3-b^3$ and $a^3+b^3$ , or in $(a-b)^2=a^2-2ab+b^2$ . Example: $a^3+b^3$ has $-ab$ in its second factor, not $+ab$ .
Multiplying out a designed number expression by hand instead of replacing the big number with a letter.
In telescoping products, cancelling correctly but losing track of which terms survive at each end. Example: in $\frac21\cdot\frac32\cdot\frac43\cdots\frac{n+1}n$ , only the last top and the first bottom survive, leaving $n+1$ .

Traps that recur

Trying to reduce 2^202 by finding the order of 2 modulo the divisor, or forgetting to check that 201 is smaller than the divisor. (2020 AMC 10B #22)
Trying to solve the system directly or subtracting the equations instead of adding them; the sum is the one combination that collapses to a sum of squares. (2012 AMC 10A #24)
Trying to solve the two equations for a, b, c individually; the system has infinitely many solutions, and the target expression is the same for all of them. (2002 AMC 10B #20)
Using 13 as a + b + c (each edge length appears 4 times, so a + b + c = 13/4) or 11/2 as ab + bc + ca instead of 11/4. (2023 AMC 10B #17)

1.2024 AMC 10A #1
What is the value of $9901\cdot101-99\cdot10101?$
1. A) $2$
2. B) $20$
3. C) $200$
4. D) $202$
5. E) $2020$
2.2024 AMC 10B #2
What is $10! - 7! \cdot 6!$
$\textbf{(A) } -120 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 600 \qquad\textbf{(E) } 720$
[ONLY FOR CERTAIN CHINESE TESTPAPERS]
What is $10! - 7! \cdot 6! - 5!$
1. A) $-120$
2. B) $0$
3. C) $120$
4. D) $600$
5. E) $720$
3.2021 AMC Fall 10A #1
What is the value of $\frac{(2112-2021)^2}{169}$ ?
1. A) $7$
2. B) $21$
3. C) $49$
4. D) $64$
5. E) $91$
4.2021 AMC Fall 10B #3
The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$
1. A) $1$
2. B) $9$
3. C) $2020$
4. D) $2021$
5. E) $4041$
5.2020 AMC 10B #1
What is the value of $1 - (-2) - 3 - (-4) - 5 - (-6)?$
1. A) $-20$
2. B) $-3$
3. C) $3$
4. D) $5$
5. E) $21$
6.2020 AMC 10A #3
Assuming $a\neq3$ , $b\neq4$ , and $c\neq5$ , what is the value in simplest form of the following expression? $\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}$
1. A) ${-}1$
2. B) $1$
3. C) $\frac{abc}{60}$
4. D) $\frac{1}{abc} - \frac{1}{60}$
5. E) $\frac{1}{60} - \frac{1}{abc}$
7.2017 AMC 10B #4
Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$ ?
1. A) $-3$
2. B) $-1$
3. C) $1$
4. D) $2$
5. E) $3$
8.2017 AMC 10A #5
The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
1. A) $1$
2. B) $2$
3. C) $4$
4. D) $8$
5. E) $12$
9.2016 AMC 10A #1
What is the value of $\dfrac{11!-10!}{9!}$ ?
1. A) $99$
2. B) $100$
3. C) $110$
4. D) $121$
5. E) $132$
10.2013 AMC 10B #5
Positive integers $a$ and $b$ are each less than $6$ . What is the smallest possible value for $2 \cdot a - a \cdot b$ ?
1. A) $-20$
2. B) $-15$
3. C) $-10$
4. D) $0$
5. E) $2$
11.2010 AMC 10B #1
What is $100(100-3)-(100\cdot100-3)$ ?
1. A) $-20,000$
2. B) $-10,000$
3. C) $-297$
4. D) $-6$
5. E) $0$
12.2008 AMC 10A #5
Which of the following is equal to the product $\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdot\cdots\cdot\frac{4n+4}{4n}\cdot\cdots\cdot\frac{2008}{2004}?$
1. A) $251$
2. B) $502$
3. C) $1004$
4. D) $2008$
5. E) $4016$
13.2003 AMC 10B #1
Which of the following is the same as
$\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?$
1. A) $-1$
2. B) $-\frac{2}{3}$
3. C) $\frac{2}{3}$
4. D) $1$
5. E) $\frac{14}{3}$
14.2002 AMC 10B #4
What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$ ?
1. A) $0$
2. B) $1$
3. C) $10$
4. D) $11$
5. E) $12$
15.2024 AMC 10B #9
Real numbers $a, b,$ and $c$ have arithmetic mean $0$ . The arithmetic mean of $a^2, b^2,$ and $c^2$ is $10$ . What is the arithmetic mean of $ab, ac,$ and $bc$ ?
1. A) $-5$
2. B) $-\dfrac{10}{3}$
3. C) $-\dfrac{10}{9}$
4. D) $0$
5. E) $\dfrac{10}{9}$
16.2022 AMC 10B #5
What is the value of $\frac{\left(1+\frac13\right)\left(1+\frac15\right)\left(1+\frac17\right)}{\sqrt{\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right)\left(1-\frac{1}{7^2}\right)}}?$
1. A) $\sqrt3$
2. B) $2$
3. C) $\sqrt{15}$
4. D) $4$
5. E) $\sqrt{105}$
17.2021 AMC 10A #9
What is the least possible value of $(xy-1)^2 + (x+y)^2$ for real numbers $x$ and $y$ ?
1. A) $0$
2. B) $\frac{1}{4}$
3. C) $\frac{1}{2}$
4. D) $1$
5. E) $2$
18.2021 AMC 10A #10
Which of the following is equivalent to $(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?$
1. A) $3^{127} + 2^{127}$
2. B) $3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}$
3. C) $3^{128}-2^{128}$
4. D) $3^{128} + 2^{128}$
5. E) $5^{127}$
19.2019 AMC 10B #6
There is a real $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$ ?
1. A) $3$
2. B) $8$
3. C) $10$
4. D) $11$
5. E) $12$
20.2018 AMC 10A #10
Suppose that real number $x$ satisfies $\sqrt{49-x^2}-\sqrt{25-x^2}=3.$ What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$ ?
1. A) $8$
2. B) $\sqrt{33}+8$
3. C) $9$
4. D) $2\sqrt{10}+4$
5. E) $12$
21.2015 AMC 10A #12
Points $(\sqrt{\pi}, a)$ and $(\sqrt{\pi}, b)$ are distinct points on the graph of $y^2+x^4=2x^2y+1$ . What is $|a-b|$ ?
1. A) $1$
2. B) $\dfrac{\pi}{2}$
3. C) $2$
4. D) $\sqrt{1+\pi}$
5. E) $1+\sqrt{\pi}$
22.2014 AMC 10B #9
For real numbers $w$ and $z$ , $\frac{\frac{1}{w} + \frac{1}{z}}{\frac{1}{w} - \frac{1}{z}} = 2014.$ What is $\frac{w+z}{w-z}$ ?
1. A) $-2014$
2. B) $\frac{-1}{2014}$
3. C) $\frac{1}{2014}$
4. D) $1$
5. E) $2014$
23.2013 AMC 10B #14
Define $a\clubsuit b=a^2b-ab^2$ . Which of the following describes the set of points $(x, y)$ for which $x\clubsuit y=y\clubsuit x$ ?
1. A) $\text{a finite set of points}$
2. B) $\text{one line}$
3. C) $\text{two parallel lines}$
4. D) $\text{two intersecting lines}$
5. E) $\text{three lines}$
24.2011 AMC 10B #14
A rectangular parking lot has a diagonal of $25$ meters and an area of $168$ square meters. In meters, what is the perimeter of the parking lot?
1. A) $52$
2. B) $58$
3. C) $62$
4. D) $68$
5. E) $70$
25.2010 AMC 10B #9
Lucky Larry's teacher asked him to substitute numbers for $a$ , $b$ , $c$ , $d$ , and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for $a$ , $b$ , $c$ , and $d$ were $1$ , $2$ , $3$ , and $4$ , respectively. What number did Larry substitute for $e$ ?
1. A) $-5$
2. B) $-3$
3. C) $0$
4. D) $3$
5. E) $5$
26.2009 AMC 10A #11
One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?
1. A) $8$
2. B) $27$
3. C) $64$
4. D) $125$
5. E) $216$
27.2002 AMC 10B #11
The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?
1. A) $50$
2. B) $77$
3. C) $110$
4. D) $149$
5. E) $194$
28.2002 AMC 10B #13
Find the value(s) of $x$ such that $8xy - 12y + 2x - 3 = 0$ is true for all values of $y$ .
1. A) $\frac23$
2. B) $\frac32 \text{ or } -\frac14$
3. C) $-\frac23 \text{ or } -\frac14$
4. D) $\frac32$
5. E) $-\frac32 \text{ or } -\frac14$
29.2023 AMC 10B #17
A rectangular box $P$ has distinct edge lengths $a$ , $b$ , and $c$ . The sum of the lengths of all $12$ edges of $P$ is $13$ , the sum of the areas of all $6$ faces of $P$ is $\dfrac{11}{2}$ , and the volume of $P$ is $\dfrac{1}{2}$ . What is the length of the longest interior diagonal connecting two vertices of $P$ ?
1. A) $2$
2. B) $\frac{3}{8}$
3. C) $\frac{9}{8}$
4. D) $\frac{9}{4}$
5. E) $\frac{3}{2}$
30.2021 AMC 10B #15
The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$ . What is the value of $x^{11}-7x^{7}+x^3?$
1. A) $-1$
2. B) $0$
3. C) $1$
4. D) $2$
5. E) $\sqrt{5}$
31.2021 AMC Fall 10B #12
Which of the following conditions is sufficient to guarantee that integers $x$ , $y$ , and $z$ satisfy the equation $x(x-y)+y(y-z)+z(z-x) = 1?$
1. A) $x>y$ and $y=z$
2. B) $x=y-1$ and $y=z-1$
3. C) $x=z+1$ and $y=x+1$
4. D) $x=z$ and $y-1=x$
5. E) $x+y+z=1$
32.2020 AMC 10A #14
Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$ . What is the value of $x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?$
1. A) $360$
2. B) $400$
3. C) $420$
4. D) $440$
5. E) $480$
33.2019 AMC 10A #19
What is the least possible value of $(x+1)(x+2)(x+3)(x+4)+2019$ where $x$ is a real number?
1. A) $2017$
2. B) $2018$
3. C) $2019$
4. D) $2020$
5. E) $2021$
34.2016 AMC 10B #17
All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?
1. A) $312$
2. B) $343$
3. C) $625$
4. D) $729$
5. E) $1680$
35.2015 AMC 10A #20
A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$ . Which of the following numbers cannot equal $A+P$ ?
1. A) $100$
2. B) $102$
3. C) $104$
4. D) $106$
5. E) $108$
36.2012 AMC 10A #17
Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$ ?
1. A) $1$
2. B) $2$
3. C) $3$
4. D) $4$
5. E) $5$
37.2007 AMC 10A #20
Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$ ?
1. A) $164$
2. B) $172$
3. C) $192$
4. D) $194$
5. E) $212$
38.2000 AMC 10 #15
Two non-zero real numbers, $a$ and $b,$ satisfy $ab = a - b$ . Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$ ?
1. A) $- 2$
2. B) $\frac { -1 }{2}$
3. C) $\frac {1}{3}$
4. D) $\frac {1}{2}$
5. E) $2$
39.2000 AMC 10 #20
Let $A$ , $M$ , and $C$ be nonnegative integers such that $A+M+C=10$ . What is the maximum value of $A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$ ?
1. A) $49$
2. B) $59$
3. C) $69$
4. D) $79$
5. E) $89$
40.2020 AMC 10B #22
What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$ ?
1. A) $100$
2. B) $101$
3. C) $200$
4. D) $201$
5. E) $202$
41.2012 AMC 10A #24
Let $a$ , $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $\begin{align*}a^2-b^2-c^2+ab&=2011\text{ and}\\ a^2+3b^2+3c^2-3ab-2ac-2bc&=-1997.\end{align*}$
What is $a$ ?
1. A) $249$
2. B) $250$
3. C) $251$
4. D) $252$
5. E) $253$
42.2002 AMC 10B #20
Let $a$ , $b$ , and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$ . Then $a^2-b^2+c^2$ is
1. A) $0$
2. B) $1$
3. C) $4$
4. D) $7$
5. E) $8$

Algebraic Manipulation — answersAMC 10 Step by Step

Name: Date: October 28, 2026

Keep this page separate from the problems.

Answers

#	Answer	The idea it turns on
1	A	Multiplying by 101 means times 100 plus one more copy, and multiplying by 99 means times 100 minus one copy. Both products then come out in one line.
2	B	10 times 9 times 8 is 720, which is also 6!. So 10! and 7! times 6! are the same number, and their difference is zero.
3	C	91 is 7 times 13, and 169 is 13 times 13. So the fraction is 7 squared over 1, which is 49. Simplify before squaring.
4	E	Over a common denominator the top becomes 2021 squared minus 2020 squared, which factors as (1)(4041). Nothing cancels, so p is 4041.
5	D	Taking away a negative number is the same as adding it, so the expression becomes 1 + 2 - 3 + 4 - 5 + 6.
6	A	Each top is a bottom with its sign flipped, like a - 3 and 3 - a. Matched up, each fraction is -1, so the product is -1.
7	D	Clear the fraction in the given equation: it simplifies to x = y. Then the fraction asked for is 4x over 2x, which is 2.
8	C	Adding the two reciprocals gives the sum of the numbers divided by their product. The problem says that sum is 4 times the product, so the answer is 4.
9	B	11! is just 11 times 10!, so the top is 10 copies of 10!. Then 10! divided by 9! is 10, giving 10 times 10.
10	B	Write 2a - ab as a(2 - b). Use b = 5 to make (2 - b) as negative as possible, then the largest a, which is 5.
11	C	Multiply out 100(100 - 3) to get 10000 - 300. The 10000 shows up in both parts and cancels, leaving -300 + 3.
12	B	The top of each fraction equals the bottom of the next one, so almost everything cancels, leaving the last top over the first bottom: 2008 over 4.
13	C	Each number on top is 2 times a counting number and the number right under it is 3 times the same counting number, so the fraction is 2/3.
14	D	The first two parts both contain (3x - 2). Pull it out and the rest cancels, leaving 3x - 1, which is 11 when x = 4.
15	A	Squaring the sum of the numbers gives their squares plus twice their pairwise products. So 0 = 30 plus twice the products, and the products add to -15.
16	B	Each 1 − 1/k² splits into (1 − 1/k)(1 + 1/k). Then the top of the fraction cancels with part of the square root, leaving a small product.
17	D	Multiply out both squares. The middle terms cancel, leaving three squares plus 1, so the value is never below 1, and x = y = 0 gives exactly 1.
18	C	Multiply by (3 - 2), which equals 1. Each factor then combines with the one before by difference of squares, and the whole product collapses to 3^128 - 2^128.
19	C	Both factorials on the left contain n! as a factor. Divide it out to get (n+1)(n+3) = 440, which is 20 times 22.
20	A	The two numbers under the roots differ by 24. Big minus small, times big plus small, equals that difference, so 3 times the sum is 24.
21	C	The equation rearranges to (y minus x squared) squared equals 1. So y is x squared plus 1 or minus 1, and the two heights always differ by 2.
22	A	Multiply top and bottom by wz: the given fraction becomes (z + w)/(z - w), which is the asked fraction with its sign flipped.
23	E	The equation becomes 2xy(x - y) = 0, which is true when x = 0, when y = 0, or when x = y: three lines.
24	C	The diagonal gives l squared plus w squared = 625, and the area gives lw = 168. So (l + w) squared = 625 + 336 = 961.
25	D	Plug in the known numbers first. The correct value works out to -2 - e and Larry's value is e - 8; setting them equal gives e = 3.
26	D	The new box has volume s(s+1)(s-1), which is s cubed minus s, so it is always s less than the cube. A drop of 5 means s = 5.
27	B	Call the middle number n. The sum is 3n and the product is (n-1)n(n+1), so the condition becomes (n-1)(n+1) = 24, giving n = 5.
28	D	Write it as (8x-12)y + (2x-3). For it to be zero for every y, the part multiplying y and the leftover part must both be zero.
29	D	The diagonal squared is a squared plus b squared plus c squared. That equals (a+b+c) squared minus 2(ab+bc+ca), and the edge total and face total give both parts.
30	B	Square twice to get x⁴ + 1/x⁴ = 7. Multiply that by x⁷ and you get x¹¹ + x³ = 7x⁷, so the expression is 0.
31	D	Doubled, the left side is the sum of the squared differences of the three numbers. That equals 2 only when two numbers are equal and the third is one away.
32	D	Combine the fractions over (xy) squared, which is 4. Then build the sum of fifth powers from the sum 4 and product -2, never finding x or y.
33	B	Pair outside and inside factors: both give x squared plus 5x, plus a number. Call that u; the product is (u + 4)(u + 6), never below minus 1.
34	D	Each corner uses one face from each opposite pair, so the eight products add to (a+b)(c+d)(e+f). The pair sums total 27, so the best is 9 x 9 x 9.
35	B	Area plus perimeter plus 4 always factors as (width + 2) times (length + 2), both at least 3. But 106 is only 2 times 53.
36	C	Cancel a - b and call it d, giving 9ab = 70d^2. Since a and b share no factor, d shares none with ab, so d divides 3.
37	D	Square a + 1/a. Since a times 1/a is 1, the middle term is just 2. Square twice: 16 - 2 = 14, then 196 - 2 = 194.
38	E	Write a/b + b/a as one fraction, (a squared + b squared)/(ab). Then use the given ab = a - b to rewrite the top, and everything cancels to 2.
39	C	Adding A + M + C + 1 = 11 turns the expression into (A+1)(M+1)(C+1). Three numbers adding to 13 have the biggest product when nearly equal: 4, 4, 5.
40	D	Multiply the divisor by 2^101 - 2^51 + 1 and you get exactly 2^202 + 1. So 2^202 + 202 is a multiple of the divisor plus 201.
41	E	Add the equations: the sum is (a-b)^2 + (b-c)^2 + (a-c)^2 = 14, so the gaps a - b and b - c are 1 and 2.
42	B	Move b to the right, square both equations and add. The mixed terms cancel, leaving 65(a^2 + c^2) = 65(1 + b^2).

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