---
title: "Algebra and Equations"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 66
exercises: 10
source: https://one-course.com/books/math/1/en/chapter/66-algebra-and-equations
---

# Chapter 66 — Algebra and Equations

Letters stand for numbers, and computing with letters proves facts about *all* numbers at once. This chapter practices [expanding](#def-g9-algebra-expand) and factoring — including the three identities that will follow you through all of mathematics — and solves first-degree [equations](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def), [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) [equations](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def), and simple inequalities, always one justified step at a time.

## 66.1 Expanding

**Definition 66.1 (Expanding).**

*Expanding* means transforming a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) into a [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def), using distributivity:

$$
k(a + b) = ka + kb,
\qquad
(a + b)(c + d) = ac + ad + bc + bd .
$$

**Example 66.2.**

Every term of the first bracket multiplies every term of the second:

$$
\begin{align*}
(x + 3)(2x + 5)
&= x \times 2x + x \times 5 + 3 \times 2x + 3 \times 5 \\
&= 2x^2 + 5x + 6x + 15 \\
&= 2x^2 + 11x + 15 .
\end{align*}
$$

Signs travel with the terms:

$$
\begin{align*}
(3x - 2)(x - 4)
&= 3x^2 - 12x - 2x + 8
= 3x^2 - 14x + 8 .
\end{align*}
$$

**Theorem 66.3 (The three identities).**

For all numbers $a$ and $b$:

$$
(a+b)^2 = a^2 + 2ab + b^2,
\qquad
(a-b)^2 = a^2 - 2ab + b^2,
\qquad
(a+b)(a-b) = a^2 - b^2 .
$$

**Proof.** Expand each left-hand side. For the first: $(a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$. For the second, replace $b$ by $-b$ in the first. For the third: $(a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2$ (the cross terms cancel). ∎

![The identity (a+b)2 = a2 + 2ab + b2, seen as areas: the big square of side a+b is made of two squares and two equal rectangles. Forgetting the “2ab” is forgetting the two rectangles!](https://one-course.com/images/onecourse/chapters/math-1/g9-algebra/fig-3f1377230717.svg)

*The identity $(a+b)^2 = a^2 + 2ab + b^2$, seen as [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area): the big square of side $a+b$ is made of two squares and two equal rectangles. Forgetting the “$2ab$” is forgetting the two rectangles!*

**Example 66.4.**

$$
(x + 5)^2 = x^2 + 10x + 25, \qquad
(3x - 1)^2 = 9x^2 - 6x + 1, \qquad
(2x + 7)(2x - 7) = 4x^2 - 49 .
$$

Mental arithmetic: $102^2 = (100 + 2)^2 = 10000 + 400 + 4 = 10404$, and $98 \times 102 = (100-2)(100+2) = 10000 - 4 = 9996$.

## 66.2 Factoring

**Method 66.5 (Factoring).**

To factor an expression, try in order:

1. *common factor* : $ka + kb = k(a + b)$ , where $k$ may itself be a bracket;
2. *[difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of squares* : $a^2 - b^2 = (a+b)(a-b)$ ;
3. *perfect square* : $a^2 + 2ab + b^2 = (a+b)^2$ (and with $-2ab$ , $(a-b)^2$ ).

Always check by [expanding](#def-g9-algebra-expand) the result.

**Example 66.6.**

*Common factor:* $15x^2 + 10x = 5x \times 3x + 5x \times 2 = 5x(3x + 2)$.

*Common bracket:*

$$
(x+2)(x-1) + (x+2)(3x+4) = (x+2)\bigl[(x-1) + (3x+4)\bigr]
= (x+2)(4x+3).
$$

*[Difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of squares:* $x^2 - 81 = (x+9)(x-9)$, and

$$
(2x+1)^2 - 9 = (2x+1)^2 - 3^2 = (2x+1+3)(2x+1-3) = (2x+4)(2x-2).
$$

Each factor still has a common factor: $(2x+4)(2x-2) = 2(x+2) \times
2(x-1) = 4(x+2)(x-1)$.

## 66.3 Equations

**Method 66.7 (First-degree equations).**

To solve an [equation](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) like $5x - 3 = 2x + 9$:

1. gather the $x$ -terms on one side, the numbers on the other, by adding the same quantity to both sides;
2. reduce each side;
3. divide both sides by the coefficient of $x$ ;
4. check the solution in the original [equation](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) .

**Example 66.8.**

$$
\begin{align*}
5x - 3 &= 2x + 9 \\
5x - 2x &= 9 + 3 && \text{(add $3 - 2x$ to both sides)}\\
3x &= 12 \\
x &= 4 && \text{(divide by 3).}
\end{align*}
$$

Check: $5 \times 4 - 3 = 17$ and $2 \times 4 + 9 = 17$. The solution is $4$.

With brackets, expand first: $3(x - 2) = x + 8$ becomes $3x - 6 = x + 8$, so $2x = 14$ and $x = 7$.

**Theorem 66.9 (Zero-product rule).**

A [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) is [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) if and only if at least one factor is [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero):

$$
A \times B = 0
\quad\Longleftrightarrow\quad
A = 0 \ \text{ or } \ B = 0 .
$$

**Proof.** If a factor is [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) clearly is. Conversely if $A B = 0$ with $A \neq 0$, dividing both sides by $A$ gives $B = 0$. ∎

**Example 66.10.**

Solve $(x - 3)(2x + 10) = 0$: either $x - 3 = 0$ or $2x + 10 = 0$, so the solutions are $3$ and $-5$.

Solve $x^2 - 49 = 0$: factor first, $(x-7)(x+7) = 0$, solutions $7$ and $-7$.

Solve $x^2 = 3x$: never divide by $x$! Move and factor: $x^2 - 3x = x(x - 3) = 0$, solutions $0$ and $3$.

## 66.4 Inequalities

**Proposition 66.11 (Rules for inequalities).**

An inequality is preserved when adding the same number to both sides, and when multiplying both sides by the same *positive* number. Multiplying both sides by a *negative* number reverses the inequality sign.

**Proof.** If $a \leq b$ then $b - a \geq 0$. Adding $c$: $(b+c) - (a+c) = b - a
\geq 0$, so $a + c \leq b + c$. Multiplying by $c > 0$: $bc - ac = (b-a)c \geq 0$, so $ac \leq bc$. Multiplying by $c < 0$: $(b-a)c \leq 0$, so $ac \geq bc$: the order is reversed. ∎

**Example 66.12.**

Solve $7 - 2x < 15$:

$$
\begin{align*}
7 - 2x &< 15 \\
-2x &< 8 && \text{(subtract 7)} \\
x &> -4 && \text{(divide by $-2$: reverse the sign).}
\end{align*}
$$

The solutions are all numbers greater than $-4$: on a number line, a hollow dot at $-4$ and shading to the right.

![The solutions of 7 - 2x < 15: all x > -4 (hollow dot: -4 itself is not a solution).](https://one-course.com/images/onecourse/chapters/math-1/g9-algebra/fig-754390be3efb.svg)

*The solutions of $7 - 2x < 15$: all $x > -4$ (hollow dot: $-4$ itself is not a solution).*

## 66.5 Exercises

**Exercise 66.1 ★.**

Expand and reduce:

$$
4(3x - 2), \qquad
(x + 6)(x + 2), \qquad
(2x - 3)(x + 5), \qquad
(x - 1)(x - 9).
$$

**Solution of Exercise 66.1.**

$4(3x - 2) = 12x - 8$.

$(x+6)(x+2) = x^2 + 2x + 6x + 12 = x^2 + 8x + 12$.

$(2x-3)(x+5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15$.

$(x-1)(x-9) = x^2 - 9x - x + 9 = x^2 - 10x + 9$.

**Exercise 66.2 ★.**

Expand using the identities:

$$
(x + 8)^2, \qquad
(3x - 4)^2, \qquad
(x + 10)(x - 10), \qquad
(5x + 2)(5x - 2).
$$

**Solution of Exercise 66.2.**

$(x+8)^2 = x^2 + 16x + 64$.

$(3x-4)^2 = 9x^2 - 24x + 16$.

$(x+10)(x-10) = x^2 - 100$.

$(5x+2)(5x-2) = 25x^2 - 4$.

**Exercise 66.3 ★.**

Compute mentally, using an identity: $101^2$; $59^2$ (write $59 = 60 - 1$); $53 \times 47$.

**Solution of Exercise 66.3.**

$101^2 = (100+1)^2 = 10000 + 200 + 1 = 10201$.

$59^2 = (60-1)^2 = 3600 - 120 + 1 = 3481$.

$53 \times 47 = (50+3)(50-3) = 2500 - 9 = 2491$.

**Exercise 66.4 ★.**

Factor:

$$
9x + 12, \qquad
x^2 - 64, \qquad
x^2 + 14x + 49, \qquad
18x^2 - 6x .
$$

**Solution of Exercise 66.4.**

$9x + 12 = 3(3x + 4)$.

$x^2 - 64 = (x+8)(x-8)$.

$x^2 + 14x + 49 = (x+7)^2$.

$18x^2 - 6x = 6x(3x - 1)$.

**Exercise 66.5 ★.**

Solve, writing every step:

$$
4x + 7 = 19, \qquad
6x - 5 = 2x + 11, \qquad
5(x - 3) = 3x + 1 .
$$

**Solution of Exercise 66.5.**

$4x + 7 = 19$: $4x = 12$, so $x = 3$.

$6x - 5 = 2x + 11$: $4x = 16$, so $x = 4$.

$5(x-3) = 3x + 1$: $5x - 15 = 3x + 1$, so $2x = 16$ and $x = 8$.

**Exercise 66.6 ★★.**

Solve using the [zero-product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) rule:

$$
(x + 4)(3x - 12) = 0, \qquad
x^2 - 121 = 0, \qquad
(2x - 1)^2 = 0, \qquad
x^2 = 8x .
$$

**Solution of Exercise 66.6.**

$(x+4)(3x-12) = 0$: $x = -4$ or $x = 4$.

$x^2 - 121 = (x-11)(x+11) = 0$: $x = 11$ or $x = -11$.

$(2x-1)^2 = 0$: the single solution $x = \frac12$.

$x^2 = 8x$: $x(x - 8) = 0$, so $x = 0$ or $x = 8$.

**Exercise 66.7 ★★.**

Factor $E = (3x + 2)^2 - 25$, then solve $E = 0$.

**Solution of Exercise 66.7.**

[Difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of squares with $a = 3x + 2$ and $b = 5$:

$$
E = (3x + 2)^2 - 5^2 = (3x + 2 + 5)(3x + 2 - 5) = (3x + 7)(3x - 3).
$$

[Zero-product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) rule: $3x + 7 = 0$ gives $x = -\frac73$, and $3x - 3 = 0$ gives $x = 1$.

**Exercise 66.8 ★★.**

Solve the inequalities and represent the solutions on a number line:

$$
3x - 4 \geq 8, \qquad
5 - 2x > 1, \qquad
4(x + 1) \leq x - 5 .
$$

**Solution of Exercise 66.8.**

$3x - 4 \geq 8$: $3x \geq 12$, so $x \geq 4$ (filled dot at $4$, shade right).

$5 - 2x > 1$: $-2x > -4$, dividing by $-2$ reverses the sign: $x < 2$ (hollow dot at $2$, shade left).

$4(x+1) \leq x - 5$: $4x + 4 \leq x - 5$, so $3x \leq -9$, so $x \leq -3$ (filled dot at $-3$, shade left).

**Exercise 66.9 ★★.**

I think of a number, multiply it by $3$, add $7$, and get the same result as if I had multiplied it by $5$ and subtracted $9$. What is my number? (Set up an [equation](https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations#def-g8-equations-def) and solve it.)

**Solution of Exercise 66.9.**

Let $x$ be the number: $3x + 7 = 5x - 9$. Then $16 = 2x$, so $x = 8$. Check: $3 \times 8 + 7 = 31$ and $5 \times 8 - 9 = 31$.

**Exercise 66.10 ★★★.**

Let $n$ be a whole number.

1. Expand $(n + 1)(n - 1)$ and deduce that the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of the two neighbors of $n$ is always $n^2 - 1$ (e.g. $24 \times 26 =  25^2 - 1 = 624$ ).
2. Show that the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of three consecutive integers is always a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $3$ .

**Solution of Exercise 66.10.**

*1.* $(n+1)(n-1) = n^2 - 1$ (third identity). The neighbors of $n$ are $n - 1$ and $n + 1$, so their [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) is $n^2 - 1$: e.g. $24 \times 26 = 625 - 1 = 624$.

*2.* Three consecutive integers can be written $n - 1$, $n$, $n + 1$. Their [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is

$$
(n-1) + n + (n+1) = 3n ,
$$

a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $3$ (three times the middle one).

## 66.6 Problem: The algebra of digits

**Problem 66.1.**

Weekend problem — why the divisibility rules work, the 1089 trick unmasked, and mental arithmetic that looks like sorcery

Grade school taught you *rules* (a number is [divisible](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $9$ when its digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) is); magicians perform *tricks* (everyone ends on $1089$); markets know *shortcuts* ($35^2 = 1225$, instantly). All of it is the same secret: a number with digits $a$, $b$, $c$ is the expression $100a + 10b + c$, and this chapter’s algebra ([Theorem 66.3](#thm-g9-algebra-identities), [expanding](#def-g9-algebra-expand), factoring) can take it apart. By the end of this problem you will have proved the rules, unmasked the trick, and learned the sorcery.

**Part I — Digits under the microscope.**

1. A two-digit number with tens digit $a$ and units digit $b$ is $10a + b$ . Expand and factor the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) between the number and its reversal, $(10a + b) - (10b + a)$ , and state the discovery. Check it on $72 - 27$ .
2. Prove the rule of nine for three-digit numbers: verify the identity $$100a + 10b + c = 9\,(11a + b) + (a + b + c),$$ and explain why it shows that a number and its digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) leave the *same [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder)* in the division by $9$ — so one is [divisible](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $9$ exactly when the other is ([Proposition 64.3](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#prop-g9-arith-rules), now proved).
3. Deduce the rule of three from the same identity, in one sentence.
4. “Casting out nines”, the accountants’ check: to test $47 \times 86 = 4\,042$ , replace each number by the digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of its digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) : $47 \to 11 \to 2$ , $86 \to 14 \to 5$ , and $2 \times 5 = 10 \to 1$ ; then $4\,042 \to 10 \to 1$ : consistent. Explain why a correct [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) must pass the test (write $47 = 9k + 2$ , $86 = 9l + 5$ and expand), and find a wrong result that the test fails to catch (why are swapped digits invisible to it?).
5. Prove the rule of eleven for four-digit numbers: verify $$1000a + 100b + 10c + d  = 11\,(91a + 9b + c) + (- a + b - c + d),$$ and use it to decide whether $2\,926$ is [divisible](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $11$.

**Part II — The 1089 trick.** The trick: think of a three-digit number whose first and last digits differ by at least $2$; reverse it; subtract the smaller from the larger; reverse the result; add the last two numbers. The magician announces: *1089*.

6. Perform the trick on $742$ and on one number of your choice.
7. Let the number have digits $a$ , $b$ , $c$ with $a > c$ . Factor the first subtraction, $(100a + 10b + c) - (100c + 10b + a)$ , and list all the values it can take when $a - c$ runs from $2$ to $9$ .
8. Each of those [multiples](https://one-course.com/books/math/1/en/chapter/32-division-and-multiples#def-g5-division-multiple) of $99$ has a striking digit pattern. Writing $m = a - c$, verify the identity $$99m = 100\,(m - 1) + 90 + (10 - m),$$ and read off the three digits of $99m$.
9. Add $99m$ to its own reversal, using question 8’s digits, and prove that the total is $1089$ — whatever $m$ is. The trick is dead; long live the algebra.
10. Why did the magician require the first and last digits to differ by at least $2$ ? Examine the cases $a = c$ and $a - c = 1$ (where the subtraction gives $99$ , which must be written $099$ for the reversal to work), and state the small print that keeps the trick honest.

**Part III — Sorcery for daily use.**

11. The birthday trick: “multiply your birth month by $5$ , add $7$ , double, add your birth day, subtract $14$ .” Show that the result is always $10m + d$ — month and day, readable at a glance. (Try it on someone.)
12. You redesign the trick as: “multiply the month by $5$ , add $9$ , double, add the day.” What number must the spectator now subtract at the end for the same month–day readout?
13. The missing-digit trick: the spectator picks any number, subtracts its digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) (question 2 says the result is always a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$ — explain why for a three-digit number), then crosses out one *nonzero* digit and reads you the remaining digits. Explain how you recover the crossed-out digit, and why $0$ had to be excluded.
14. Market sorcery: prove the identity $(10t + 5)^2 = 100\,t(t+1) + 25$ , state the rule for squaring any number ending in $5$ , and compute $45^2$ , $85^2$ and $105^2$ in your head.
15. Finale, extending [Exercise 66.10](#exo-g9-algebra-10) : prove that the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of five consecutive integers is always a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $5$ , but that the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of *four* consecutive integers is *never* a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $4$ . (Two short computations — and one moral about what algebra does to the words “always” and “never”.)

**Solution of Problem 66.1.**

**1.** $(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$: the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) between a two-digit number and its reversal is always a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$. Check: $72 - 27 = 45 = 9 \times 5$, and $a - b = 7 - 2 = 5$.

**2.** [Expanding](#def-g9-algebra-expand) the right side: $9(11a + b) + a + b + c = 99a + 9b + a + b + c
= 100a + 10b + c$. So the number equals a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$ plus its digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def): dividing by $9$, both leave the same [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder). In particular [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $0$ for one means [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $0$ for the other: the rule of nine, proved.

**3.** $9(11a + b)$ is also a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $3$, so the same identity shows number and digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) leave the same [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) in the division by $3$: the rule of three.

**4.** Write $47 = 9k + 2$ and $86 = 9l + 5$. Then

$$
47 \times 86 = (9k + 2)(9l + 5)
= 9\,(9kl + 5k + 2l) + 10 :
$$

the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) leaves the same [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) as $2 \times 5 = 10$, namely $1$. A correct result must therefore also leave [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $1$ — which $4\,042$ does. But the test only sees [remainders](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder): $4\,042$ and $4\,402$ (digits swapped) pass identically, so passing proves nothing — failing is what convicts.

**5.** [Expanding](#def-g9-algebra-expand): $11(91a + 9b + c) - a + b - c + d
= 1001a + 99b + 11c - a + b - c + d
= 1000a + 100b + 10c + d$. So the number is [divisible](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $11$ exactly when its alternating [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $-a + b - c + d$ is. For $2\,926$: $-2 + 9 - 2 + 6 = 11$, [divisible](https://one-course.com/books/math/1/en/chapter/37-whole-numbers#def-g6-wholes-divisible) by $11$: yes, $2\,926 = 11 \times 266$.

**6.** $742 - 247 = 495$; $495 + 594 = 1\,089$. Any valid example lands on $1\,089$ too.

**7.** $(100a + 10b + c) - (100c + 10b + a) = 99(a - c)$. For $a - c = 2, 3, \dots, 9$:

$$
198,\ 297,\ 396,\ 495,\ 594,\ 693,\ 792,\ 891 .
$$

**8.** $100(m - 1) + 90 + (10 - m) = 100m - 100 + 90 + 10
- m = 99m$. So the three digits of $99m$ are $m - 1$, then $9$, then $10 - m$ (check $495$: $m = 5$ gives $4$, $9$, $5$).

**9.** The reversal of $99m$ has digits $10 - m$, $9$, $m - 1$, so it equals $100(10 - m) + 90 + (m - 1)$. Adding:

$$
100\,(m - 1 + 10 - m) + 180 + (10 - m + m - 1)
= 100 \times 9 + 180 + 9 = 1\,089 ,
$$

independent of $m$. Whatever the starting number, the magician is safe.

**10.** If $a = c$, the subtraction gives $0$ and the trick dies at once. If $a - c = 1$, it gives $99$, which must be treated as the three-digit string $099$ (reversal $990$, $99 + 990 = 1\,089$: saved!). The small print: [differences](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of $1$ work only if the spectator writes the subtraction result with three digits, leading [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) included — requiring a gap of at least $2$ avoids the argument.

**11.** $2(5m + 7) + d - 14 = 10m + 14 + d - 14 =
10m + d$: the tens (and hundreds, for October–December) show the month, the units (and tens) the day.

**12.** $2(5m + 9) + d = 10m + 18 + d$: subtract $18$.

**13.** By question 2, $n - (\text{digit sum of } n) =
9(11a + b)$ for a three-digit $n$: always a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$ (similar identities work for any length). A [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$ has digit [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$ (question 2 again), so the digits the spectator reads out must be completed to the next [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$: the missing digit is that complement. Ambiguity: if the read digits already sum to a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $9$, the crossed digit could be $0$ or $9$ — excluding $0$ removes the doubt.

**14.** $(10t + 5)^2 = 100t^2 + 100t + 25 =
100\,t(t + 1) + 25$: write $t(t+1)$, then glue $25$ behind it. $45^2$: $4 \times 5 = 20$, so $2\,025$. $85^2$: $8 \times 9 = 72$, so $7\,225$. $105^2$: $10 \times 11 = 110$, so $11\,025$.

**15.** Five consecutive: $n + (n+1) + (n+2) + (n+3) +
(n+4) = 5n + 10 = 5(n + 2)$: a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor) of $5$, always. Four consecutive: $n + (n+1) + (n+2) + (n+3) = 4n + 6 =
4(n + 1) + 2$: [remainder](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-remainder) $2$ in the division by $4$, never a [multiple](https://one-course.com/books/math/1/en/chapter/64-arithmetic-divisors-and-prime-numbers#def-g9-arith-divisor). Algebra converts “it seems to work” into *always*, and “I found no example” into *never* — two words no amount of testing can reach ([Method 48.8](https://one-course.com/books/math/1/en/chapter/48-literal-expressions#met-g7-literal-test)).