1.8: Chapter 1- Comments and Solutions

Last updated
Save as PDF

Page ID: 24417

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

(a) Assuming that the $2 \times$, $3 \times$, $4 \times$, and $5 \times$ tables are known, and that one has understood that the order of the factors does not matter, all that remains to be learned is $6 \times 6$, $6 \times 7$, $6 \times 8$, $6 \times 9$; $7 \times 7$, $7 \times 8$, $7 \times 9$; $8 \times 8$, $8 \times 9$; and $9 \times 9$.

(b) (i) $(4 \times 10^{-3}) \times (2 \times 10^{-2}) = 8 \times 10^{-5} = 0.00008$

(ii) $(8 \times 10^{-4}) \times (7 \times 10^{-2}) = 56 \times 10^{-6} = 0.000056$

(iii) $(7 \times 10^{-3}) \times (12 \times 10^{-2}) = 84 \times 10^{-5} = 0.00084$

(iv) $(7 \times 10^{-3}) \times (12 \times 10^{-2}) = 84 \times 10^{-5} = 0.00084$

(v) $(8 \times 10^{-2})^{2} = 64 \times 10^{-4} = 0.0064$

(a) (i) $1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144; 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961$

(ii) $1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331$

(iii) $1 = 2^{0}, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024$

(b) (i) $31 (31^{2} = 96^{1}, 32^{2} = 21^{0} = 1024)$

(ii) $99 (100^{2} = 10^{4} = 10000)$

(iii) $316 (310^{2} = 96100 \lt 100000 \lt 320^{2}$; so look more carefully between $310$ and $320$)

(ii) $21 (20^{3} = 8000, 22^{3} = 10648)$

(iii) $99 (100^{3} = 10^{6} = 1000000)$

(d) (i) Those powers $2^{e}$ of the form $2^{2n}$ for which the exponent $e$ is a multiple of $2$: i.e. $e \equiv 0$ (mod $2$).

(ii) Those powers $2^{e}$ of the form $2^{3n}$ for which the exponent $e$ is a multiple of $3$: i.e. $e \equiv 0$ (mod $3$).

(e) $64 = 2^{6} = 4^{3} = 8^{2}. 729 = 3^{6} = 9^{3} = 27^{2}$.

(a) (i) $7$; (ii) $12$; (iii) $21$; (iv) $13$; (v) $14$; (vi) $31$; (vii) $10^{31} = 31$

(b) (i) $2\sqrt{2}$; (ii) $2\sqrt{3}$; (iii) $5\sqrt{2}$; (iv) $7\sqrt{3}$; (v) $12\sqrt{2}$; (vi) $21\sqrt{2}$

(c) (i) $\angle ABC = 108^{\circ}$. $\triangle BAC$ is isosceles $(BA = BC)$, so $\angle BAC = \angle BCA = 36^{\circ}$.

$\therefore \angle{CAE} = \angle{BAE} - \angle{BAC} = 72^{\circ} = 180^{\circ} - \angle{AED}$.

So $AC$ is parallel to $ED$ (since corresponding angles add to $180^{\circ}$).

(ii) $AX$ is parallel to $ED$; similarly $DX$ is parallel to $EA$. Hence $AXDE$ is a parallelogram, with $EA = ED$.

(iii) The two triangles are both isosceles and $\angle AXD = \angle CXB = 108^{\circ}$ (vertically opposite angles). Hence $\angle XAD = \angle XCB = 36^{\circ}$, and $ \angle XDA = \angle XBC = 36^{\circ}$.

(iv) $AD : CB = DX : BX$, so $x : 1 = 1 : (x - 1)$; hence $x^{2} - x - 1 = 0$ and $x = \frac{1 + \sqrt{5}}{2} - \text{the} \ Golden \ Ratio$ usually denoted by the Greek letter $\tau$ (tau), with approximate value $0.6180339887...$

(v) $BX = x -1 = \frac{\sqrt{5} -1}{2} \ (= \frac{1}{\tau} = \tau -1)$ with approximate value $0.6180339887...$

(vi) We may either check that corresponding angles are equal in pairs $(36^{\circ}, 72^{\circ}, 72^{\circ})$, or that corresponding sides are in the same ratio $x : 1 = 1 : (x -1)$.

(vii) $\cos{36^{\circ}} = \frac{\sqrt{5} + 1}{4}; \cos{72^{\circ}} = \frac{\sqrt{5} -1}{4}$ (drop perpendiculars from $D$ to $AB$ and from $X$ to $BC$; or use the Cosine Rule).

(viii) $\sin^{2}{36^{\circ}} + \cos^{2}{36^{\circ}} = 1: \sin{36^{\circ}} = \frac{\sqrt{10 -2\sqrt{5}}}{4}; \ \sin{72^{\circ}} = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$

Note: The Golden Ratio crops up in many unexpected places (including the regular pentagon, and the Fibonacci numbers). Unfortunately much that is written about its ubiquity is pure invention. One of the better popular treatments, that highlights the number’s significance, while taking a sober view of spurious claims, is the book The Golden Ratio by Mario Livio.

(a) $12345 = 5 \times 2469 = 3 \times 5 \times 823$. But is $823$ a prime number? It is easy to check that $823$ is not divisible by $2$, or $3$, or $5$, or $7$, or $11$. The Square Root Test (displayed below) tells us that it is only necessary to check four more potential prime factors.

Square Root Test: If $N = a \times b$ with $a \le b$, then $a \times a \le a \times b = N$, so the smaller factor $a \le \sqrt{N}$.

Hence, if $N (= 823$ say) is not prime, its smallest factor $\gt 1$ is at most equal to $\sqrt{N} (= \sqrt{823} \lt 29)$. Checking $a = 13, 17, 19, 23$ shows that the required prime factorisation is $12345 = 3 \times 5 \times 823$.

(b) There are $25 \ (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97)$.

Notes:

(i) For small primes, mental arithmetic should suffice. But one should also be aware of the general Sieve of Eratosthenes (a Greek polymath from the 3rd century BC). Start with the integers $1–100$ arranged in ten columns, and proceed as follows:

$\begin{array}{cccccccccc}{1} & {2} & {3} & {4} & {5} & {6} & {7} & {8} & {9} & {10} \\ {11} & {12} & {13} & {14} & {15} & {16} & {17} & {18} & {19} & {20} \\ {21} & {22} & {23} & {24} & {25} & {26} & {27} & {28} & {29} & {30} \\ {31} & {32} & {33} & {34} & {35} & {36} & {37} & {38} & {39} & {40} \\ {41} & {42} & {43} & {44} & {45} & {46} & {47} & {48} & {49} & {50} \\ {61} & {62} & {63} & {64} & {65} & {66} & {67} & {68} & {69} & {50} \\ {71} & {72} & {73} & {74} & {75} & {76} & {77} & {78} & {79} & {80} \\ {81} & {82} & {83} & {84} & {85} & {86} & {87} & {88} & {89} & {90} \\ {91} & {92} & {93} & {94} & {95} & {96} & {97} & {98} & {99} & {100}\end{array}$

Delete $1$ (which is not a prime: see (ii) below).

Circle the first undeleted integer; remove all other multiples of $2$.

Circle the first undeleted integer; remove all other multiples of $3$.

Circle the first undeleted integer; remove all other multiples of $5$.

Circle the first undeleted integer; remove all other multiples of $7$.

All remaining undeleted integers $\lt 100$ must be prime (by the Square Root Test (see part (a)).

(ii) The multiplicative structure of integers is surprisingly subtle. The first thing to notice is that $1$ has a special role, in that it is the multiplicative identity: for each integer $n$, we have $1 \times n = n$. Hence $1$ is “multiplicatively neutral” – it has no effect.

The “multiplicative building blocks” for integers are the prime numbers: every integer $\gt 1$ can be broken down, or factorised as a product of prime numbers, in exactly one way. The integer $1$ has no proper factors, and has no role to play in breaking down larger integers by factorisation. So $1$ is not a prime. (If we made the mistake of counting $1$ as a prime number, then we would have to make all sorts of silly exceptions – for example, to allow for the fact that $2 = 2 \times 1 = 2 \times 1 \times 1 = ...$, so $2$ could then be factorised in infinitely many ways.)

(iii) Notice that $91 = 7 \times 13$ is not a prime; so there is exactly one prime in the $90$s – namely $97$.

How many primes are there in the next run of $10$ (from $100–109$)?

How many primes are there from $190–199$? How many from $200–210$?

(ii) Many students struggle with this, and may suggest $143$, or $323$, or even $63$. The problem conceals a (very thinly) disguised message:

One cannot calculate with words.

To make use of mathematics, we must routinely translate words into symbols. As soon as “one less than a square” is translated into symbols, bells should begin to ring. For you know that $x^{2} - 1 = (x - 1)(x + 1)$, so $x^{2} - 1$ can only be prime if the smaller factor $(x - 1)$ is equal to $1$.

(a)(i) If we try to avoid such a “relatively prime pair”, then we must not choose any of $11, 13, 17, 19$ (since they are prime, and have no multiples in the given range). So we are forced to choose the other six integers: $10, 12, 14, 15, 16, 18$ – and there are then exactly two pairs which are relatively prime, namely $14, 15$ and $15, 16$.

(ii) If we try to avoid such a pair, then we can choose at most one even integer. So we are then forced to choose all five available odd integers, and our list will be: “unknown even, $11, 13, 15, 17, 19$”. If the even integer is chosen to be $14$, or $16$, then every pair in my list has $\text{LCM} = 1$. So the answer is “No”.

(b)(i) If we try to avoid such a pair, then we must not choose $97$ (the only prime number in the nineties). And we must not choose $95 = 5 \times 19$ (which is relatively prime to all other integers in the given range – except for $90$); and we must not choose $91 = 7 \times 13$ (which is relatively prime to all other integers in the given range – except for $98$). So we are forced to choose six integers from $90, 92, 93, 94, 96, 98, 99$. Whichever integer we then omit leaves a pair which is relatively prime.

(ii) If we try to avoid such a pair, then we can choose at most one even integer. So we are then forced to choose all five available odd integers, and our list will be: “unknown even, $91, 93, 95, 97, 99$”, and so must include the pair $93, 99$ – with common factor $3$. So the answer is “Yes”.

(c) In parts (a) and (b), the possible integers are limited (in (a) to the “teens”, and in (b) to the “nineties”); so it is natural to reach for ad hoc arguments as we did above. But in part (c) you know nothing about the numbers chosen.

Note: The question says that “I choose”, and asks whether “you” can be sure. So you have to find either a general argument that works for any $n$, or a counterexample. And the theme of this chapter indicates that it should not require any extended calculation.

The relevant “general idea” is the Pigeon Hole Principle which we may meet in the second part of this collection. So this problem may be viewed as a gentle introduction.

(i) Group the $2n$ consecutive integers

$a + 1, a + 2, …, a +2n$

into $n$ pairs of consecutive integers

$\{a+1, a+2\},\{a+3, a+4\}, \ldots,\{a+(2 n-1), a+2 n\}$

If we choose at most one integer from each pair, then we never get more than $n$ integers.
So as soon as we choose $n + 1$ integers from $2n$ consecutive integers, we are forced to choose both integers in some pair $k$, $k + 1$, and this pair of consecutive integers is always relatively prime (see Problem 6(b)(i)).

(ii) We saw in part (a)(ii) that, if $n = 5$ and the $2n$ integers start at $10$, then we can choose six integers (either $11, 13, 14, 15, 17, 19$, or $11, 13, 15, 16, 17, 19$), and in each case every pair has $\text{LCM} = 1$. So for $n = 5$ the answer is “No” (because there is at least one case where one cannot be sure).

However, as soon as $n$ is at least $6$, we show that the argument in part (a)(ii) breaks down. As before, if we try to choose a subset in which no pair has a common factor, then we can choose at most one even integer. So we are forced to choose all the odd integers. But any run of at least six consecutive odd integers includes two multiples of $3$. So for $n \ge 6$, the answer is “Yes”.

(a)(i) Suppose $k$ is a factor $of$ $m$ and $n$. Then we can write $m = kp$ and $n = kq$ for some integers $p, q$. Hence $m-n = k(p-q)$, so $k$ is a factor of $m-n$. Also $m +n = k(p + q)$, so $k$ is a factor of $m + n$.

(ii) Any factor of $m$ and $n$ is also a factor of their difference $m + n$; so the set of common factors of $m$ and $n$ is a subset of the set of common factors of $m-n$ and $n$.

And any factor of $m-n$ and $n$ is also a factor of their sum $m$; so the set of common factors of $m - n$ and $n$ is a subset of the set of common factors of $m$ and $n$.

Hence the two sets of common factors are identical. In particular, the two “highest common factors” are equal.

(iii) Subtract $91$ from $1001$ ten times to see that

$HCF(1001, 91) = HCF(1001 - 910, 91) = 91$.

(b)(i) Subtract $m$ from $m + 1$ once to see that

$HCF( m + 1, m) = HCF(1,m) = 1$

(ii) Subtract $m$ from $2m + 1$ twice to see that

$HCF(2m +1, m) = HCF(m+1, m) = HCF(1,m) = 1$

(iii) Subtract $m -1$ from $m^{2} +1$ “$m +1$ times” to see that

$HCF(m^{2}+1, m-1) = HCF((m^2+1) - (m^{2} -1), m -1) = HCF(2, m-1)$

Hence, if $m$ is odd, the $\text{HCF} = 2$; if $m$ is even, the $\text{HCF} = 1$.

7. They are equal. (The first is

$\frac{17}{100} \times 19000000$

the second is

$\frac{19}{100} \times 17000000$

which are equal since multiplication is commutative and associative.)

(a)

$\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} = \frac{6}{2} = 3$

(b)

$\sqrt{\frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \frac{6}{5} \times \frac{7}{6} \times \frac{8}{7}} = \sqrt{\frac{8}{2}} = 2$

(c)

$10 \times 9 \times8 \times7 \times6 \times5 \times4 \times3 \times2 \times1 \ \text{seconds}$

$\begin {array} {align} = \frac{10 \times 9 \times8 \times7 \times6 \times5 \times4 \times3 \times2 \times1 }{60} \ \text{minutes} \\ = \frac{10 \times 9 \times8 \times7 \times6 \times5 \times4 \times3 \times2 \times1 }{60 \times 60} \ \text{hours} \\ = \frac{10 \times 9 \times8 \times7 \times6 \times5 \times4 \times3 \times2 \times1 }{60 \times 60 \times 24} \ \text{days} \\ = \frac{10 \times 9 \times8 \times7 \times6 \times5 \times4 \times3 \times2 \times1 }{60 \times 60 \times 24 \times 7} \ \text{weeks} \\ = 6 \ \text{weeks (after cancelling).} \end {array}$

Note: These three questions underline what we mean by structural arithmetic. Fractions should never be handled by evaluating numerators and denominators. Instead one should always be on the lookout for structural features which simplify calculation – such as cancellation.

(a) Suppose a rectangle in the “DIN A” series has dimensions $a$ by $b$, with $a \lt b$. Folding in half produces a rectangle of size $\frac{b}{2}$ by $a$. Hence $b : a = a : \frac{b}{2}$, so $b^{2} = 2a^{2}$, and $b : a = \sqrt{2}:1$.

(b) (i) $\frac{1}{r}$

(ii) $r$

10.

(a) “$15\%$ discount” means the price actually charged is “$85\%$ of the marked price”. Hence each marked price needs to be multiplied by $0.85$.

The distributive law says we may add the marked prices first and then multiply the total (exactly $£80$) by $0.85$ to get

$£\left(\frac{85}{100} \times 80\right) = £(17 \times 4) = £68$

Note: The context (shopping, sales tax, and discount) is mathematically uninteresting. What matters here is the underlying multiplicative structure of the solution, which arises in many different contexts.

(b) “Add $20\%$ VAT” means multiplying the discounted pre-VAT total ($£68$) by $1.2$, or $\frac{6}{5}$. Hence the final price, with VAT added, is $£(1.2 \times 0.85 \times 80)$.

If the VAT were added first, the price before discount would be $£(1.2 \times 80)$, and the final price after allowing for the discount would be $£(0.85 \times 1.2 \times 80)$. Since multiplication is commutative, the two calculations have the same result, so the order does not matter (just as the final result in Problem 9 is the same whether one first enlarges A4 to A3 and then reduces A3 to A4, or first reduces A4 to A5 and then enlarges A5 to A4).

Note: Notice that we did not evaluate the two answers to see that they gave the same output $£81.60$. If we had, then the equality might have been a fluke due to the particular numbers chosen. Instead we left the answer unevaluated, in structured form, which showed that the equality would hold for any input.

(c) To cancel out multiplying by $\frac{6}{5}$ we need to multiply by $\frac{5}{6}$ – a discount of $\frac{1}{6}$, or $16\frac{2}{3}\%$.

Note: This question has nothing to do with financial applications. It is included to underline the fact that although percentage change questions use the language of addition and subtraction (“increase”, or “decrease”), the mathematics suggests they should be handled multiplicatively.

11.

(a)(i) $2 \times 5 \lt 3 \times 4$, so

$\frac{7}{2 \times 5} \gt \frac{7}{3 \times 4}$

Hence

$\frac{1}{2} + \frac{1}{5} \gt \frac{1}{3} + \frac{1}{4}$

(ii) At first sight, “$10 \lt 12$” may not seem related to “$\frac{1}{2} + \frac{1}{5} \gt \frac{1}{3} + \frac{1}{4}$”. Yet the crucial fact we started from in part (i) was “$2 \times 5 = 10 \lt 12 = 3 \times 4$”.

(b) $10 \lt 12$ so

$(x+ 2)(x + 5) = x^{2} + 7x + 10 \lt x^{2} + 7x + 12 = (x +3)(x+4)$

(i) If all four brackets are positive (i.e. if $x \gt -2$), then we also have $2x + 7 \gt 0$, and it follows that

$\begin{array} {align} \frac{1}{x + 2} + \frac{1}{x +5} &= \frac{2x + 7}{(x+2)(x+5)} \\ &\gt \frac{2x + 7}{(x+3)(x+4)} \\ &= \frac{1}{x + 3} + \frac{1}{x+4} \end{array}$

(ii) When calculating with the given algebraic expression, the values

$x = -2, -3, -4, -5$

are "forbidden values."

If $x \gt -2$, then (as in part (i)) we have

$\begin{array} {align} \frac{1}{x + 2} + \frac{1}{x +5} &= \frac{2x + 7}{(x+2)(x+5)} \\ &\gt \frac{2x + 7}{(x+3)(x+4)} \\ &= \frac{1}{x + 3} + \frac{1}{x+4} \end{array}$

For permitted values of $x \lt -2$, one or more of the brackets $(x + 2), (x + 5), (x + 3), (x + 4)$ will be negative. However, one can still carry out the algebra to simplify

$\frac{1}{x + 2} + \frac{1}{x +5} = \frac{2x + 7}{(x+2)(x+5)}$ and $\frac{1}{x + 3} + \frac{1}{x +4} = \frac{2x + 7}{(x+3)(x+4)}$

When $x = -\frac{7}{2}$ both expressions are equal, and equal to $0$. The simplified numerators are both positive if $x \gt -\frac{7}{2}$, and both negative if $x \lt -\frac{7}{2}$; and the sign of the denominators changes as one moves through the four intervals $-3 \lt x \lt -2, -4 \lt x \lt -3, -5 \lt x \lt -4, x \lt -5$, with the inequality switching from

$\begin{array}{align} “\gt” \ (\text{for} \ x \gt -2) &\text{to} \ “\lt” \ (\text{for} -3 \lt x\lt -2), \\ &\text{to} \ “\gt” \ (\text{for} -3.5 \lt x\lt -3), \\ &\text{to} \ “\lt” \ (\text{for} -4 \lt x\lt -3.5), \\ &\text{to} \ “\gt” \ (\text{for} -5 \lt x\lt -4), \\ &\text{to} \ “\lt” \ (\text{for}\ x\lt -5)
\end{array}$

12.

(a)(i) $3 + 2\sqrt{2}$;

(ii) $3- 2\sqrt{2}$;

(iii) $7+5\sqrt{2}$

Note: Notice that you can write down the answer to (ii) as soon as you have finished (i), without doing any further calculation.

(b)(i) $ 2+ \sqrt{6}$;

(ii) $\sqrt{2} + \sqrt{3}$;

(iii) $\frac{1+\sqrt{5}}{2}$ (the Golden Ratio $\frac{1+\sqrt{5}}{2} = \tau $ is the larger root of the quadratic equation $x^{2} - x -1 = 0$. Hence $\frac{3+\sqrt{5}}{2} = \tau + 1 = \tau^{2}$;

(iv) $\sqrt{10-2\sqrt{5}}$: this does not simplify further.

Note: Some readers may think an apology is in order for part (iv). The lesson here is that, while one should always try to simplify, there is no way of knowing in advance whether a simplification is possible. And there is no way out of this dilemma. So one is reduced to thinking: any simplification would involve $\sqrt{5}$, and if one tries to solve (a + b\sqrt{5})^{2} = 10 -2\sqrt{5}, then the solutions for $a$ and $b$ do not lead to anything “simpler”. (This repeated surd should perhaps have rung bells, as it was equal to the exact expression for $4\sin{36^{\circ}}$ in Problem 3(c). It was included here partly because the question of its simplification should already have arisen when it featured in that context.)

13.

In reconstructing the missing digits the number of possible solutions is determined by the highest common factor of the multiplier and 10. At the first step (in the units column):

because $HCF(6, 10) = 2, \square \times 6 = 8 \ (\text{mod} \ 10)$ has two solutions which differ by $5$ – namely $3$ and $8$.

The first possibility then requires us to solve $(\square \times 3) + 1 = 2 \ (\text{mod} \ 10)$: because $HCF(3, 10) = 1$, this has just one solution – namely $7$. This gives rise to the solution $76 \times 3 = 228$.

The second possibility requires us to solve $(\square \times 8) + 4 = 2 \ (\text{mod} \ 10)$: because $HCF(8, 10) = 2$, this has two solutions which differ by $5$ – namely $1$ and $6$. This gives rise to two further solutions: $16 \times 8 = 128$, and $66 \times 8 = 528$.

14.

(a) The solutions are entirely elementary, with no trickery. But they can be surprisingly elusive. And since this elusiveness is the only reason for including the problem, we hesitate to relieve any frustration by giving the solution. The whole thrust of the “$24$ game” is to underline the scope for “getting to know” the many faces of a number like $24$: for example, as $24 = 12 + 12 \ (= 3 \times 4 + 3 \times 4$ for $3, 3, 4, 4)$; as $24 = 25 - 1 \ (= 5 \times 5 -3 \div 3$ for $3, 3, 5, 5)$; and as $24 = 27 -3 \ (= 3 \times 3 \times 3 - 3$ for $3, 3, 3, 3)$. So one should be looking for ways of exploiting other important arithmetical aspects of $24$ – in particular, as $4 \times 6$ and as $3 \times 8$.

(b)(i)

$\begin{array} {align} 0 = (4-4) + (4-4); 1 = (4 \div 4) \times (4 \div 4); 2 = (4 \div 4) + (4 \div 4); 3 = (4 + 4 + 4) \div 4; \\ 4 = ((4-4) \times 4) + 4; 5 = (( 4 \times 4) + 4) \div 4); 6 = 4 + ((4 + 4) \div 4); 7 = 4 + 4 - (4 \div 4); \\ 8 = (4 + 4) \times (4 \div 4); 9 = (4+ 4) + (4 \div 4). \end{array}$

The output $10$ seems to be impossible with the given restrictions.

(ii) With squaring and $\sqrt{\ \ \ \ }$ allowed we can manage $10 = 4 + 4 + 4 - \sqrt{4}$. Indeed, one can make everything up to $40$ except (perhaps) $39$.

15.

(a)(i) $a^{2} + 2ab + b^{2}; a^{3} + 3a^{2}b + 3ab^{2} + b^{3}$

(ii) Replace $b$ by $(b)$: $a^{2} - 2ab + b^{2}; a^{3} - 3a^{2}b + 3ab^{2} - b^{3}$

(b)(i) $(x +1)^{2}$

(ii) $(x^{2} - 1)^{2}$

(iii) $(x^{2} - 1)^{3}$

(c)(i) $a^{2} - b^{2}$

(ii) Replace "$b$" by "$b+ c$": $a^{2}-(b+c)^{2} = a^{2} - b^{2} - c^{2}- 2bc$

Replace "$b$" by "$b- c$": $a^{2}-(b-c)^{2} = a^{2} - b^{2} - c^{2} + 2bc$

(d) One way is to rewrite this expression as a difference of two squares:

$\begin{array} {align} (2x)^{2} - (x^{2} - 2x +1) &= (2x)^{2} - (x-1)^{2} \\ &= (2x -(x-1))(2x +(x -1)) \\ &= (x +1)(3x -1)
\end{array}$

Note: As so often, the messages here are largely implicit. In part (a)(ii) we explicitly highlight the intention to use what you already know (by simply substituting "$-b$” in place of “$b$”. In part (b), you are expected to recognise (i), and then to see (ii) and (iii) as mild variations on the expansions of $(a - b)^{2}$ and $(a- b)^{3}$ in part (a). Part (c) repeats (in silence) the message of (a)(ii): think – don’t slog it out. And part (d) encourages you to keep an eye out for thinly disguised instances of “a difference of two squares”.

16.

(a) Final digits: "block" $4$, $6$ of length $2$;

Leading digits: “block” $4$, $1$, $6$, $2$, $1$ of length $5$.

(b) Claim The sequence of “units digits” really does recur.

Proof Given a power of $4$ that has units digit $4$, the usual multiplication algorithm for multiplying by $4$ produces a number with units digit $6$.

Given this new power of $4$ with units digit $6$, the usual multiplication algorithm for multiplying by $4$ produces a number with units digit $4$.

At this stage the sequence of units digits begins a new cycle.

[Alternatively: The units digit is simply equal to the relevant power of $4 \ (\text{mod} \ 10)$. Multiplying by $4$ changes $4$ to $6 \ (\text{mod} \ 10)$; multiplying by $4$ changes $6$ to $4 \ (\text{mod} \ 10)$; – and the cycle repeats.]

(c) The sequence of leading digits seems to recur every $5$ terms, because $4^{5} = 2^{10} = 1024$ is almost exactly equal to $1000$. Each time we move on $5$ steps in the sequence, we multiply by $4^{5} = 1024$. As far as the leading digit is concerned, this has the same effect as multiplying the initial term $(4)$ by slightly more than $1.024$ (then adding any ‘carries’), which is very like multiplying by $1$ – and so does not change the leading digit (yet).

However, each time we move on $10$ steps in the sequence, we multiply by $4^{10} = 1024^{2} = 1048576$. As far as the leading digit is concerned, this has the same effect as multiplying by slightly more than $1.048576$.

When we move on $25$ steps, we multiply by $4^{25} = 1125899906842624$. And as far as the leading digit is concerned, this has the same effect as multiplying by slightly more than $1.12599906842624$. And so on.

Eventually, the multiplier becomes large enough to change one of the leading digits.

17. The total is $100$.

Having found this by direct calculation, we should think indirectly and notice that $100 = 10^{2}$.

And we should then ask: “Why $10$? What has $10$ got to do with the $4 \times$ multiplication table?”

A quick check of the $1 \times$ multiplication table (total $= 1$), the $2 \times$ multiplication table (total $= 9$), etc. may suggest what we should have seen immediately.

$\begin{array} {align} \text{The first row has sum:} &&&& \ \ \ \ \ \ \ (1 +2+3+4). \\ \text{The second row has total} &&&& 2 \times (1 +2+3+4). \\ \text{The third row has total} &&&& 3 \times (1 +2+3+4). \\ \text{The fourth row has total} &&&& 4 \times (1 +2+3+4). \end{array} $

$\therefore \text{The total is} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1+2+3+4) \times (1+2+3+4) = (1+2+3+4)^{2}$

19.

(a) $\sin{45^{\circ}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} = \cos{45^{\circ}}; \tan{45^{\circ}} = 1$

(b) $AM = \sqrt{3}; \ \sin{30^{\circ}} = \frac{1}{2}, \cos{30^{\circ}} = \frac{\sqrt{3}}{2}, \tan{30^{\circ}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}; \ \sin{60^{\circ}} = \frac{\sqrt{3}}{2}, \cos{60^{\circ}} = \frac{1}{2}, \tan{60^{\circ}} = \sqrt{3}$

(c)(iii) $\cos{315^{\circ}} = \cos{45^{\circ}} = \frac{\sqrt{2}}{2}; \sin{225^{\circ}} = -\sin{45^{\circ}} = -\frac{\sqrt{2}}{2}; \tan{210^{\circ}} = \tan{30^{\circ}} = \frac{\sqrt{3}}{3}; \cos{120^{\circ}} = -\cos{60^{\circ}} = -\frac{1}{2}; \sin{960^{\circ}} = \sin{240^{\circ}} = -\sin{60^{\circ}} = -\frac{\sqrt{3}}{2}; \tan{(-135^{\circ})} = \tan{45^{\circ}} = 1$

(d) Cut the $n$-gon into $n$ “cake slices”, and use the formula “$\frac{1}{2}ab\sin{C}$” for each slice.

(i) $\frac{3\sqrt{3}}{4}$

(ii) $2$

(iii) $\frac{3\sqrt{3}}{2}$

(iv) $2\sqrt{2}$

(v) $3$

(e) Work out the side length of the $n$-gon, then cut the $n$-gon into $n$ "slices," and use the formula "\frac{1}{2}(\text{base} \times \text{height})" for each slice.

(i) $3\sqrt{3}$

(ii) $4$

(iii) $2\sqrt{3}$

(iv) $8(\sqrt{2} - 1)$

(v) $12(2 -\sqrt{3})$

Note: There is no hidden trig here: all you need is Pythagoras’ Theorem. For example, in part (e)(iv) we can extend alternate sides of the regular octagon to form the circumscribed $2$ by $2$ square. The four corner triangles are isosceles right angled triangles with hypotenuse of length $s$ (the side of the octagon). Hence each side of the square is equal to $s + 2 \cdot \frac{s}{\sqrt{2}} = 2$ whence $s = 2(\sqrt{2} -1)$.

20.

(a) $\sqrt{8}$

(b) $\sqrt{2}$, $\sqrt{2}$

21. Construct the perpendicular from $A$ to $BC$ (possibly extended); let this meet the line $BC$ at $X$. There are four possibilities:

(i) Either $X = C$, in which case $\angle{BCA}$ is a right angle as required; or $X = B$, in which case $b^{2} = a^{2} + c^{2}$, contradicting $a^{2} + b^{2} = c^{2}$;

(ii) $X \ne B,C$, and $C$ lies between $B$ and $X$;

(iii) $X \ne B,C$, and $X$ lies between $B$ and $C$;

(iv) $X \ne B,C$, and $B$ lies between $X$ and $C$.

We analyse case (ii) and leave cases (iii) and (iv) to the reader.

(ii) $\triangle{AXC}$ and $\triangle{AXB}$ are both right angled triangles; so by Pythagoras’ Theorem we know that

$\begin{array}{align} AC^{2} &= AX^{2} + XC^{2}, \text{and} \\ AB^{2} &= AX^{2} + XB^{2} \\ &= AX^{2} + (XC + CB)^{2} \\ &=AX^{2} + XC^{2} + CB^{2} + 2XC \cdot CB \\ &= AC^{2} + CB^{2} + 2XC \cdot CB
\end{array}$

Since we are told that $AC^{2} + CB^{2} = AB^{2}$, it follows that $2XC \cdot CB = 0$, contrary to $X \ne C$.

Note: Notice that the proof of the converse of Pythagoras’ Theorem makes use of Pythagoras’ Theorem itself.

$page61image23648$ 22.

(a) $ c = b+1$, so $a^{2} = c^{2} - b^{2} = 2b + 1$. Hence $a$ is odd, and we can write $a = 2m + 1$.

(b) Suppose $b = 2n-1$ is also odd. Then $c^{2} = 4n^{2}$ is divisible by $4$ – which contradicts the fact that $b^{2} = 4(n^{2} -n) +1$, and $a^{2} = 4(m^{2} +m) +1$, so $a^{2} + b^{2}$ leaves remainder $2$ on division by $4$.

Hence $b = 2n$ is even and $c = 2n+1$ is odd. But then

$(2m + 1)^{2} + (2n)^{2} = a^{2} + b^{2} = c^{2} = (2n +1)^{2}$

so $4(m^{2} + m) = 4n$, and $n = m(m+1)$.

23.

(a) If $a$ and $b$ are both even, then $HCF(a,b) \ne 1$, so the triple would not be primitive.

If $a$ and $b$ are both odd, we use the idea from Problem 22(b). Suppose $a = 2m +1, \ b = 2n +1$; then $a^{2} = 4(m^{2} + m) +1$, and $b^{2} = 4(n^{2} + n) +1$, so $a^{2} + b^{2} = 2 \times (2(m^{2} + m + n^{2} +n) +1)$. But this is “twice an odd number”, so cannot be equal to $c^{2}$ (since $c$ would have to be even, and any even square must be a multiple of $4$).

Hence we may assume that $a$ is odd and $b$ is even: so $c$ is odd.

(b) Then $a^{2} + b^{2} = c^{2}$ yields $b^{2} = c^{2} - a^{2} = (c - a)(c +a)$, so

$\left(\frac{b}{2}\right)^{2} = \left( \frac{c -a}{2} \right) \left( \frac{c +a}{2} \right)$

Any common factor of $\frac{c+a}{2}$ and $\frac{c-a}{2}$ divides their sum $c$ and their difference $a$, so $HCF\left( \frac{c-a}{2}, \frac{c+a}{2}\right) = 1$. Since the difference of these two factors is $a$, which is odd, they have opposite parity.

(c) If two integers are relatively prime, and their product is a square, then each of the factors has to be a square (consider their prime factorisations). Hence $\frac{c+a}{2} = p^2$ and $\frac{c-a}{2} = q^2$, where $HCF(p,q) = 1$ and $p$ and $q$ have opposite parity.

Therefore

$c = p^{2} + q^{2}, \ \ \ a = p^{2} - q^{2} , \ \ \ b = 2pq$

(d) It is easy to check that any triple of the given form is (i) primitive, and (ii) satisfies $a^{2} + b^{2} = c^{2}$.