Elementary Mathematics 2020 (I)

Set A

Questions 1–20 of 100

1Arithmetic • Number System • Prime Numbers

HardIdeal time: about 2.5 min

The number $2 \times 3 \times 5 \times 7 \times 11 + 1$ is

Aa prime numberBnot a prime, but power of a primeCnot a power of a prime, but a composite even numberDnot a power of a prime, but a composite odd number

Show answer & solution

Option A

a prime number

Approach

Multiply out to get $2311$ , then settle its status by trial division against every prime up to $2311$ .

Prime, composite, and the status of

1

Step-by-step solution

1
$2 \times 3 \times 5 \times 7 \times 11 = 2310$ , so the number is $2311$ .
Why: The four options are claims about one particular integer, so the first job is to name it.
2
$2311$ is odd, so it is not even; and it leaves remainder $1$ on division by each of $2, 3, 5, 7, 11$ .
Why: Every one of those primes divides $2310$ exactly, so each leaves remainder $1$ in $2310 + 1$ . This is why option C is impossible before any work is done.
Dividend $=$ divisor $\times$ quotient $+$ remainder
3
$2311 \approx 48.1$ , so only the primes $13, 17, 19, 23, 29, 31, 37, 41, 43, 47$ remain to be tested.
Why: If $2311$ had a factor above its square root it would also have the matching factor below it, so testing up to $48$ is enough.
Prime, composite, and the status of $1$ Primality by trial division up to $n$
4
$2311 = 13 (177) + 10 = 17 (135) + 16 = 19 (121) + 12 = 23 (100) + 11 = 29 (79) + 20 = 31 (74) + 17 = 37 (62) + 17 = 41 (56) + 15 = 43 (53) + 32 = 47 (49) + 8$ .
Why: No remainder is zero, so $2311$ has no divisor other than $1$ and itself: it has exactly two positive divisors and is prime.

Shortcut

The $+ 1$ forces the number to be odd and coprime to $2, 3, 5, 7, 11$ , which kills options B and C at once; only the trial division from $13$ to $47$ then separates A from D.

Common mistakes

Option B

Guessed that $2310 + 1$ must be a power of the next prime, $13$ ; but $1 3^{2} = 169$ and $1 3^{3} = 2197$ , neither of which is $2311$ .

Option C

Read the number as the product $2310$ itself and so called it even and composite, dropping the $+ 1$ .

Option D

Assumed that one more than a product of primes cannot itself be prime (a misremembering of Euclid's argument), or abandoned the trial division before reaching $47$ .

Skills used

Prime, composite, and the status of

1

Dividend

=

divisor

\times

quotient

+

remainder Primality by trial division up to

n

2Arithmetic • Number System • LCM and HCF

ModerateIdeal time: about 2 min

Two unequal pairs of numbers satisfy the following conditions :

(i): The product of the two numbers in each pair is $2160$

(ii): The HCF of the two numbers in each pair is $12$ .

If $x$ is the mean of the numbers in the first pair and $y$ is the mean of the numbers in the second pair, then what is the mean of $x$ and $y$ ?

60

72

75

78

Show answer & solution

Option B

72

Approach

Write each pair as $12 a$ and $12 b$ with $a$ and $b$ coprime, so that the product condition fixes $ab$ , then list the two coprime factorisations.

Coprime means no shared prime factor

Step-by-step solution

1
Let a pair be $12 a$ and $12 b$ with $HCF (a, b) = 1$ .
Why: An HCF of $12$ means $12$ is the largest common factor, so after taking it out nothing common is left: $a$ and $b$ must be coprime.
Coprime means no shared prime factor
2
$(12 a) (12 b) = 2160 \Rightarrow 144 ab = 2160 \Rightarrow ab = 15$ .
Why: The product condition applies to both pairs, so both pairs give the same $ab$ .
HCF $\times$ LCM = product of the two numbers
3
$ab = 15$ with $a < b$ coprime gives $(a, b) = (1, 15)$ or $(3, 5)$ , so the pairs are $(12, 180)$ and $(36, 60)$ .
Why: $15 = 1 \times 15 = 3 \times 5$ , and both factorisations happen to be coprime, which is exactly why two unequal pairs exist.
Prime factorisation
4
$x = \frac{12 + 180}{2} = 96$ and $y = \frac{36 + 60}{2} = 48$ .
Why: These are the two means the question names.
Mean of a frequency distribution
5
$\frac{x + y}{2} = \frac{96 + 48}{2} = 72$ .
Why: The quantity asked for is the mean of those two means.

Shortcut

The mean of $x$ and $y$ is $\frac{1}{4}$ of the sum of all four numbers, $\frac{12 + 180 + 36 + 60}{4} = \frac{288}{4} = 72$ , so the two pair-means never need to be written down separately.

Common mistakes

Option A

Stopped at the second pair and reported the larger of its members, $60$ , instead of going on to the mean of the means.

Option C

No route reaches $75$ : it is neither a pair mean ( $96$ or $48$ ) nor an average of any two of the four numbers.

Option D

Averaged $96$ with $60$ -- one member of the second pair rather than that pair's mean $48$ -- giving $\frac{96 + 60}{2} = 78$ .

Skills used

Coprime means no shared prime factor HCF

\times

LCM = product of the two numbers Prime factorisation Mean of a frequency distribution

3Algebra • Logarithm • Characteristic and Mantissa

EasyIdeal time: about 1 min

How many digits are there in $(54)^{10}$ ? (Given that $lo g_{10} 2 = 0.301$ and $lo g_{10} 3 = 0.477$ )

16

18

19

27

Show answer & solution

Option B

18

Approach

Take $lo g_{10}$ of the power, split $54$ into $2 \times 3^{3}$ to use the two given logarithms, then read the digit count off the characteristic.

Number of digits from a common logarithm

Step-by-step solution

1
$lo g_{10} (54)^{10} = 10 lo g_{10} 54$
Why: The power rule converts an awkward tenth power into a multiplication.
$lo g a^{n} = n lo g a$
2
$54 = 2 \times 27 = 2 \times 3^{3}$ , so $lo g_{10} 54 = lo g_{10} 2 + 3 lo g_{10} 3$ .
Why: Only $lo g_{10} 2$ and $lo g_{10} 3$ are given, so $54$ has to be rebuilt out of $2$ s and $3$ s.
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$
3
$lo g_{10} 54 = 0.301 + 3 (0.477) = 0.301 + 1.431 = 1.732$ , so $lo g_{10} (54)^{10} = 17.32$ .
Why: Substituting the given values finishes the logarithm.
4
Digits $= ⌊ 17.32 ⌋ + 1 = 17 + 1 = 18$ .
Why: A positive integer $N$ has $⌊ lo g_{10} N ⌋ + 1$ digits; the integer part is taken, never the nearest integer.
Number of digits from a common logarithm

Common mistakes

Option A

No route reaches $16$ from $lo g_{10} 54 = 1.732$ .

Option C

Rounded $17.32$ up to $18$ and then still added $1$ , so the $+ 1$ was applied on top of a rounding instead of on top of the integer part.

Option D

No route reaches $27$ ; it looks like the $27$ in $54 = 2 \times 27$ carried through by mistake.

Skills used

Number of digits from a common logarithm

lo g a^{n} = n lo g a

lo g (ab) = lo g a + lo g b

and

lo g \frac{a}{b} = lo g a - lo g b

4Algebra • Surds

ModerateIdeal time: about 1.5 min

Which one of the following is a set of solutions of the equation $x^{x} = n x^{x}$ , if $n$ is a positive integer ?

{1, n^{2}}

{1, n}

{1, n}

{n, n^{2}}

Show answer & solution

Option A

{1, n^{2}}

Approach

Write the $n$ -th root as a fractional index, equate the two indices of the common base $x$ , and treat $x = 1$ as the separate case where the base makes the indices irrelevant.

Power of a power, and the move to a common base

Step-by-step solution

1
$n x^{x} = (x^{x})^{1/ n} = x^{x / n}$ , so the equation is $x^{x} = x^{x / n}$ .
Why: A radical of order $n$ is the index $\frac{1}{n}$ , and powers of powers multiply, which puts both sides on the one base $x$ .
Power of a power, and the move to a common base
2
If $x = 1$ : both sides are $1^{anything} = 1$ , so $x = 1$ is a solution for every $n$ .
Why: Base $1$ satisfies the equation whatever the indices are, so it must be collected before the indices are compared.
3
For $x > 0$ , $x \neq = 1$ : $x^{x} = x^{x / n} \Rightarrow x = \frac{x}{n} \Rightarrow n x = x$ .
Why: With equal bases other than $1$ , equal powers force equal indices.
4
$n x = x = (x)^{2} \Rightarrow x = n \Rightarrow x = n^{2}$ .
Why: Dividing by $x$ (non-zero, since $x \neq = 0$ ) leaves $x = n$ ; squaring gives $x$ .
Square both sides to clear a radical
5
The solution set is ${1, n^{2}}$ .
Why: Both cases are genuine solutions, so the set holds both.

Common mistakes

Option B

Solved $x = \frac{x}{n}$ by reading it as $x = n$ instead of squaring $x = n$ .

Option C

From $n x = x$ cancelled wrongly to $n = x$ rather than $n = x$ .

Option D

Found $x = n^{2}$ but dropped the $x = 1$ root, padding the set with $n$ instead.

Skills used

Power of a power, and the move to a common base Square both sides to clear a radical

5Algebra • Sets • Operations

Very easyIdeal time: about 40 s

In a competitive examination, $250$ students have registered. Out of these, $50$ students have registered for Physics, $75$ students for Mathematics and $35$ students for both Mathematics and Physics. What is the number of students who have registered neither for Physics nor for Mathematics ?

90

100

150

160

Show answer & solution

Option D

160

Approach

Count the students in at least one subject by inclusion-exclusion, then subtract from the total.

Inclusion-exclusion for two sets

Step-by-step solution

1
$n (P) = 50$ , $n (M) = 75$ , $n (P \cap M) = 35$ , $n (U) = 250$
Why: Naming the four counts first keeps the $35$ from being used twice.
Inclusion-exclusion for two sets
2
$n (P \cup M) = 50 + 75 - 35 = 90$
Why: Students registered for both have been counted in each of the $50$ and the $75$ , so the overlap is removed once.
3
Neither $= n (U) - n (P \cup M) = 250 - 90 = 160$
Why: 'Neither' is the complement of the union inside the $250$ registered students.

Shortcut

Count the three disjoint blocks instead: Physics only $15$ , Mathematics only $40$ , both $35$ , total $90$ ; the rest is $160$ .

Common mistakes

Option A

Stopped at $n (P \cup M) = 90$ , the number registered for at least one subject, instead of subtracting it from $250$ .

Option B

No route reaches $100$ from $250$ , $50$ , $75$ and $35$ .

Option C

No route reaches $150$ from the four given counts.

Skills used

Inclusion-exclusion for two sets

6Arithmetic • Number System • Digit Problems

EasyIdeal time: about 50 s

If the sum of the digits of a number $1 0^{n} - 1$ , where $n$ is a natural number, is equal to $3798$ , then what is the value of $n$ ?

421

422

423

424

Show answer & solution

Option B

422

Approach

Recognise $1 0^{n} - 1$ as the number made of $n$ nines, so its digit sum is $9 n$ .

Closed form of a repeated-digit number

Step-by-step solution

1
$1 0^{1} - 1 = 9$ , $1 0^{2} - 1 = 99$ , $1 0^{3} - 1 = 999$ , so $1 0^{n} - 1$ is $n$ copies of the digit $9$ .
Why: Subtracting $1$ from a $1$ followed by $n$ zeros borrows all the way along, turning every digit into a $9$ .
Closed form of a repeated-digit number
2
Digit sum $= 9 n$
Why: All $n$ digits are $9$ , so the sum is $9$ taken $n$ times.
3
$9 n = 3798 \Rightarrow n = \frac{3798}{9} = 422$
Why: One linear equation in $n$ .
Turn a worded condition into a linear equation in one unknown

Common mistakes

Option A

Took $1 0^{n} - 1$ to have $n - 1$ nines, so solved $9 (n - 1) = 3798$ and reported $n = 423$ 's neighbour $421$ -- a count that is one short.

Option C

Added $1$ to $422$ , as though the number of nines were one more than the exponent.

Option D

No route reaches $424$ ; $9 \times 424 = 3816$ .

Skills used

Closed form of a repeated-digit number Turn a worded condition into a linear equation in one unknown

7Algebra • Surds

ModerateIdeal time: about 1.5 min

Which one of the following is the largest number among $222 2^{2}$ , $22 2^{22}$ , $2 2^{222}$ , $2^{2222}$ ?

2^{2222}

2 2^{222}

22 2^{22}

222 2^{2}

Show answer & solution

Option A

2^{2222}

Approach

Bring every number to base $2$ -- replacing each base by a convenient power of $2$ that bounds it -- and then compare the indices.

Power of a power, and the move to a common base

Step-by-step solution

1
$222 2^{2} = 4937284$ , a seven-digit number.
Why: The first candidate is small enough to evaluate outright, which removes it from the contest immediately.
2
$222 < 256 = 2^{8}$ , so $22 2^{22} < (2^{8})^{22} = 2^{176}$ .
Why: Raising a base to a larger power of $2$ can only increase it, so this bounds the second candidate above.
Power of a power, and the move to a common base
3
$22 < 32 = 2^{5}$ , so $2 2^{222} < (2^{5})^{222} = 2^{1110}$ .
Why: The same bounding move on the third candidate.
4
$2^{2222} > 2^{1110} > 2 2^{222}$ and $2^{2222} > 2^{176} > 22 2^{22} > 222 2^{2}$ , so $2^{2222}$ is the largest.
Why: With one common base, the largest index wins; both bounds fall well short of $2^{2222}$ .

Shortcut

Compare the number of binary digits: $2^{2222}$ has $2223$ of them, while $2 2^{222} = 2^{222} \cdot 1 1^{222}$ has fewer than $222 + 4 (222) = 1110$ . The gap is too wide to need care.

Common mistakes

Option B

Judged $2 2^{222}$ largest on the grounds that it has both a sizeable base and a huge index, without converting to a common base; in fact $2 2^{222} < 2^{1110}$ .

Option C

No route makes $22 2^{22}$ largest: it is below $2^{176}$ , smaller than two of the others.

Option D

Chose the largest base, $2222$ , ignoring that its index is only $2$ .

Skills used

Power of a power, and the move to a common base

8Arithmetic • Number System • Prime Numbers

ModerateIdeal time: about 1.5 min

If $m$ is the number of prime numbers between $0$ and $50$ ; and $n$ is the number of prime numbers between $50$ and $100$ , then what is $(m - n)$ equal to ?

4

5

6

7

Show answer & solution

Option B

5

Approach

List the primes in each half of the range by sieving out multiples of $2, 3, 5, 7$ , count them, and subtract.

Prime, composite, and the status of

1

Step-by-step solution

1
Primes below $50$ : $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$ , so $m = 15$ .
Why: $1$ is neither prime nor composite, so it is not counted; every other survivor of the multiples of $2, 3, 5, 7$ below $50$ is prime, since $7^{2} = 49$ .
Prime, composite, and the status of $1$
2
Primes between $50$ and $100$ : $53, 59, 61, 67, 71, 73, 79, 83, 89, 97$ , so $n = 10$ .
Why: Here a candidate under $100$ needs testing only against $2, 3, 5, 7$ , because $1 1^{2} = 121 > 100$ ; that removes $51 = 3 (17)$ , $57$ , $63$ , $69$ , $77 = 7 (11)$ , $87$ , $91 = 7 (13)$ , $93$ and $99$ .
3
$m - n = 15 - 10 = 5$
Why: The quantity asked for.

Common mistakes

Option A

Counted a composite such as $91 = 7 \times 13$ or $51 = 3 \times 17$ as prime, making $n = 11$ .

Option C

Counted $1$ as a prime, making $m = 16$ .

Option D

No route reaches $7$ ; it would need $m$ and $n$ to be out by two between them.

Skills used

Prime, composite, and the status of

1

9Arithmetic • Ratio and Proportion • Proportion

EasyIdeal time: about 1.5 min

If $\frac{a}{b} = \frac{1}{3}$ , $\frac{b}{c} = 2$ , $\frac{c}{d} = \frac{1}{2}$ , $\frac{d}{e} = 3$ and $\frac{e}{f} = \frac{1}{4}$ , then what is the value of $\frac{ab c}{d e f}$ ?

\frac{1}{4}

\frac{3}{4}

\frac{3}{8}

\frac{27}{4}

Show answer & solution

Option C

\frac{3}{8}

Approach

Assign a value to one letter and walk the chain of ratios to get all six, then compute the required quotient.

Join two ratios on their shared term

Step-by-step solution

1
Take $a = 1$ . Then $\frac{a}{b} = \frac{1}{3}$ gives $b = 3$ ; $\frac{b}{c} = 2$ gives $c = \frac{3}{2}$ .
Why: The five ratios fix the six letters only up to one overall scale, and the asked quotient has three letters above and three below, so the scale cancels and any convenient $a$ will do.
Join two ratios on their shared term
2
$\frac{c}{d} = \frac{1}{2}$ gives $d = 3$ ; $\frac{d}{e} = 3$ gives $e = 1$ ; $\frac{e}{f} = \frac{1}{4}$ gives $f = 4$ .
Why: Each ratio links one new letter to the one before it, so the chain runs straight through.
3
$\frac{ab c}{d e f} = \frac{1 \times 3 \times \frac{3}{2}}{3 \times 1 \times 4} = \frac{\frac{9}{2}}{12} = \frac{9}{24} = \frac{3}{8}$
Why: Both products are now plain numbers.

Shortcut

Group the quotient as $\frac{a}{d} \cdot \frac{b}{e} \cdot \frac{c}{f}$ and read each factor off the chain: $\frac{a}{d} = \frac{1}{3} \cdot 2 \cdot \frac{1}{2} = \frac{1}{3}$ , $\frac{b}{e} = 2 \cdot \frac{1}{2} \cdot 3 = 3$ , $\frac{c}{f} = \frac{1}{2} \cdot 3 \cdot \frac{1}{4} = \frac{3}{8}$ , and the product is $\frac{1}{3} \cdot 3 \cdot \frac{3}{8} = \frac{3}{8}$ .

Common mistakes

Option A

Multiplied all five given ratios, $\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d} \cdot \frac{d}{e} \cdot \frac{e}{f} = \frac{1}{4}$ , which is $\frac{a}{f}$ and not $\frac{ab c}{d e f}$ .

Option B

Reached $\frac{3}{8}$ and then doubled it, typically by using $f = 2 e$ in place of $f = 4 e$ .

Option D

No route reaches $\frac{27}{4}$ ; it is larger than $1$ , whereas the chain makes the numerator the smaller product.

Skills used

Join two ratios on their shared term

10Arithmetic • Number System • Divisibility

EasyIdeal time: about 1 min

Which one of the following is the largest divisor of $3^{x} + 3^{x + 1} + 3^{x + 2}$ , if $x$ is any natural number ?

3

13

39

117

Show answer & solution

Option C

39

Approach

Factor out the lowest power of $3$ to put the expression in the form $13 \cdot 3^{x}$ , then ask what must divide it for every natural $x$ -- which means for $x = 1$ as well.

Power of a power, and the move to a common base

Step-by-step solution

1
$3^{x} + 3^{x + 1} + 3^{x + 2} = 3^{x} (1 + 3 + 3^{2}) = 3^{x} \times 13$
Why: Taking out the smallest power leaves a bracket of plain numbers; $1 + 3 + 9 = 13$ .
Power of a power, and the move to a common base
2
For $x = 1$ the value is $3 \times 13 = 39$ .
Why: A divisor that must work for every natural $x$ must in particular divide the smallest value, and $x = 1$ is the smallest natural number.
Dividend $=$ divisor $\times$ quotient $+$ remainder
3
$39 = 3 \times 13$ divides $3^{x} \times 13$ for every $x \geq 1$ , so $39$ is the largest such divisor.
Why: Every value carries at least one factor $3$ and exactly one factor $13$ , so $3 \times 13$ always divides it, while nothing larger can -- it would have to divide $39$ .
Prime factorisation

Common mistakes

Option A

Took out only the common factor $3^{x}$ and reported the base $3$ , without noticing the bracket contributes $13$ .

Option B

Reported the bracket $1 + 3 + 9 = 13$ alone, dropping the factor of $3$ that $3^{x}$ always carries for natural $x$ .

Option D

Tested only $x = 2$ , where the value is $117$ , and missed that at $x = 1$ the value is just $39$ , which $117$ does not divide.

Skills used

Power of a power, and the move to a common base Dividend

=

divisor

\times

quotient

+

remainder Prime factorisation

11Algebra • Equations • Linear Equations

ModerateIdeal time: about 1.5 min

A two-digit number is $9$ more than four times of the number obtained by interchanging its digits. If the product of digits in the two-digit number is $8$ , then what is the number ?

81

42

24

18

Show answer & solution

Option A

81

Approach

Write the number in place-value form, turn the stated relation into an equation in the two digits, and use the product condition to finish.

Place-value expansion of a numeral

Step-by-step solution

1
Let the number be $10 a + b$ with $a \in {1, \dots, 9}$ , $b \in {0, \dots, 9}$ ; the interchanged number is $10 b + a$ .
Why: Place-value form is what lets 'interchanging its digits' be written algebraically at all.
Place-value expansion of a numeral
2
$10 a + b = 4 (10 b + a) + 9$
Why: ' $9$ more than four times' means four times the interchanged number, plus $9$ .
Turn a worded condition into a linear equation in one unknown
3
$10 a + b = 40 b + 4 a + 9 \Rightarrow 6 a - 39 b = 9 \Rightarrow 2 a - 13 b = 3$
Why: Collecting terms and dividing through by $3$ keeps the numbers small.
4
$ab = 8$ with single digits gives $(a, b) \in {(1, 8), (2, 4), (4, 2), (8, 1)}$ ; only $(8, 1)$ satisfies $2 a - 13 b = 3$ , since $16 - 13 = 3$ .
Why: The product condition leaves four candidates, which is quicker to test than solving $2 a = 13 b + 3$ in integers.
5
The number is $10 (8) + 1 = 81$ .
Why: Rebuilding the numeral from its digits.

Shortcut

$2 a = 13 b + 3$ forces $b$ to be odd and small: $b = 1$ gives $a = 8$ , and $b = 3$ already gives $a = 21$ . So $(a, b) = (8, 1)$ before the product condition is even used -- the product then confirms it.

Common mistakes

Option B

Used only the digit-product condition: $4 \times 2 = 8$ , but $4 (24) + 9 = 105 \neq = 42$ .

Option C

Used only the digit-product condition: $2 \times 4 = 8$ , but $4 (42) + 9 = 177 \neq = 24$ .

Option D

Solved correctly and then reported the interchanged number $18$ rather than the original $81$ .

Skills used

Place-value expansion of a numeral Turn a worded condition into a linear equation in one unknown

12Algebra • Quadratic Equations • Roots

EasyIdeal time: about 1 min

If $α$ and $β$ are the roots of the quadratic equation $x^{2} + k x - 15 = 0$ such that $α - β = 8$ , then what is the positive value of $k$ ?

2

3

4

5

Show answer & solution

Option A

2

Approach

Use the sum and product of the roots, then the identity $(α - β)^{2} = (α + β)^{2} - 4 α β$ to bring in the given difference.

Sum and product of roots (Vieta)

Step-by-step solution

1
$α + β = - k$ and $α β = - 15$
Why: For $x^{2} + k x - 15 = 0$ the sum of roots is $- \frac{k}{1}$ and the product is $\frac{- 15}{1}$ .
Sum and product of roots (Vieta)
2
$(α - β)^{2} = (α + β)^{2} - 4 α β$
Why: This is the identity that connects a difference of roots to the two symmetric functions the equation gives directly, so the roots never have to be found.
Square of a sum and of a difference
3
$8^{2} = (- k)^{2} - 4 (- 15) \Rightarrow 64 = k^{2} + 60 \Rightarrow k^{2} = 4$
Why: Substituting the given $α - β = 8$ .
4
$k = \pm 2$ , so the positive value is $k = 2$ .
Why: The question asks for the positive value, which settles the sign.

Shortcut

$α β = - 15$ and $α - β = 8$ are satisfied by $3$ and $- 5$ on sight; their sum is $- 2$ , so $k = 2$ .

Common mistakes

Option B

Found the roots $3$ and $- 5$ and reported the positive root $3$ instead of $k$ .

Option C

Stopped at $k^{2} = 4$ and reported $4$ without taking the square root.

Option D

Reported the size of the negative root, $5$ , instead of $k$ .

Skills used

Sum and product of roots (Vieta)Square of a sum and of a difference

13Algebra • Polynomial • Factor and Remainder Theorem

EasyIdeal time: about 1 min

What are the values of $p$ and $q$ respectively, if $(x - 1)$ and $(x + 2)$ divide the polynomial $x^{3} + 4 x^{2} + p x + q$ ?

1, - 6

2, - 6

1, 6

2, 6

Show answer & solution

Option A

1, - 6

Approach

Each divisor gives a root, so apply the factor theorem twice and solve the resulting pair of linear equations in $p$ and $q$ .

Factor and remainder theorem

Step-by-step solution

1
$(x - 1)$ divides $f (x) = x^{3} + 4 x^{2} + p x + q \Rightarrow f (1) = 0$ ; $(x + 2)$ divides $f (x) \Rightarrow f (- 2) = 0$ .
Why: A linear factor $(x - c)$ is a factor exactly when $f (c) = 0$ ; the factor $(x + 2)$ is tested at $x = - 2$ .
Factor and remainder theorem
2
$f (1) = 1 + 4 + p + q = 0 \Rightarrow p + q = - 5$
Why: The first condition written out.
3
$f (- 2) = - 8 + 16 - 2 p + q = 0 \Rightarrow - 2 p + q = - 8$
Why: The second condition; note $(- 2)^{3} = - 8$ and $4 (- 2)^{2} = 16$ .
4
Subtracting: $(p + q) - (- 2 p + q) = - 5 - (- 8) \Rightarrow 3 p = 3 \Rightarrow p = 1$ , hence $q = - 6$ .
Why: The two equations share the coefficient $+ 1$ on $q$ , so subtracting removes $q$ at once.
Eliminate one variable from a linear pair

Common mistakes

Option B

Reached the right $q$ but mis-solved $3 p = 3$ as $p = 2$ , or slipped a coefficient while eliminating.

Option C

Got $p = 1$ but took $q = + 6$ , dropping the sign when reading $q = - 5 - p$ .

Option D

No route reaches $p = 2, q = 6$ : it satisfies neither $p + q = - 5$ nor $- 2 p + q = - 8$ .

Skills used

Factor and remainder theorem Eliminate one variable from a linear pair

14Algebra • Polynomial • HCF and LCM

Very easyIdeal time: about 40 s

If $(x + k)$ is the HCF of $x^{2} + 5 x + 6$ and $x^{2} + 8 x + 15$ , then what is the value of $k$ ?

5

3

2

1

Show answer & solution

Option B

3

Approach

Factorise both quadratics and read off the one factor they share.

HCF and LCM of polynomials from their complete factorisations

Step-by-step solution

1
$x^{2} + 5 x + 6 = (x + 2) (x + 3)$
Why: Splitting the middle term: $2 + 3 = 5$ and $2 \times 3 = 6$ .
Factorise a quadratic by splitting the middle term
2
$x^{2} + 8 x + 15 = (x + 3) (x + 5)$
Why: Splitting again: $3 + 5 = 8$ and $3 \times 5 = 15$ .
3
The only common irreducible factor is $(x + 3)$ , so $HCF = x + 3$ and $k = 3$ .
Why: The HCF is the product of the factors common to both, each to the lowest power present; $(x + 2)$ and $(x + 5)$ appear in only one polynomial each.
HCF and LCM of polynomials from their complete factorisations

Shortcut

The HCF is a factor of the difference $(x^{2} + 8 x + 15) - (x^{2} + 5 x + 6) = 3 x + 9 = 3 (x + 3)$ , which points straight at $x + 3$ .

Common mistakes

Option A

Took $(x + 5)$ , a factor of the second polynomial only -- at $x = - 5$ the first gives $25 - 25 + 6 = 6 \neq = 0$ .

Option C

Took $(x + 2)$ , a factor of the first polynomial only -- at $x = - 2$ the second gives $4 - 16 + 15 = 3 \neq = 0$ .

Option D

No route reaches $k = 1$ : $(x + 1)$ divides neither polynomial.

Skills used

HCF and LCM of polynomials from their complete factorisations Factorise a quadratic by splitting the middle term

15Algebra • Equations • Exponential Equations

EasyIdeal time: about 1 min

If $5^{x + 1} - 5^{x - 1} = 600$ , then what is the value of $1 0^{2 x}$ ?

1

1000

100000

1000000

Show answer & solution

Option D

1000000

Approach

Factor out the lowest power of $5$ so the bracket becomes a number, solve for $x$ , then substitute.

Power of a power, and the move to a common base

Step-by-step solution

1
$5^{x + 1} - 5^{x - 1} = 5^{x - 1} (5^{2} - 1) = 24 \cdot 5^{x - 1}$
Why: Both terms carry $5^{x - 1}$ ; taking it out leaves a bracket of plain numbers, since $5^{x + 1} = 5^{x - 1} \cdot 5^{2}$ .
Power of a power, and the move to a common base
2
$24 \cdot 5^{x - 1} = 600 \Rightarrow 5^{x - 1} = 25$
Why: Dividing by $24$ keeps the arithmetic in small numbers.
3
$5^{x - 1} = 5^{2} \Rightarrow x - 1 = 2 \Rightarrow x = 3$
Why: With equal bases other than $1$ , equal powers force equal indices.
4
$1 0^{2 x} = 1 0^{6} = 1000000$
Why: Substituting $x = 3$ into the quantity actually asked for.

Common mistakes

Option A

No route reaches $1$ ; it would need $x = 0$ , and $5^{1} - 5^{- 1} = 4.8 \neq = 600$ .

Option B

Found $x = 3$ but evaluated $1 0^{x}$ instead of $1 0^{2 x}$ .

Option C

Lost one from the final index, writing $1 0^{2 x - 1} = 1 0^{5}$ .

Skills used

Power of a power, and the move to a common base

16Arithmetic • Number System • LCM and HCF

HardIdeal time: about 2.5 min

A number divides $12288$ , $28200$ and $44333$ so as to leave the same remainder in each case. What is that number ?

272

232

221

120

Show answer & solution

Option C

221

Approach

Equal remainders mean the divisor divides every difference of the three numbers, so take the differences and find their greatest common factor.

Dividend

=

divisor

\times

quotient

+

remainder

Step-by-step solution

1
If $12288 = n q_{1} + r$ , $28200 = n q_{2} + r$ and $44333 = n q_{3} + r$ , then $n$ divides each difference, e.g. $28200 - 12288 = n (q_{2} - q_{1})$ .
Why: Subtracting two expressions with the same remainder cancels $r$ , which is what turns an unknown remainder into a divisibility condition.
Dividend $=$ divisor $\times$ quotient $+$ remainder
2
$28200 - 12288 = 15912$ and $44333 - 28200 = 16133$
Why: Two differences are enough; the third, $44333 - 12288 = 32045$ , is their sum and carries no new information.
3
$n$ divides $16133 - 15912 = 221$
Why: A number dividing both differences divides their difference too, and $221$ is small enough to factorise.
4
$221 = 13 \times 17$ , and $15912 = 221 \times 72$ , so $221$ itself divides both differences; being the largest divisor of $221$ , it is the number.
Why: The candidates are the divisors $1, 13, 17, 221$ of $221$ ; since $221$ works, nothing larger can.
Prime factorisation

Shortcut

Subtract the two differences first: $16133 - 15912 = 221$ . That one subtraction pins the answer down to a divisor of $221$ , with no factorisation of five-digit numbers needed.

Common mistakes

Option A

$272 = 2^{4} \times 17$ picks up the factor $17$ but not $13$ , and $15912 \div 272$ is not a whole number.

Option B

$232 = 2^{3} \times 29$ divides none of the differences.

Option D

$120$ divides none of the differences; $15912 \div 120 = 132.6$ .

Skills used

Dividend

=

divisor

\times

quotient

+

remainder Prime factorisation

17Algebra • Expressions • Algebraic Identities

EasyIdeal time: about 45 s

If $x^{2} + 9 y^{2} = 6 x y$ , then what is $y : x$ equal to ?

1 : 3

1 : 2

2 : 1

3 : 1

Show answer & solution

Option A

1 : 3

Approach

Move everything to one side and recognise a perfect square.

Square of a sum and of a difference

Step-by-step solution

1
$x^{2} + 9 y^{2} = 6 x y \Rightarrow x^{2} - 6 x y + 9 y^{2} = 0$
Why: Collecting on one side is what exposes the square; $x^{2}$ , $9 y^{2}$ and $6 x y$ are exactly the three terms of $(x - 3 y)^{2}$ .
Square of a sum and of a difference
2
$x^{2} - 6 x y + 9 y^{2} = (x - 3 y)^{2} = 0$
Why: With $a = x$ and $b = 3 y$ : $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ , and $2 ab = 2 (x) (3 y) = 6 x y$ matches.
3
$(x - 3 y)^{2} = 0 \Rightarrow x = 3 y$
Why: A square is zero only when the quantity squared is zero -- so this gives a single relation, not two.
A product is zero only when a factor is zero
4
$\frac{y}{x} = \frac{y}{3 y} = \frac{1}{3}$ , so $y : x = 1 : 3$ .
Why: The question asks for $y : x$ , not $x : y$ .
Introduce a common factor $k$ for a ratio

Common mistakes

Option B

No route reaches $1 : 2$ ; it would need $x = 2 y$ , which gives $4 y^{2} + 9 y^{2} = 13 y^{2}$ against $6 x y = 12 y^{2}$ .

Option C

No route reaches $2 : 1$ ; $y = 2 x$ gives $x^{2} + 36 x^{2} = 37 x^{2}$ against $12 x^{2}$ .

Option D

Found $x = 3 y$ and reported $x : y = 3 : 1$ instead of the $y : x$ that was asked for.

Skills used

Square of a sum and of a difference A product is zero only when a factor is zero Introduce a common factor

k

for a ratio

18Arithmetic • Number System

EasyIdeal time: about 50 s

If $m$ and $n$ are positive integers such that $m^{n} = 1331$ , then what is the value of $(m - 1)^{n - 1}$ ?

1

100

121

125

Show answer & solution

Option B

100

Approach

Factorise $1331$ into primes to identify $m$ and $n$ , then substitute.

Prime factorisation

Step-by-step solution

1
$1331 = 11 \times 121 = 11 \times 11 \times 11 = 1 1^{3}$
Why: $1331$ ends in $1$ and is not divisible by $3$ or $7$ ; dividing by $11$ twice settles it.
Prime factorisation
2
$m^{n} = 1 1^{3}$ with $m, n$ positive integers and $m$ a genuine power gives $m = 11$ , $n = 3$ .
Why: Since $11$ is prime, $m$ must itself be a power of $11$ ; $m = 1 1^{3}$ with $n = 1$ is the degenerate reading, which would make the question empty.
3
$(m - 1)^{n - 1} = (11 - 1)^{3 - 1} = 1 0^{2} = 100$
Why: Substituting; note it is $m - 1$ raised to $n - 1$ , not $m$ raised to $n - 1$ .
Power of a power, and the move to a common base

Common mistakes

Option A

Took the degenerate solution $m = 1331$ , $n = 1$ , giving $133 0^{0} = 1$ ; defensible on the literal wording, but it makes the question vacuous.

Option C

Computed $m^{n - 1} = 1 1^{2} = 121$ , using $m$ where the expression asks for $m - 1$ .

Option D

No route reaches $125$ ; it is $5^{3}$ , unrelated to $m = 11$ , $n = 3$ .

Skills used

Prime factorisation Power of a power, and the move to a common base

19Algebra • Expressions • Simplification

ModerateIdeal time: about 1.5 min

What is $\frac{1}{a ^{m - n} - 1} + \frac{1}{a ^{n - m} - 1}$ equal to ?

1

- 1

0

2 a^{m - n}

Show answer & solution

Option B

- 1

Approach

Substitute a single letter for $a^{m - n}$ ; the second denominator is then its reciprocal, and the two fractions collapse.

Negative exponent is a reciprocal

Step-by-step solution

1
Put $t = a^{m - n}$ . Then $a^{n - m} = a^{- (m - n)} = \frac{1}{t}$ .
Why: The two exponents are negatives of each other, so one substitution covers both terms -- a negative exponent is a reciprocal.
Negative exponent is a reciprocal
2
The expression becomes $\frac{1}{t - 1} + \frac{1}{\frac{1}{t} - 1}$ .
Why: Rewriting with the single unknown $t$ .
3
$\frac{1}{\frac{1}{t} - 1} = \frac{1}{\frac{1 - t}{t}} = \frac{t}{1 - t}$
Why: Combining the inner level over the common denominator $t$ and then inverting the single fraction so formed.
Evaluate a nested fraction from the inside out
4
$\frac{1}{t - 1} + \frac{t}{1 - t} = \frac{1}{t - 1} - \frac{t}{t - 1} = \frac{1 - t}{t - 1} = - 1$
Why: $1 - t = - (t - 1)$ , so the two fractions share a denominator and the numerator is the negative of it.

Shortcut

The expression is symmetric under swapping $m$ and $n$ , which swaps the two terms; so it is a constant in $m - n$ , and a single test value settles which.

Common mistakes

Option A

Reached $\frac{1 - t}{t - 1}$ and read it as $+ 1$ , missing that numerator and denominator are negatives of each other.

Option C

Assumed the two terms must cancel because their exponents are opposite; they do not -- it is the $- 1$ in each denominator that breaks the symmetry.

Option D

No route reaches $2 a^{m - n}$ ; the expression is a constant, independent of $a$ , $m$ and $n$ .

Skills used

Negative exponent is a reciprocal Evaluate a nested fraction from the inside out

20Algebra • Surds • Simplification

EasyIdeal time: about 1 min

If $x = 2$ , $y = 33$ and $z = 66$ , then which one of the following is correct ?

y < x < z

z < x < y

z < y < x

x < y < z

Show answer & solution

Option B

z < x < y

Approach

Rewrite all three as sixth roots -- the common index -- and then compare the radicands as plain integers.

Power of a power, and the move to a common base

Step-by-step solution

1
$x = 2^{1/2}$ , $y = 3^{1/3}$ , $z = 6^{1/6}$
Why: Radicals of different orders cannot be compared directly; fractional indices make the orders visible.
Power of a power, and the move to a common base
2
$LCM (2, 3, 6) = 6$ , so $x = 2^{3/6} = 68$ , $y = 3^{2/6} = 69$ , $z = 66$ .
Why: Raising each index to the common denominator $6$ turns all three into sixth roots of integers.
$ab = a b$ and $a / b = a / b$
3
$6 < 8 < 9$ , and $t \mapsto 6 t$ preserves order for $t > 0$ , so $z < x < y$ .
Why: With one common index, the larger radicand is the larger number.
Comparing two fractions

Common mistakes

Option A

Got the radicands $8$ , $9$ , $6$ right but wrote the chain from largest to smallest, reversing the inequality signs.

Option C

Compared $2 = 1.414$ with $33 = 1.442$ by eye and put $x$ above $y$ ; the two differ only in the second decimal place.

Option D

Assumed a higher root order gives a larger value, so ordered them by index as $2 < 33 < 66$ .

Skills used

Power of a power, and the move to a common base

ab = a b

and

a / b = a / b

Comparing two fractions