Elementary Mathematics 2018 (II)

Set A

Questions 1–20 of 100

1Arithmetic • Number System • LCM and HCF

EasyIdeal time: about 1.5 min

The highest four-digit number which is divisible by each of the numbers $16$ , $36$ , $45$ , $48$ is

9180

9360

9630

9840

Show answer & solution

Option B

9360

Approach

A number divisible by all four is divisible by their LCM, so find the LCM and then the largest four-digit multiple of it.

Prime factorisation

Step-by-step solution

1
$16 = 2^{4}$ , $36 = 2^{2} \times 3^{2}$ , $45 = 3^{2} \times 5$ , $48 = 2^{4} \times 3$
Why: Prime factorisation is what lets the LCM be read off; the four numbers share factors, so multiplying them would overshoot badly.
Prime factorisation
2
$LCM = 2^{4} \times 3^{2} \times 5 = 720$
Why: The LCM takes the highest power of each prime that appears anywhere: $2^{4}$ from $16$ and $48$ , $3^{2}$ from $36$ and $45$ , $5$ from $45$ .
3
$9999 \div 720 = 13$ remainder $639$ , so the largest multiple is $720 \times 13 = 9360$
Why: The largest four-digit number is $9999$ ; dividing it by $720$ and keeping only the whole part gives the last multiple of $720$ that still fits in four digits.

Common mistakes

Option A

No route reaches $9180$ : it is divisible by $36$ and $45$ but not by $16$ , and it is not the largest multiple of anything in sight.

Option C

No route reaches $9630$ : it fails divisibility by $16$ and by $36$ .

Option D

Took the LCM as $2^{4} \times 3 \times 5 = 240$ , keeping only one factor of $3$ ; the largest four-digit multiple of $240$ is $240 \times 41 = 9840$ , which is not divisible by $9$ and so not by $36$ or $45$ .

Skills used

Prime factorisation

2Algebra • Indices

EasyIdeal time: about 45 s

If $x = y^{a}$ , $y = z^{b}$ and $z = x^{c}$ , then the value of $ab c$ is

1

2

- 1

0

Show answer & solution

Option A

1

Approach

Substitute the three relations into one another until $x$ is expressed as a power of itself, then equate indices.

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

Step-by-step solution

1
$x = y^{a} = (z^{b})^{a} = z^{ab}$
Why: Replacing $y$ by $z^{b}$ removes $y$ ; a power raised to a power multiplies the indices.
Power of a power, and the move to a common base
2
$x = z^{ab} = (x^{c})^{ab} = x^{ab c}$
Why: Replacing $z$ by $x^{c}$ closes the loop, so both sides are now powers of the same base.
3
$x^{1} = x^{ab c} \Rightarrow ab c = 1$
Why: For a base that is positive and not $1$ , equal powers force equal indices.

Shortcut

Take logarithms instead: $lo g x = a lo g y$ , $lo g y = b lo g z$ , $lo g z = c lo g x$ ; multiplying the three gives $lo g x lo g y lo g z = ab c lo g x lo g y lo g z$ , so $ab c = 1$ .

Common mistakes

Option B

No route reaches $2$ ; it is not produced by any misreading of the index chain.

Option C

Wrote the closing equation as $x^{ab c + 1} = 1$ instead of $x^{ab c} = x^{1}$ , so the index came out as $ab c = - 1$ .

Option D

Reached $ab c - 1 = 0$ correctly and then reported the left-hand side, $0$ , instead of solving for $ab c$ .

Skills used

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

3Algebra • Surds • Simplification

HardIdeal time: about 2 min

If $x = 2 + 2^{2/3} + 2^{1/3}$ , then the value of the expression $x^{3} - 6 x^{2} + 6 x$ will be

2

1

0

- 2

Show answer & solution

Option A

2

Approach

Move the $2$ to the left, cube both sides using $(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)$ , and use $2^{1/3} \times 2^{2/3} = 2$ to fold the surds back into $x$ .

Cube of a sum and of a difference

Step-by-step solution

1
$x - 2 = 2^{1/3} + 2^{2/3}$
Why: Isolating the two surd terms is what makes the cube worth taking: their cubes and their product are both rational.
Power of a power, and the move to a common base
2
$(2^{1/3})^{3} = 2$ , $(2^{2/3})^{3} = 4$ , $2^{1/3} \times 2^{2/3} = 2^{1} = 2$
Why: These are the three numbers the cube identity needs. The product is the key one: the indices $\frac{1}{3}$ and $\frac{2}{3}$ add to $1$ .
3
$(x - 2)^{3} = 2 + 4 + 3 \times 2 \times (x - 2) = 6 + 6 (x - 2) = 6 x - 6$
Why: Using $(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)$ keeps $a + b$ whole, and $a + b$ is exactly $x - 2$ - so the surds never have to be written again.
Cube of a sum and of a difference
4
$x^{3} - 6 x^{2} + 12 x - 8 = 6 x - 6 \Rightarrow x^{3} - 6 x^{2} + 6 x = 2$
Why: Expanding the left side and collecting gives precisely the expression the question asks for, so no further work is needed.

Common mistakes

Option B

No route reaches $1$ ; it does not arise from any of the natural slips in the cube identity.

Option C

Assumed the expression was built to vanish, i.e. read the cube as $(x - 2)^{3} = 6 x - 8$ and concluded $x^{3} - 6 x^{2} + 6 x = 0$ .

Option D

Carried the $- 8$ from expanding $(x - 2)^{3}$ to the wrong side, giving $x^{3} - 6 x^{2} + 6 x = - 2$ .

Skills used

Cube of a sum and of a difference Power of a power, and the move to a common base

4Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

How many five-digit numbers of the form XXYXX is/are divisible by $33$ ?

1

3

5

DInfinite

Show answer & solution

Option B

3

Approach

Expand XXYXX by place value, split $33$ into $11$ and $3$ , and let each divisibility test pin down one digit.

Divisibility by

11

from the alternating digit sum

Step-by-step solution

1
$X X Y X X = 10000 X + 1000 X + 100 Y + 10 X + X = 11011 X + 100 Y$
Why: Writing the numeral out is what turns a digit pattern into an equation the divisibility tests can be applied to.
Place-value expansion of a numeral
2
Alternating digit sum $= X - X + Y - X + X = Y$ , so $11 ∣ X X Y X X$ forces $Y = 0$
Why: Divisibility by $11$ is tested by the alternating sum of the digits; here the four $X$ s cancel in pairs and only $Y$ survives, and the only digit that is a multiple of $11$ is $0$ .
Divisibility by $11$ from the alternating digit sum
3
With $Y = 0$ the digit sum is $4 X$ , and $3 ∣ 4 X$ needs $3 ∣ X$
Why: Divisibility by $3$ is tested by the digit sum; $4$ contributes nothing to a factor of $3$ , so the condition falls entirely on $X$ .
A number is congruent to its digit sum modulo $9$
4
$X \in {3, 6, 9}$ (not $0$ , as $X$ leads a five-digit number): $33033$ , $66066$ , $99099$
Why: $X = 0$ would make the number $00000$ , which is not a five-digit number, so the count is three.

Shortcut

$11011 = 11 \times 7 \times 11 \times 13$ already carries a factor $11$ , so only $100 Y$ has to supply one; since $100$ leaves remainder $1$ on division by $11$ , that forces $Y = 0$ at once.

Common mistakes

Option A

Found $33033$ , the first number that works, and stopped there instead of running $X$ through the rest of its allowed values.

Option C

No route reaches $5$ : there is no reading of the two tests that admits five values of $X$ .

Option D

Treated XXYXX as an unbounded pattern and forgot that a five-digit number caps the search at $X \leq 9$ .

Skills used

Divisibility by

11

from the alternating digit sum Place-value expansion of a numeral A number is congruent to its digit sum modulo

9

5Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

A five-digit number XY235 is divisible by $3$ where X and Y are digits satisfying X $+$ Y $\leq 5$ . What is the number of possible pairs of values of (X, Y) ?

5

6

7

9

Show answer & solution

Option C

7

Approach

Turn 'divisible by $3$ ' into a condition on $X + Y$ , intersect it with $X + Y \leq 5$ , then count the digit pairs case by case.

Enumerate a residue class within a bound

Step-by-step solution

1
Digit sum $= X + Y + 2 + 3 + 5 = X + Y + 10$
Why: Divisibility by $3$ depends only on the digit sum, so the three known digits can be added once and set aside.
A number is congruent to its digit sum modulo $9$
2
$3 ∣ X + Y + 10 \Rightarrow X + Y \equiv 2$ on division by $3$ , so $X + Y \in {2, 5}$ given $X + Y \leq 5$
Why: $10$ leaves remainder $1$ , so $X + Y$ must leave remainder $2$ . Within the cap $X + Y \leq 5$ only $2$ and $5$ do - and $X + Y = 8$ or more is excluded by the cap, which is what the cap is there for.
Enumerate a residue class within a bound
3
$X \geq 1$ , because X leads a five-digit number
Why: This is the constraint the question does not spell out and the one that decides the count; a leading digit is never $0$ .
Place-value expansion of a numeral
4
$X + Y = 2$ : $(1, 1), (2, 0)$ . $X + Y = 5$ : $(1, 4), (2, 3), (3, 2), (4, 1), (5, 0)$ . Total $= 2 + 5 = 7$
Why: List by $X$ , dropping $X = 0$ in each case; $Y = 0$ is allowed throughout because Y is not a leading digit.

Common mistakes

Option A

Took only $X + Y = 5$ , missing the $X + Y = 2$ case, and counted its five pairs with $X \geq 1$ .

Option B

Took only $X + Y = 5$ and allowed $X = 0$ , giving the six pairs $(0, 5)$ through $(5, 0)$ .

Option D

Found both cases but let $X = 0$ , adding the two illegal pairs $(0, 2)$ and $(0, 5)$ to reach $3 + 6 = 9$ .

Skills used

A number is congruent to its digit sum modulo

9

Enumerate a residue class within a bound Place-value expansion of a numeral

6Algebra • Quadratic Equations • Quadratic Inequalities

EasyIdeal time: about 45 s

If $x^{2} - 6 x - 27 > 0$ , then which one of the following is correct ?

- 3 < x < 9

x < 9

x > - 3

x > 9

x < - 3

x < - 3

only

Show answer & solution

Option C

x > 9

x < - 3

Approach

Factorise the quadratic and read off where a product of two factors is positive.

Locate roots by the sign of

f

at test points

Step-by-step solution

1
$x^{2} - 6 x - 27 = (x - 9) (x + 3)$
Why: $- 27 = - 9 \times 3$ and $- 9 + 3 = - 6$ , so the middle term splits cleanly; a factorised quadratic is the only form whose sign is easy to read.
Factorise a quadratic by splitting the middle term
2
Roots are $x = - 3$ and $x = 9$ , and the leading coefficient is positive
Why: An upward parabola is negative strictly between its roots and positive outside them, so the two roots are all the information the sign needs.
Locate roots by the sign of $f$ at test points
3
$(x - 9) (x + 3) > 0 ⟺ x < - 3$ or $x > 9$
Why: Outside the roots both factors carry the same sign, so their product is positive; between them the signs differ.

Common mistakes

Option A

Solved $x^{2} - 6 x - 27 < 0$ instead: the interval between the roots is where the expression is negative, not positive.

Option B

Wrote each factor's condition separately and joined them with 'or' without checking the signs: $x < 9$ or $x > - 3$ is every real number, so it cannot be the solution of a strict inequality with real roots.

Option D

Found the branch $x < - 3$ and stopped, missing the second branch $x > 9$ .

Skills used

Factorise a quadratic by splitting the middle term Locate roots by the sign of

f

at test points

7Arithmetic • Number System • Factors and Divisors

ModerateIdeal time: about 1.5 min

The number of divisors of the number $38808$ , exclusive of the divisors $1$ and itself, is

74

72

70

68

Show answer & solution

Option C

70

Approach

Prime-factorise $38808$ , count all its divisors from the exponents, then remove $1$ and the number itself.

Prime factorisation

Step-by-step solution

1
$38808 = 8 \times 4851 = 8 \times 9 \times 539 = 8 \times 9 \times 7 \times 77$
Why: Strip the obvious factors first: $38808$ is even three times over, and $4851$ has digit sum $18$ so it carries $9$ .
Prime factorisation
2
$38808 = 2^{3} \times 3^{2} \times 7^{2} \times 11$
Why: $539 = 7 \times 77 = 7 \times 7 \times 11$ , which finishes the factorisation into primes.
3
Number of divisors $= (3 + 1) (2 + 1) (2 + 1) (1 + 1) = 4 \times 3 \times 3 \times 2 = 72$
Why: Each divisor chooses an exponent independently, from $0$ up to the exponent in the factorisation - so add one to each exponent and multiply.
4
$72 - 2 = 70$
Why: The question excludes $1$ and $38808$ itself, and those are two of the $72$ .

Common mistakes

Option A

Added the two excluded divisors instead of subtracting them, giving $72 + 2 = 74$ .

Option B

Stopped at the total divisor count $72$ and ignored the words 'exclusive of the divisors $1$ and itself'.

Option D

Subtracted twice over - once for $1$ and the number, once again - landing on $72 - 4 = 68$ .

Skills used

Prime factorisation

8Algebra • Polynomial • HCF and LCM

ModerateIdeal time: about 2 min

HCF and LCM of two polynomials are $(x + 3)$ and $(x^{3} - 9 x^{2} - x + 105)$ respectively. If one of the two polynomials is $x^{2} - 4 x - 21$ , then the other is

x^{2} + 2 x - 21

x^{2} + 2 x + 15

x^{2} - 2 x - 15

x^{2} - x - 15

Show answer & solution

Option C

x^{2} - 2 x - 15

Approach

Use HCF $\times$ LCM $=$ product of the two polynomials, then divide out the polynomial that is given.

HCF and LCM of polynomials from their complete factorisations

Step-by-step solution

1
$x^{2} - 4 x - 21 = (x - 7) (x + 3)$
Why: $- 21 = - 7 \times 3$ and $- 7 + 3 = - 4$ ; factorising the known polynomial is what makes the division at the end a cancellation rather than long division.
Factorise a quadratic by splitting the middle term
2
$x^{3} - 9 x^{2} - x + 105$ at $x = 7$ : $343 - 441 - 7 + 105 = 0$ , so $(x - 7)$ is a factor
Why: The LCM must contain every factor of the given polynomial, so $(x - 7)$ is worth testing first; the factor theorem tests it with one substitution.
Factor and remainder theorem
3
$x^{3} - 9 x^{2} - x + 105 = (x - 7) (x^{2} - 2 x - 15) = (x - 7) (x - 5) (x + 3)$
Why: Dividing by $(x - 7)$ leaves $x^{2} - 2 x - 15$ , which splits as $- 15 = - 5 \times 3$ with $- 5 + 3 = - 2$ .
4
Other $= \frac{HCF \times LCM}{first} = \frac{( x + 3 ) ( x - 7 ) ( x - 5 ) ( x + 3 )}{( x - 7 ) ( x + 3 )} = (x - 5) (x + 3) = x^{2} - 2 x - 15$
Why: For two expressions the product of HCF and LCM equals the product of the expressions, so the third one follows by dividing.
HCF and LCM of polynomials from their complete factorisations

Shortcut

Read it straight off the factorisations: the LCM is $(x + 3) (x - 5) (x - 7)$ and the given polynomial is $(x + 3) (x - 7)$ , so the missing polynomial must supply $(x - 5)$ and share $(x + 3)$ to make the HCF $(x + 3)$ - that is $(x + 3) (x - 5)$ .

Common mistakes

Option A

$x^{2} + 2 x - 21 = (x + 7) (x - 3)$ - the signs of both roots flipped, which destroys the shared factor $(x + 3)$ and so cannot give HCF $(x + 3)$ .

Option B

Kept the $+ 15$ from the constant of the LCM without the matching sign change, giving a quadratic $x^{2} + 2 x + 15$ that does not factorise at all.

Option D

Slipped the middle term of the quotient while dividing $x^{3} - 9 x^{2} - x + 105$ by $(x - 7)$ , getting $x^{2} - x - 15$ instead of $x^{2} - 2 x - 15$ .

Skills used

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

9Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 2 min

If $α$ and $β$ are two real numbers such that $α + β = - \frac{q}{p}$ and $α β = \frac{r}{p}$ , where $1 < p < q < r$ , then which one of the following is the greatest ?

\frac{1}{α + β}

\frac{1}{α} + \frac{1}{β}

- \frac{1}{α β}

\frac{α β}{α + β}

Show answer & solution

Option C

- \frac{1}{α β}

Approach

Rewrite all four expressions in terms of $p$ , $q$ and $r$ ; they are all negative, so the greatest is the one closest to zero.

Comparing two fractions

Step-by-step solution

1
$α + β = - \frac{q}{p}$ and $α β = \frac{r}{p}$ are given
Why: These are the only facts available about $α$ and $β$ , so every option must be reduced to them before anything can be compared.
Sum and product of roots (Vieta)
2
A: $\frac{1}{α + β} = - \frac{p}{q}$ ; B: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = - \frac{q}{r}$
Why: $\frac{1}{α} + \frac{1}{β}$ is the sum over the product, which is exactly the pair of quantities that are given.
3
C: $- \frac{1}{α β} = - \frac{p}{r}$ ; D: $\frac{α β}{α + β} = - \frac{r}{q}$
Why: The factor $p$ cancels in both of the last two, leaving four quantities built from $p$ , $q$ and $r$ alone.
4
All four are negative, and $\frac{p}{r} < \frac{p}{q}$ , $\frac{p}{r} < \frac{q}{r}$ , $\frac{p}{r} < 1 < \frac{r}{q}$ , so $- \frac{p}{r}$ is the greatest
Why: Among negative numbers the greatest is the one of smallest size. $\frac{p}{r}$ beats $\frac{p}{q}$ because $r > q$ enlarges the denominator, beats $\frac{q}{r}$ because $p < q$ shrinks the numerator, and beats $\frac{r}{q}$ which is above $1$ .
Comparing two fractions

Common mistakes

Option A

Compared $\frac{p}{q}$ with $\frac{p}{r}$ the wrong way round, forgetting that a larger denominator makes the fraction smaller, so picked $- \frac{p}{q}$ as the one nearest zero.

Option B

No route reaches $- \frac{q}{r}$ : it is neither the smallest nor the largest in size under $1 < p < q < r$ .

Option D

Ignored that every option is negative and picked the one of largest size, $- \frac{r}{q}$ , as though 'greatest' meant greatest magnitude.

Skills used

Sum and product of roots (Vieta)Comparing two fractions

10Arithmetic • Time and Work

ModerateIdeal time: about 1.5 min

Two workers 'A' and 'B' working together completed a job in $5$ days. Had 'A' worked twice as efficiently as he actually did and 'B' worked one-third as efficiently as he actually did, the work would have completed in $3$ days. In how many days could 'A' alone complete the job ?

3 \frac{1}{2}

daysB

4 \frac{1}{6}

daysC

5 \frac{1}{2}

daysD

6 \frac{1}{4}

days

Show answer & solution

Option D

6 \frac{1}{4}

days

Approach

Work in one-day shares rather than days: write both scenarios as equations in $\frac{1}{x}$ and $\frac{1}{y}$ and eliminate.

Linearise a system by substituting reciprocals

Step-by-step solution

1
Let A alone take $x$ days and B alone $y$ days: $\frac{1}{x} + \frac{1}{y} = \frac{1}{5}$
Why: Rates add when two people work together; times do not, which is why the equation is written in reciprocals.
Take total work as the LCM of the times
2
$\frac{2}{x} + \frac{1}{3 y} = \frac{1}{3}$
Why: Twice as efficient doubles the daily share, one-third as efficient divides it by three; the altered pair finishes in $3$ days, so their combined rate is $\frac{1}{3}$ .
3
Put $u = \frac{1}{x}$ , $v = \frac{1}{y}$ : $u + v = \frac{1}{5}$ and $2 u + \frac{v}{3} = \frac{1}{3}$
Why: In $u$ and $v$ the pair is an ordinary linear system, which eliminates in one line.
Linearise a system by substituting reciprocals
4
$\frac{u + v}{3} = \frac{1}{15}$ ; subtracting from the second, $2 u - \frac{u}{3} = \frac{1}{3} - \frac{1}{15} = \frac{4}{15}$ , so $\frac{5 u}{3} = \frac{4}{15}$ and $u = \frac{4}{25}$
Why: Dividing the first equation by $3$ matches the $\frac{v}{3}$ in the second, so $v$ disappears on subtraction and only $u$ - the quantity asked about - is left.
5
$x = \frac{1}{u} = \frac{25}{4} = 6 \frac{1}{4}$ days
Why: $u$ is A's share of the job per day, so its reciprocal is the number of days A needs alone.

Common mistakes

Option A

No route reaches $3 \frac{1}{2}$ days; it is below the $5$ days the pair needs together, which A alone cannot beat.

Option B

Added instead of subtracted when eliminating $v$ , getting $\frac{5 u}{3} = \frac{1}{3} + \frac{1}{15} = \frac{2}{5}$ and so $u = \frac{6}{25}$ - which would make B's rate negative.

Option C

No route reaches $5 \frac{1}{2}$ days; it does not satisfy either equation.

Skills used

Take total work as the LCM of the times Linearise a system by substituting reciprocals

11Algebra • Expressions • Simplification

ModerateIdeal time: about 1.5 min

If $x^{6} + \frac{1}{x ^{6}} = k (x^{2} + \frac{1}{x ^{2}})$ , then $k$ is equal to

(x^{2} - 1 + \frac{1}{x ^{2}})

(x^{4} - 1 + \frac{1}{x ^{4}})

(x^{4} + 1 + \frac{1}{x ^{4}})

(x^{4} - 1 - \frac{1}{x ^{4}})

Show answer & solution

Option B

(x^{4} - 1 + \frac{1}{x ^{4}})

Approach

Treat $x^{2}$ as a single letter, so the left side is a sum of two cubes and factorises.

Sum and difference of two cubes

Step-by-step solution

1
Put $u = x^{2}$ : the left side is $u^{3} + \frac{1}{u ^{3}}$ and the bracket on the right is $u + \frac{1}{u}$
Why: $x^{6} = (x^{2})^{3}$ , so the substitution turns sixth powers into cubes - and the right-hand bracket is then the matching first power.
Power of a power, and the move to a common base
2
$u^{3} + \frac{1}{u ^{3}} = (u + \frac{1}{u}) (u^{2} - u \cdot \frac{1}{u} + \frac{1}{u ^{2}}) = (u + \frac{1}{u}) (u^{2} - 1 + \frac{1}{u ^{2}})$
Why: Apply $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ with $a = u$ and $b = \frac{1}{u}$ . The middle term is $- ab = - 1$ because $u$ and $\frac{1}{u}$ multiply to $1$ .
Sum and difference of two cubes
3
$k = u^{2} - 1 + \frac{1}{u ^{2}} = x^{4} - 1 + \frac{1}{x ^{4}}$
Why: Comparing with $k (x^{2} + \frac{1}{x ^{2}})$ , the factor multiplying the bracket is $k$ ; restoring $u = x^{2}$ gives fourth powers.

Common mistakes

Option A

Factorised $x^{3} + \frac{1}{x ^{3}}$ instead of $x^{6} + \frac{1}{x ^{6}}$ - one power level too low, so the answer came out in squares rather than fourth powers.

Option C

Used $a^{2} + ab + b^{2}$ as the quadratic factor, giving $+ 1$ for the middle term; the sum of cubes takes $- ab$ , not $+ ab$ .

Option D

Carried the minus sign onto the last term as well, writing $x^{4} - 1 - \frac{1}{x ^{4}}$ ; only the middle term is negative.

Skills used

Power of a power, and the move to a common base Sum and difference of two cubes

12Algebra • Expressions • Algebraic Identities

EasyIdeal time: about 1.5 min

If the sum of the squares of three consecutive natural numbers is $110$ , then the sum of their cubes is

625

654

684

725

Show answer & solution

Option C

684

Approach

Centre the three numbers on the middle one so the squares collapse, find it, then cube.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
Let the numbers be $n - 1$ , $n$ , $n + 1$
Why: Naming the middle number rather than the first makes the cross terms cancel in the next line; that is the whole point of centring.
Turn a worded condition into a linear equation in one unknown
2
$(n - 1)^{2} + n^{2} + (n + 1)^{2} = 3 n^{2} + 2$
Why: Expanding, the $- 2 n$ and $+ 2 n$ cancel and the two $1$ s add, so only $3 n^{2} + 2$ survives.
Square of a sum and of a difference
3
$3 n^{2} + 2 = 110 \Rightarrow n^{2} = 36 \Rightarrow n = 6$
Why: $n$ is a natural number, so only the positive root counts; the three numbers are $5$ , $6$ , $7$ .
4
$5^{3} + 6^{3} + 7^{3} = 125 + 216 + 343 = 684$
Why: The question asks for the cubes, so the numbers themselves have to be found first - there is no shortcut from the sum of squares to the sum of cubes.

Common mistakes

Option A

No route reaches $625$ ; it is not the sum of the cubes of any three consecutive naturals.

Option B

Slipped when adding the cubes: $125 + 216 = 341$ , not $311$ , and the error of $30$ carries straight to the total.

Option D

Mis-cubed one of the numbers - $7^{3}$ as $384$ rather than $343$ is the usual version - inflating the total by $41$ .

Skills used

Turn a worded condition into a linear equation in one unknown Square of a sum and of a difference

13Arithmetic • Number System • LCM and HCF

ModerateIdeal time: about 1.5 min

The product of two integers $p$ and $q$ , where $p > 60$ and $q > 60$ , is $7168$ and their HCF is $16$ . The sum of these two integers is

256

184

176

164

Show answer & solution

Option C

176

Approach

Write both integers as $16$ times a coprime co-factor, find the co-factor pair from the product, then apply the $> 60$ condition.

Coprime means no shared prime factor

Step-by-step solution

1
HCF $= 16$ , so $p = 16 a$ and $q = 16 b$ with $a$ and $b$ coprime
Why: The HCF divides both numbers, and pulling it out leaves co-factors with no common prime - if $a$ and $b$ shared a factor the HCF would be larger than $16$ .
Coprime means no shared prime factor
2
$pq = 256 ab = 7168 \Rightarrow ab = 28$
Why: $16 \times 16 = 256$ and $7168 \div 256 = 28$ , which reduces the problem to one small product.
HCF $\times$ LCM = product of the two numbers
3
Coprime pairs with $ab = 28$ : $(1, 28)$ and $(4, 7)$ - $(2, 14)$ is rejected because $2$ divides both
Why: $28 = 2^{2} \times 7$ , so the only way to split it with no shared prime is to give one side the whole $2^{2}$ and the other the $7$ , or to give one side everything.
Prime factorisation
4
$(1, 28)$ gives $16$ and $448$ , and $16 < 60$ ; $(4, 7)$ gives $64$ and $112$ , both above $60$ . Sum $= 64 + 112 = 176$
Why: The condition $p > 60$ , $q > 60$ is what selects between the two coprime pairs - without it the question would have two answers.

Common mistakes

Option A

Used the pair $32$ and $224$ , which does multiply to $7168$ , but its HCF is $32$ and $32$ is not above $60$ - both stated conditions fail.

Option B

Used the pair $128$ and $56$ : the product and the $> 60$ condition both hold, but their HCF is $8$ , not $16$ .

Option D

No pair of integers multiplies to $7168$ and sums to $164$ .

Skills used

Coprime means no shared prime factor HCF

\times

LCM = product of the two numbers Prime factorisation

14Algebra • Logarithm • Laws of Logarithms

HardIdeal time: about 2 min

If $lo g_{10} 2 = 0.3010$ and $lo g_{10} 3 = 0.4771$ , then the value of $lo g_{100} (0.72)$ is equal to

0.9286

1 ˉ .9286

1.8572

1 ˉ .8572

Show answer & solution

Option B

1 ˉ .9286

Approach

Break $0.72$ into powers of $2$ , $3$ and $10$ , evaluate the common logarithm, then halve it to change the base from $10$ to $100$ .

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

and

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

Step-by-step solution

1
Let $lo g_{100} 0.72 = t$ , so $10 0^{t} = 0.72$ , i.e. $1 0^{2 t} = 0.72$ and $2 t = lo g_{10} 0.72$
Why: Only base- $10$ values are tabulated, so the base- $100$ logarithm has to be turned into a base- $10$ one; $100 = 1 0^{2}$ makes that a single halving.
Logarithm to exponential form
2
$0.72 = \frac{72}{100} = \frac{2 ^{3} \times 3 ^{2}}{1 0 ^{2}}$
Why: $72 = 8 \times 9$ ; splitting it into powers of $2$ and $3$ is what lets the two tabulated values be used.
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$
3
$lo g_{10} 0.72 = 3 lo g_{10} 2 + 2 lo g_{10} 3 - 2$
Why: A product becomes a sum of logarithms and a quotient a difference; each exponent comes down in front by the power rule.
$lo g a^{n} = n lo g a$
4
$= 3 (0.3010) + 2 (0.4771) - 2 = 0.9030 + 0.9542 - 2 = 1.8572 - 2 = - 0.1428$
Why: Substituting the two given values and subtracting the $2$ from $lo g_{10} 1 0^{2}$ .
5
$t = \frac{- 0.1428}{2} = - 0.0714 = - 1 + 0.9286 = 1 ˉ .9286$
Why: The bar notation keeps the mantissa positive and puts the minus sign on the characteristic alone, so $- 0.0714$ is written as $1 ˉ .9286$ - the same number, in the form the options use.
Bar notation for a negative characteristic

Common mistakes

Option A

Took $lo g_{10} 0.72$ as $+ 1.8572$ , dropping the $- 2$ , and then halved to get $0.9286$ - a positive answer, when the logarithm of a number below $1$ must be negative.

Option C

Stopped at $3 lo g_{10} 2 + 2 lo g_{10} 3 = 1.8572$ , forgetting both the $- 2$ from dividing by $100$ and the halving for base $100$ .

Option D

Computed $lo g_{10} 0.72 = 1 ˉ .8572 = - 0.1428$ correctly but forgot the base is $100$ , so never divided by $2$ .

Skills used

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

and

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

Logarithm to exponential form

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

Bar notation for a negative characteristic

15Algebra • Logarithm

ModerateIdeal time: about 1.5 min

If $a^{x} = b^{y} = c^{z}$ and $ab c = 1$ , then the value of $\frac{1}{x} + \frac{1}{y} + \frac{1}{z}$ will be equal to

- 1

0

1

3

Show answer & solution

Option B

0

Approach

Name the common value $k$ , express $a$ , $b$ and $c$ as powers of $k$ , and read the index off $ab c = 1$ .

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

Step-by-step solution

1
Let $a^{x} = b^{y} = c^{z} = k$ , so $a = k^{1/ x}$ , $b = k^{1/ y}$ , $c = k^{1/ z}$
Why: Naming the common value is the only way to get the three separate equations onto one base; taking the $x$ -th root of $a^{x} = k$ inverts the index.
Power of a power, and the move to a common base
2
$ab c = k^{1/ x} \times k^{1/ y} \times k^{1/ z} = k^{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$
Why: Multiplying powers of the same base adds the indices - and the sum of indices is exactly the quantity the question asks for.
3
$ab c = 1 = k^{0}$
Why: The given condition is now a statement about a power of $k$ , and $1$ is $k$ to the power zero.
4
$k^{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} = k^{0} \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$
Why: Equal powers of the same base force equal indices, provided $k$ is positive and not $1$ - which it is, since $k = 1$ would make $a = b = c = 1$ and leave $x, y, z$ undetermined.

Shortcut

Take logarithms: $x lo g a = y lo g b = z lo g c = lo g k$ , so $lo g a + lo g b + lo g c = lo g k (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$ . The left side is $lo g (ab c) = lo g 1 = 0$ , and $lo g k \neq = 0$ , so the bracket vanishes.

Common mistakes

Option A

Read $ab c = 1$ as $k^{- 1}$ rather than $k^{0}$ , so the sum of indices came out as $- 1$ .

Option C

Carried the value $1$ from $ab c = 1$ straight across as the answer, instead of turning it into the index $0$ .

Option D

Assumed $a = b = c$ forces $x = y = z$ with each reciprocal equal to $1$ , and added three of them - an assumption the question never makes.

Skills used

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

16Algebra • Quadratic Equations • Roots

ModerateIdeal time: about 1.5 min

If $α$ and $β$ are the roots of the equation $a x^{2} + b x + c = 0$ , then the value of $\frac{1}{a α + b} + \frac{1}{a β + b}$ is

\frac{a}{b c}

\frac{b}{a c}

\frac{c}{ab}

\frac{1}{ab c}

Show answer & solution

Option B

\frac{b}{a c}

Approach

Add the two fractions over a common denominator, then replace $α + β$ and $α β$ by their values in $a$ , $b$ , $c$ .

Sum and product of roots (Vieta)

Step-by-step solution

1
$α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$
Why: The expression is symmetric in $α$ and $β$ , so it can be written using the sum and the product alone - the roots never have to be found.
Sum and product of roots (Vieta)
2
$\frac{1}{a α + b} + \frac{1}{a β + b} = \frac{( a β + b ) + ( a α + b )}{( a α + b ) ( a β + b )}$
Why: One common denominator turns two fractions into one, and both the new numerator and the new denominator are symmetric.
Clear the denominators of an equation
3
Numerator $= a (α + β) + 2 b = a (- \frac{b}{a}) + 2 b = - b + 2 b = b$
Why: The $a$ cancels against the $a$ in $- \frac{b}{a}$ , leaving a single $b$ .
4
Denominator $= a^{2} α β + ab (α + β) + b^{2} = a^{2} \cdot \frac{c}{a} + ab (- \frac{b}{a}) + b^{2} = a c - b^{2} + b^{2} = a c$
Why: The two $b^{2}$ terms cancel exactly, which is what makes the answer as clean as $\frac{b}{a c}$ .
5
Sum $= \frac{b}{a c}$
Why: Numerator over denominator, with nothing left to simplify.

Shortcut

Since $α$ is a root, $a α^{2} + b α + c = 0$ , so $α (a α + b) = - c$ and $a α + b = - \frac{c}{α}$ ; likewise for $β$ . The sum is then $- \frac{α + β}{c} = \frac{b}{a c}$ in one line.

Common mistakes

Option A

No route reaches $\frac{a}{b c}$ ; it is a permutation of the three letters with no derivation behind it.

Option C

No route reaches $\frac{c}{ab}$ ; it is a permutation of the three letters with no derivation behind it.

Option D

No route reaches $\frac{1}{ab c}$ ; it is a permutation of the three letters with no derivation behind it.

Skills used

Sum and product of roots (Vieta)Clear the denominators of an equation

17Arithmetic • Number System • Divisibility

EasyIdeal time: about 1 min

Consider the following statements in respect of three 3-digit numbers XYZ, YZX and ZXY :

1: The sum of the numbers is not divisible by (X $+$ Y $+$ Z).

2: The sum of the numbers is divisible by $111$ .

Which of the above statements is/are correct ?

1

onlyB

2

onlyCBoth

1

and

2

DNeither

1

nor

2

Show answer & solution

Option B

2

only

Approach

Expand the three numerals by place value and add; the sum factorises, and both statements are then decided by reading the factors.

Place-value expansion of a numeral

Step-by-step solution

1
$X Y Z = 100 X + 10 Y + Z$ , $Y Z X = 100 Y + 10 Z + X$ , $Z X Y = 100 Z + 10 X + Y$
Why: Each digit occupies a different place in each of the three numbers, so writing them out is what makes the pattern visible.
Place-value expansion of a numeral
2
Sum $= (100 + 10 + 1) X + (100 + 10 + 1) Y + (100 + 10 + 1) Z = 111 (X + Y + Z)$
Why: Every digit appears once in the hundreds place, once in the tens and once in the units, so each is multiplied by $111$ - that symmetry is the whole content of the question.
3
Statement 1 is false: $111 (X + Y + Z)$ is plainly divisible by $(X + Y + Z)$ . Statement 2 is true: it is $111$ times an integer.
Why: The single factorised form settles both statements at once, and it shows they cannot both hold - statement 1 denies exactly what the factor $(X + Y + Z)$ asserts.

Common mistakes

Option A

Got the sum $111 (X + Y + Z)$ but then tested statement 1 as though $(X + Y + Z)$ were unrelated to the sum, marking the non-divisibility claim as correct.

Option C

Verified statement 2, which is easy, and accepted statement 1 without testing it - missing that the factorisation makes the two statements contradict each other.

Option D

Mis-added the place values, so the sum did not come out as a multiple of $111$ and both statements looked false.

Skills used

Place-value expansion of a numeral

18Algebra • Equations

HardIdeal time: about 2.5 min

The number of all pairs (m, n), where m and n are positive integers, such that $\frac{1}{m} + \frac{1}{n} - \frac{1}{mn} = \frac{2}{5}$ is

6

5

4

2

Show answer & solution

Option C

4

Approach

Clear the fractions, solve for $n$ in terms of $m$ , and use the fact that $n$ must be a positive integer to close the search down to a handful of values of $m$ .

Clear the denominators of an equation

Step-by-step solution

1
Multiply through by $5 mn$ : $5 n + 5 m - 5 = 2 mn$
Why: Nothing can be counted while the unknowns sit in denominators; clearing them first makes the equation a relation between two integers.
Clear the denominators of an equation
2
$5 m - 5 = 2 mn - 5 n = n (2 m - 5) \Rightarrow n = \frac{5 m - 5}{2 m - 5}$
Why: Collecting the $n$ terms on one side and factoring gives $n$ explicitly, so each value of $m$ can simply be tested.
3
$n > 0$ needs $2 m - 5 > 0$ , so $m \geq 3$ ; and $n \geq 3$ needs $5 m - 5 \geq 6 m - 15$ , so $m \leq 10$
Why: $n = \frac{5 m - 5}{2 m - 5}$ always exceeds $\frac{5}{2}$ , so $n \geq 3$ ; feeding that back bounds $m$ above. Bounding the search is what turns an infinite question into eight tests. ( $m = 1$ gives $n = 0$ and $m = 2$ gives $n = - 5$ , both rejected.)
Enumerate a residue class within a bound
4
$m = 3 \to n = \frac{10}{1} = 10$ ; $m = 4 \to \frac{15}{3} = 5$ ; $m = 5 \to \frac{20}{5} = 4$ ; $m = 6, 7, 8, 9 \to \frac{25}{7}, \frac{30}{9}, \frac{35}{11}, \frac{40}{13}$ all non-integer; $m = 10 \to \frac{45}{15} = 3$
Why: Eight divisions settle it; only four leave a whole number.
5
The pairs are $(3, 10)$ , $(4, 5)$ , $(5, 4)$ , $(10, 3)$ - four in all
Why: The question asks for pairs $(m, n)$ , which are ordered, so $(3, 10)$ and $(10, 3)$ both count.

Shortcut

Rearrange to $2 mn - 5 m - 5 n + 5 = 0$ , double it and complete the product: $(2 m - 5) (2 n - 5) = 15$ . With $m, n$ positive both brackets are odd and at least $- 3$ , so only $1 \times 15$ , $3 \times 5$ , $5 \times 3$ , $15 \times 1$ survive - four pairs, read off without any division.

Common mistakes

Option A

Counted the negative factorisations $(- 3) (- 5)$ and $(- 5) (- 3)$ of $15$ as well; they give $(m, n) = (1, 0)$ and $(0, 1)$ , which are not pairs of positive integers.

Option B

Admitted one boundary case - usually $m = 1$ , $n = 0$ - alongside the four genuine pairs.

Option D

Counted ${3, 10}$ and ${4, 5}$ as unordered pairs, so the two reversals were not counted separately.

Skills used

Clear the denominators of an equation Enumerate a residue class within a bound

19Algebra • Indices

HardIdeal time: about 2.5 min

If $a = x y^{p - 1}$ , $b = y z^{q - 1}$ , $c = z x^{r - 1}$ , then $a^{q - r} b^{r - p} c^{p - q}$ is equal to

ab c

x y z

0

DNone of the above

Show answer & solution

Option D

None of the above

Approach

Substitute and collect the index of each of $x$ , $y$ , $z$ in the product, then compare those indices against the ones the options claim.

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

Step-by-step solution

1
$a^{q - r} b^{r - p} c^{p - q} = (x y^{p - 1})^{q - r} (y z^{q - 1})^{r - p} (z x^{r - 1})^{p - q}$
Why: There is nothing to use but the three definitions, so substitute them and let the index laws do the rest.
Power of a power, and the move to a common base
2
Index of $x$ : $(q - r) + (r - 1) (p - q)$ ; of $y$ : $(p - 1) (q - r) + (r - p)$ ; of $z$ : $(q - 1) (r - p) + (p - q)$
Why: Each bracket contributes to two of the three letters; raising a power to a power multiplies the indices, and powers of the same letter then add.
3
With $p = 2$ , $q = 3$ , $r = 4$ these indices are $- 4$ , $1$ , $3$ , so the product is $x^{- 4} y z^{3}$
Why: The indices are not identically zero, so the product is not $1$ ; putting in one convenient set of values makes that concrete and gives something to compare.
4
$ab c = x^{1 + r - 1} y^{p - 1 + 1} z^{q - 1 + 1} = x^{r} y^{p} z^{q} = x^{4} y^{2} z^{3}$ for the same $p, q, r$ , and $x y z = x y z$ ; neither matches $x^{- 4} y z^{3}$ , and a product of powers of non-zero quantities is never $0$
Why: Options A, B and C are each ruled out by a single comparison of indices, so the only remaining answer is D.

Common mistakes

Option A

Assumed the product must collapse to the three quantities multiplied together, without checking that the index of $x$ in $ab c$ is $r$ while in the product it is $(q - r) + (r - 1) (p - q)$ .

Option B

Expected the cyclic pattern to cancel down to $x y z$ , which happens only in the special case $x = y = z$ - there the product is $1$ , not $x y z$ .

Option C

Read the index sum $p (q - r) + q (r - p) + r (p - q) = 0$ , which is the cancellation that occurs when $x = y = z$ , and reported the index $0$ as the value of the expression instead of $x^{0} = 1$ .

Skills used

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

20Geometry • Polygon • Interior and Exterior Angles

HardIdeal time: about 2 min

The number of sides of two regular polygons are in the ratio $5 : 4$ . The difference between their interior angles is $9^{\circ}$ . Consider the following statements :

1: One of them is a pentagon and the other is a rectangle.

2: One of them is a decagon and the other is an octagon.

3: The sum of their exterior angles is $72 0^{\circ}$ .

Which of the above statements is/are correct ?

1

onlyB

2

onlyC

1

and

3

2

and

3

Show answer & solution

Option D

2

and

3

Approach

Write the two side counts as $5 k$ and $4 k$ , form the difference of interior angles, solve for $k$ , then test each statement.

Interior and exterior angle sums of a polygon

Step-by-step solution

1
Let the numbers of sides be $5 k$ and $4 k$
Why: A ratio fixes the two counts only up to a common factor, and the $9^{\circ}$ condition is what pins that factor down - so $k$ must be carried, not assumed to be $1$ .
Introduce a common factor $k$ for a ratio
2
Interior angle of a regular $n$ -gon $= 18 0^{\circ} - \frac{36 0 ^{\circ}}{n}$
Why: The interior angles sum to $(n - 2) 18 0^{\circ}$ , so each of the $n$ equal angles is $18 0^{\circ} - \frac{36 0 ^{\circ}}{n}$ ; written this way the difference is easy to take.
Interior and exterior angle sums of a polygon
3
Difference $= \frac{36 0 ^{\circ}}{4 k} - \frac{36 0 ^{\circ}}{5 k} = \frac{9 0 ^{\circ}}{k} - \frac{7 2 ^{\circ}}{k} = \frac{1 8 ^{\circ}}{k} = 9^{\circ} \Rightarrow k = 2$
Why: The $18 0^{\circ}$ cancels, leaving only the two correction terms; the polygon with fewer sides has the smaller interior angle, so subtract that way round.
Clear the denominators of an equation
4
The polygons have $10$ and $8$ sides - a decagon and an octagon
Why: $5 k = 10$ and $4 k = 8$ . So statement 2 is correct and statement 1, which describes $k = 1$ , is not.
5
Exterior angles of each polygon sum to $36 0^{\circ}$ , so for the two together $36 0^{\circ} + 36 0^{\circ} = 72 0^{\circ}$
Why: The exterior-angle sum is $36 0^{\circ}$ whatever the number of sides, so both polygons contribute the same amount and statement 3 is correct.

Common mistakes

Option A

Read the ratio $5 : 4$ as the side counts themselves, so found a pentagon and a quadrilateral, and also rejected statement 3 by taking $36 0^{\circ}$ for the two polygons together rather than for each.

Option B

Solved for $k = 2$ correctly but read statement 3 as a claim about one polygon, where the exterior sum is $36 0^{\circ}$ , instead of about both.

Option C

Read the ratio as the side counts themselves - ignoring the $9^{\circ}$ condition, which exists precisely to fix $k = 2$ - while getting statement 3 right.

Skills used

Introduce a common factor

k

for a ratio Interior and exterior angle sums of a polygon Clear the denominators of an equation