Elementary Mathematics 2019 (I)

Set A

Questions 1–20 of 100

1Arithmetic • Number System • Divisibility

EasyIdeal time: about 1 min

What is the remainder when $(1 7^{29} + 1 9^{29})$ is divided by $18$ ?

6

2

1

0

Show answer & solution

Option D

0

Approach

Reduce each base modulo $18$ first: $17$ and $19$ sit one on either side of $18$ , so both powers collapse without being computed.

Reduce a large power by the cycle of its base modulo

n

Step-by-step solution

1
$17 \equiv - 1 (mod 18)$ and $19 \equiv 1 (mod 18)$
Why: Both bases are within $1$ of the divisor, so replacing each by its residue is the cheapest possible simplification.
Reduce a large power by the cycle of its base modulo $n$
2
$1 7^{29} \equiv (- 1)^{29} = - 1 (mod 18)$
Why: $29$ is odd, so the negative sign survives the power — this is the step the whole question turns on.
3
$1 9^{29} \equiv 1^{29} = 1 (mod 18)$
Why: Every power of $1$ is $1$ .
4
$1 7^{29} + 1 9^{29} \equiv - 1 + 1 = 0 (mod 18)$
Why: The two residues cancel, so the sum is an exact multiple of $18$ and the remainder is $0$ .

Shortcut

$a + b$ always divides $a^{n} + b^{n}$ when $n$ is odd. Here $17 + 19 = 36$ , a multiple of $18$ , so $18$ divides $1 7^{29} + 1 9^{29}$ and the remainder is $0$ .

Common mistakes

Option A

Filler — no route reaches $6$ .

Option B

Read $(- 1)^{29}$ as $+ 1$ out of even-exponent habit, giving $1 + 1 = 2$ .

Option C

Reduced only $19$ and set $1 7^{29}$ aside, reporting the $1$ that $1 9^{29}$ contributes.

Skills used

Reduce a large power by the cycle of its base modulo

n

2Arithmetic • Number System • Divisibility

ModerateIdeal time: about 2 min

What is the largest value of n such that $1 0^{n}$ divides the product $2^{5} \times 3^{3} \times 4^{8} \times 5^{3} \times 6^{7} \times 7^{6} \times 8^{12} \times 9^{9} \times 1 0^{6} \times 1 5^{12} \times 2 0^{14} \times 2 2^{11} \times 2 5^{15}$ ?

65

55

50

45

Show answer & solution

Option A

65

Approach

Each factor of $10$ needs one $2$ and one $5$ , so count the exponent of $2$ and of $5$ in the product and take the smaller.

Prime factorisation

Step-by-step solution

1
$4^{8} = 2^{16}$ , $6^{7} = 2^{7} 3^{7}$ , $8^{12} = 2^{36}$ , $1 0^{6} = 2^{6} 5^{6}$ , $1 5^{12} = 3^{12} 5^{12}$ , $2 0^{14} = 2^{28} 5^{14}$ , $2 2^{11} = 2^{11} 1 1^{11}$ , $2 5^{15} = 5^{30}$
Why: Every composite base has to be broken into primes before its contribution of $2$ s and $5$ s can be read off.
Prime factorisation Power of a power, and the move to a common base
2
exponent of $5 = 3 + 6 + 12 + 14 + 30 = 65$
Why: Collect the fives from $5^{3}$ , $1 0^{6}$ , $1 5^{12}$ , $2 0^{14}$ and $2 5^{15}$ ; no other base is a multiple of $5$ .
3
exponent of $2 = 5 + 16 + 7 + 36 + 6 + 28 + 11 = 109$
Why: Collect the twos from $2^{5}$ , $4^{8}$ , $6^{7}$ , $8^{12}$ , $1 0^{6}$ , $2 0^{14}$ and $2 2^{11}$ .
4
$n = min (109, 65) = 65$
Why: $1 0^{n} = 2^{n} 5^{n}$ , so the scarcer prime caps $n$ ; the fives run out long before the twos do.

Shortcut

Only the fives can possibly be the binding constraint here, since seven of the thirteen factors contribute twos and only five contribute fives. Counting the fives alone gives $65$ , and one glance at $8^{12} = 2^{36}$ confirms the twos are in no danger of running short.

Common mistakes

Option B

Filler — no route reaches $55$ .

Option C

Counted $2 5^{15}$ as fifteen fives rather than thirty, giving $3 + 6 + 12 + 14 + 15 = 50$ .

Option D

Counted only the bases written with a $5$ in them ( $5^{3}$ , $1 5^{12}$ , $2 5^{15}$ ) for $3 + 12 + 30 = 45$ , missing the fives hidden inside $1 0^{6}$ and $2 0^{14}$ .

Skills used

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

3Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

How many pairs (A, B) are possible in the number $479865 A B$ if the number is divisible by $9$ and it is given that the last digit of the number is odd ?

5

6

9

11

Show answer & solution

Option A

5

Approach

Divisibility by $9$ fixes $A + B$ modulo $9$ ; the digit range then leaves only two possible sums, and the odd- $B$ condition is applied to each.

A number is congruent to its digit sum modulo

9

Step-by-step solution

1
$4 + 7 + 9 + 8 + 6 + 5 = 39$ , so the digit sum of $479865 A B$ is $39 + A + B$
Why: A number is divisible by $9$ exactly when its digit sum is, so only the sum of the six known digits is needed.
A number is congruent to its digit sum modulo $9$
2
$39 + A + B \equiv 0 (mod 9)$ and $39 \equiv 3 (mod 9)$ , so $A + B \equiv 6 (mod 9)$
Why: $39 = 36 + 3$ , and $A + B$ must supply the missing $6$ .
3
$0 \leq A + B \leq 18$ , so $A + B = 6$ or $A + B = 15$
Why: Two digits cannot sum beyond $18$ , so the residue class $6, 15, 24, \dots$ stops after two values.
Enumerate a residue class within a bound
4
$A + B = 6$ with $B$ odd: $(A, B) = (5, 1), (3, 3), (1, 5)$
Why: $B = 7$ or $B = 9$ would force $A$ negative, so the list closes at three.
5
$A + B = 15$ with $B$ odd: $(A, B) = (8, 7), (6, 9)$ . Total $3 + 2 = 5$ pairs.
Why: $A = 15 - B \leq 9$ needs $B \geq 6$ , so only $B = 7$ and $B = 9$ survive; the two sums are mutually exclusive, so the counts add.

Common mistakes

Option B

Kept one pair in which $A$ falls outside $0$ to $9$ , such as $(A, B) = (10, 5)$ from $A + B = 15$ .

Option C

Filler — no route reaches $9$ .

Option D

Counted every pair with $A + B \equiv 6 (mod 9)$ — seven with sum $6$ and four with sum $15$ — ignoring the condition that the last digit is odd.

Skills used

A number is congruent to its digit sum modulo

9

Enumerate a residue class within a bound

4Arithmetic • Number System • Divisibility

ModerateIdeal time: about 1.5 min

Consider the multiplication $999 \times abc = def 132$ in decimal notation, where a, b, c, d, e and f are digits. What are the values of a, b, c, d, e and f respectively ?

6, 6, 8, 6, 8, 7

8, 6, 8, 6, 7, 8

6, 8, 8, 7, 8, 6

8, 6, 8, 8, 6, 7

Show answer & solution

Option D

8, 6, 8, 8, 6, 7

Approach

Write $999 = 1000 - 1$ , so the product becomes a shift minus the number itself; the last three digits then pin $ab c$ immediately.

Place-value expansion of a numeral

Step-by-step solution

1
Let $N = ab c$ . Then $999 N = 1000 N - N$ .
Why: $999$ is one less than a power of ten, so the multiplication turns into a subtraction and no long multiplication is needed.
Closed form of a repeated-digit number
2
$1000 N - N = 1000 (N - 1) + (1000 - N)$
Why: Borrowing one thousand makes both parts three-digit blocks: $N - 1$ occupies the top three places and $1000 - N$ the bottom three.
Place-value expansion of a numeral
3
The bottom three digits are $132$ , so $1000 - N = 132$ and $N = 868$ : $a = 8$ , $b = 6$ , $c = 8$ .
Why: The product is printed as $d e f 132$ , so its last three digits are exactly the $1000 - N$ block.
4
The top three digits are $N - 1 = 867$ , so $d = 8$ , $e = 6$ , $f = 7$ .
Why: That is the other block, and it fixes the remaining three letters.

Common mistakes

Option A

Transposed the digits of $ab c$ to $668$ ; $999 \times 668 = 667332$ does not end in $132$ .

Option B

Found $ab c = 868$ and $d e f = 867$ correctly, then wrote the second block's digits in the wrong order as $678$ .

Option C

Filler — $999 \times 688 = 687312$ , which matches neither block.

Skills used

Place-value expansion of a numeral Closed form of a repeated-digit number

5Arithmetic • Time, Speed and Distance • Relative Speed

ModerateIdeal time: about 1.5 min

Three cars A, B and C started from a point at $5$ p.m., $6$ p.m. and $7$ p.m. respectively and travelled at uniform speeds of $60$ km/hr, $80$ km/hr and $x$ km/hr respectively in the same direction. If all the three met at another point at the same instant during their journey, then what is the value of $x$ ?

120

110

105

100

Show answer & solution

Option A

120

Approach

A and B fix the meeting point between themselves; C's speed then follows from covering that same distance in its own running time.

Speed, distance and time

Step-by-step solution

1
At $6$ p.m. A is $60 \times 1 = 60$ km ahead of B
Why: A has been travelling one hour longer, so the gap B must close is one hour of A's speed.
Speed, distance and time
2
Closing speed $= 80 - 60 = 20$ km/hr, so B catches A after $\frac{60}{20} = 3$ hours, at $9$ p.m.
Why: Both travel the same way, so the gap shrinks at the difference of the speeds.
Relative speed: add when opposed, subtract when alike
3
The meeting point is $80 \times 3 = 240$ km from the start
Why: Use B's own motion; A gives the same figure, $60 \times 4 = 240$ km, which is a free check.
4
C runs from $7$ p.m. to $9$ p.m., so $x = \frac{240}{2} = 120$
Why: All three are at the same place at the same instant, so C must cover the same $240$ km in the $2$ hours available to it.

Common mistakes

Option B

Filler — no route reaches $110$ .

Option C

Filler — no route reaches $105$ .

Option D

Filler — $100$ km/hr for $2$ hours gives $200$ km, which A reaches at $8$ : $20$ p.m. and B at $8$ : $30$ p.m., so the three never meet.

Skills used

Speed, distance and time Relative speed: add when opposed, subtract when alike

6Arithmetic • Age Problems

HardIdeal time: about 2.5 min

Priya's age was cube of an integral number (different from $1$ ) four years ago and square of an integral number after four years. How long should she wait so that her age becomes square of a number in the previous year and cube of a number in the next year ?

7

yearsB

12

yearsC

14

yearsD

21

years

Show answer & solution

Option C

14

years

Approach

Pin her present age from the cube-and-square pair four years either side, then find the next age that is one more than a square and one less than a cube.

Turn a worded condition into a linear equation in one unknown

Step-by-step solution

1
Let her present age be $P$ : $P - 4$ is a cube other than $1$ , and $P + 4$ is a perfect square.
Why: Translate the two clauses directly; the exclusion of $1$ is what rules out the childhood answer $P = 5$ .
Turn a worded condition into a linear equation in one unknown
2
Cubes above $1$ are $8, 27, 64, \dots$ ; $8 + 8 = 16 = 4^{2}$ works, while $27 + 8 = 35$ and $64 + 8 = 72$ are not squares.
Why: An age bounds the search to a handful of cubes, and each candidate is settled by bracketing $P + 4$ between consecutive squares.
Enumerate a residue class within a bound Least perfect square not below a given bound
3
$P = 12$
Why: Four years ago she was $8 = 2^{3}$ , and in four years she will be $16 = 4^{2}$ .
4
Wait $t$ years. Her age $A = 12 + t$ must have $A - 1$ a perfect square and $A + 1$ a perfect cube.
Why: 'Square of a number in the previous year and cube of a number in the next year' refers to the ages either side of $A$ , not to $A$ itself — this is the reading the question hinges on.
5
$A + 1 = 27 \Rightarrow A = 26$ , and $A - 1 = 25 = 5^{2}$ . Hence $t = 26 - 12 = 14$ .
Why: Test the cubes in turn: $A + 1 = 8$ gives $A - 1 = 6$ , not a square; $27$ works; $64$ gives $62$ , not a square.

Common mistakes

Option A

Filler — no route reaches $7$ .

Option B

Reported her present age, $12$ , instead of the number of years she must wait.

Option D

Filler — no route reaches $21$ .

Skills used

Turn a worded condition into a linear equation in one unknown Enumerate a residue class within a bound Least perfect square not below a given bound

7Arithmetic • Number System

ModerateIdeal time: about 2 min

Which of the following statements is not true ?

AThe difference of two prime numbers, both greater than

2

, is divisible by

2

.BFor two different integers m, n and a prime number p, if p divides the product

m \times n

, then p divides either m or n.CIf a number is of the form

6 n - 1

(n being a natural number), then it is a prime number.DThere is only one set of three prime numbers such that there is a gap of

2

between two adjacent prime numbers.

Show answer & solution

Option C

If a number is of the form

6 n - 1

(n being a natural number), then it is a prime number.

Approach

Judge each statement in turn; the false one will fall to a single counterexample.

Prime factorisation

Step-by-step solution

1
(A) Two primes above $2$ are both odd. Writing them $2 m + 1$ and $2 n + 1$ , the difference is $2 (m - n)$ , a multiple of $2$ . True.
Why: The only even prime is $2$ , which the statement excludes, so both numbers are forced odd.
Write an odd number as $2 m + 1$
2
(B) If a prime $p$ divides $mn$ , then $p ∣ m$ or $p ∣ n$ . True — this is Euclid's lemma.
Why: The hypothesis names $p$ as prime, which is exactly the condition the lemma needs; it is what makes prime factorisation unique.
Euclid's lemma
3
(C) $n = 6$ gives $6 n - 1 = 35 = 5 \times 7$ , which is composite. False.
Why: Every prime above $3$ is of the form $6 n \pm 1$ , but the converse fails — the form supplies candidates, not primes. One counterexample settles it.
Prime factorisation
4
(D) Among $p$ , $p + 2$ , $p + 4$ one is always a multiple of $3$ , so $3, 5, 7$ is the only such triple. True.
Why: Three numbers two apart cover all three residues modulo $3$ ; all three can be prime only when the multiple of $3$ is $3$ itself.
Enumerate a residue class within a bound

Common mistakes

Option A

Read 'both greater than $2$ ' loosely and tested $3 - 2 = 1$ , which is odd, so judged (A) false.

Option B

Recalled Euclid's lemma as requiring $m$ and $n$ to be prime as well, and so judged (B) false.

Option D

Thought of $5, 7, 11$ or $11, 13, 17$ as further triples, overlooking that in those the two gaps are not both $2$ .

Skills used

Prime factorisation Write an odd number as

2 m + 1

Euclid's lemma Enumerate a residue class within a bound

8Algebra • Expressions • Maximum and Minimum Values

EasyIdeal time: about 1 min

For $x > 0$ , what is the minimum value of $x + \frac{x + 2}{2 x}$ ?

1

2

2 \frac{1}{2}

DCannot be determined

Show answer & solution

Option C

2 \frac{1}{2}

Approach

Split the fraction so that $x$ and $\frac{1}{x}$ stand together, then apply $x + \frac{1}{x} \geq 2$ .

A positive number plus its reciprocal is at least

2

Step-by-step solution

1
$x + \frac{x + 2}{2 x} = x + \frac{x}{2 x} + \frac{2}{2 x} = x + \frac{1}{2} + \frac{1}{x}$
Why: Splitting the numerator peels off a constant and leaves the reciprocal term exposed; the expression is now a sum whose minimum is a standard result.
2
For $x > 0$ , $x + \frac{1}{x} \geq 2$ , with equality exactly at $x = 1$
Why: $x + \frac{1}{x} - 2 = \frac{( x - 1 ) ^{2}}{x} \geq 0$ , and $x = 1$ lies inside the permitted range $x > 0$ , so the bound is actually attained.
A positive number plus its reciprocal is at least $2$
3
minimum $= 2 + \frac{1}{2} = \frac{5}{2} = 2 \frac{1}{2}$
Why: The constant $\frac{1}{2}$ lifts the minimum value but does not move the point where it occurs.

Common mistakes

Option A

Filler — no route reaches $1$ .

Option B

Applied $x + \frac{1}{x} \geq 2$ but dropped the $\frac{1}{2}$ left over from splitting the fraction.

Option D

Concluded no minimum exists because the domain $x > 0$ is open, overlooking that the minimum sits at the interior point $x = 1$ .

Skills used

A positive number plus its reciprocal is at least

2

9Algebra • Surds

HardIdeal time: about 2.5 min

If $\frac{1 + p x}{1 - p x} \frac{1 - q x}{1 + q x} = 1$ , then what are the non-zero solutions of x ?

\pm \frac{1}{p} \frac{2 p - q}{q}, 2 p \neq = q

\pm \frac{1}{pq} p - q, p \neq = q

\pm \frac{p}{q} p - q, p \neq = q

\pm \frac{q}{p} 2 p - q, 2 p \neq = q

Show answer & solution

Option A

\pm \frac{1}{p} \frac{2 p - q}{q}, 2 p \neq = q

Approach

Square the equation to clear the radical, cross-multiply, and the whole thing collapses to a quadratic in $x$ with $x = 0$ factored out.

Square both sides to clear a radical

Step-by-step solution

1
$\frac{1 + p x}{1 - p x} \frac{1 - q x}{1 + q x} = 1 \Rightarrow (\frac{1 + p x}{1 - p x})^{2} \cdot \frac{1 - q x}{1 + q x} = 1$
Why: Squaring is the only way to remove the radical, and squaring the whole equation keeps the two sides equal.
Square both sides to clear a radical $ab = a b$ and $a / b = a / b$
2
$(1 + p x)^{2} (1 - q x) = (1 - p x)^{2} (1 + q x)$
Why: Cross-multiplying clears both denominators; neither $1 - p x$ nor $1 + q x$ may vanish, which is what the conditions attached to the options record.
Clear the denominators of an equation
3
$(1 + 2 p x + p^{2} x^{2}) (1 - q x) - (1 - 2 p x + p^{2} x^{2}) (1 + q x) = 4 p x - 2 q x - 2 p^{2} q x^{3} = 0$
Why: Expanding both squares, every term carrying an even number of sign-changing factors cancels; only $2 p x$ , $- q x$ and $- p^{2} q x^{3}$ survive, each doubled.
Square of a sum and of a difference
4
$2 x (2 p - q - p^{2} q x^{2}) = 0$
Why: Taking out $2 x$ separates the root $x = 0$ , which the question discards, from the two that are wanted.
A product is zero only when a factor is zero
5
$x^{2} = \frac{2 p - q}{p ^{2} q} \Rightarrow x = \pm \frac{1}{p} \frac{2 p - q}{q}$
Why: Only $p^{2}$ comes out of the radical, leaving $\frac{1}{p}$ outside and $\frac{2 p - q}{q}$ inside; $2 p \neq = q$ is what keeps the root non-zero.

Common mistakes

Option B

Dropped the factor $2$ that the cross-terms $2 p x$ contribute, landing on $x^{2} = \frac{p - q}{p ^{2} q ^{2}}$ .

Option C

Lost the $2$ and also inverted $\frac{1}{p ^{2} q}$ into $\frac{p ^{2}}{q ^{2}}$ when taking the square root.

Option D

Reached $x^{2} = \frac{2 p - q}{p ^{2} q}$ correctly but pulled the $q$ out of the denominator as a factor $q$ instead of leaving it under the radical.

Skills used

Square both sides to clear a radical

ab = a b

and

a / b = a / b

Clear the denominators of an equation Square of a sum and of a difference A product is zero only when a factor is zero

10Arithmetic • Percentage • Successive Increase and Decrease

EasyIdeal time: about 40 s

In a hostel the rent per room is increased by $20%$ . If number of rooms in the hostel is also increased by $20%$ and the hostel is always full, then what is the percentage change in the total collection at the cash counter ?

30%

40%

44%

48%

Show answer & solution

Option C

44%

Approach

Total collection is rent per room times number of rooms, so the two $20%$ rises multiply rather than add.

Percentage change as a multiplier

Step-by-step solution

1
collection $=$ (rent per room) $\times$ (number of rooms)
Why: The hostel is always full, so every room contributes its full rent and nothing else enters the total.
2
new collection $= 1.2 \times 1.2 \times old = 1.44 \times old$
Why: A rise of $20%$ multiplies by $1.2$ ; two independent rises multiply their factors.
Percentage change as a multiplier
3
change $= 1.44 - 1 = 0.44$ , an increase of $44%$
Why: Subtracting $1$ from the overall multiplier reads off the percentage change directly.

Common mistakes

Option A

Filler — no route reaches $30%$ .

Option B

Added the two percentages instead of multiplying the two factors.

Option D

Added the cross term twice: $20 + 20 + 2 \times 4$ .

Skills used

Percentage change as a multiplier

11Arithmetic • Time, Speed and Distance • Relative Speed

EasyIdeal time: about 1 min

Radha and Hema are neighbours and study in the same school. Both of them use bicycles to go to the school. Radha's speed is $8$ km/hr whereas Hema's speed is $10$ km/hr. Hema takes $9$ minutes less than Radha to reach the school. How far is the school from the locality of Radha and Hema ?

5

kmB

5.5

kmC

6

kmD

6.5

Show answer & solution

Option C

6

Approach

Both cover the same distance, so write each journey time as $\frac{d}{speed}$ and set the difference equal to $9$ minutes.

Speed, distance and time

Step-by-step solution

1
$\frac{d}{8} - \frac{d}{10} = \frac{9}{60}$ hours
Why: Radha is slower, so hers is the larger time; the $9$ minutes must be turned into hours to match speeds in km/hr.
Speed, distance and time
2
$\frac{d}{8} - \frac{d}{10} = \frac{10 d - 8 d}{80} = \frac{d}{40}$
Why: Combine over the common denominator $80$ before solving, so only one fraction is left.
Clear the denominators of an equation
3
$\frac{d}{40} = \frac{3}{20} \Rightarrow d = 40 \times \frac{3}{20} = 6$ km
Why: $\frac{9}{60}$ reduces to $\frac{3}{20}$ , which keeps the arithmetic in small numbers.

Common mistakes

Option A

Filler — no route reaches $5$ km.

Option B

Filler — no route reaches $5.5$ km.

Option D

Filler — no route reaches $6.5$ km.

Skills used

Speed, distance and time Clear the denominators of an equation

12Algebra • Logarithm • Equations

EasyIdeal time: about 1 min

Which of the following pair of numbers is the solution of the equation $3^{x + 2} + 3^{- x} = 10$ ?

0, 2

0, - 2

1, - 1

1, 2

Show answer & solution

Option B

0, - 2

Approach

Write both terms in terms of the single quantity $3^{x}$ ; the equation becomes a quadratic.

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

Step-by-step solution

1
$3^{x + 2} = 9 \cdot 3^{x}$ and $3^{- x} = \frac{1}{3 ^{x}}$
Why: Separating the constant part of each index puts everything in terms of $3^{x}$ alone, which is what makes a substitution possible.
Power of a power, and the move to a common base Negative exponent is a reciprocal
2
Put $t = 3^{x}$ : $9 t + \frac{1}{t} = 10 \Rightarrow 9 t^{2} - 10 t + 1 = 0$
Why: $3^{x} > 0$ , so $t \neq = 0$ and multiplying through by $t$ is safe; it clears the reciprocal and leaves a quadratic.
Clear the denominators of an equation
3
$9 t^{2} - 9 t - t + 1 = (9 t - 1) (t - 1) = 0 \Rightarrow t = \frac{1}{9}$ or $t = 1$
Why: Split the middle term using the pair $- 9$ and $- 1$ , whose product is $9 \times 1$ and whose sum is $- 10$ .
Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero
4
$3^{x} = \frac{1}{9} = 3^{- 2} \Rightarrow x = - 2$ ; $3^{x} = 1 = 3^{0} \Rightarrow x = 0$
Why: Equal powers of the same base force equal indices. $\frac{1}{9}$ is $3^{- 2}$ , not $3^{2}$ — that sign is where the question is lost.
Logarithm to exponential form

Common mistakes

Option A

Solved $3^{x} = \frac{1}{9}$ as $x = 2$ , reading the reciprocal as a positive index.

Option C

Filler — $x = 1$ gives $27 + \frac{1}{3}$ , nowhere near $10$ .

Option D

Filler — neither $x = 1$ nor $x = 2$ satisfies the equation.

Skills used

Power of a power, and the move to a common base Negative exponent is a reciprocal Clear the denominators of an equation Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero Logarithm to exponential form

13Algebra • Logarithm • Characteristic and Mantissa

EasyIdeal time: about 1 min

It is given that $lo g_{10} 2 = 0.301$ and $lo g_{10} 3 = 0.477$ . How many digits are there in $(108)^{10}$ ?

19

20

21

22

Show answer & solution

Option C

21

Approach

Rebuild $108$ from $2$ s and $3$ s, take $lo g_{10}$ of the power, then use that a number has one more digit than the integer part of its logarithm.

Number of digits from a common logarithm

Step-by-step solution

1
$108 = 4 \times 27 = 2^{2} \times 3^{3}$
Why: Only $lo g_{10} 2$ and $lo g_{10} 3$ are supplied, so $108$ has to be written with those two primes.
Prime factorisation
2
$lo g_{10} 108 = 2 lo g_{10} 2 + 3 lo g_{10} 3 = 2 (0.301) + 3 (0.477) = 0.602 + 1.431 = 2.033$
Why: The product and power rules turn the factorisation into a sum of multiples of the two tabulated values.
$lo g (ab) = lo g a + lo g b$ and $lo g \frac{a}{b} = lo g a - lo g b$ $lo g a^{n} = n lo g a$
3
$lo g_{10} (10 8^{10}) = 10 \times 2.033 = 20.33$
Why: The power rule brings the exponent $10$ out in front, which is the whole point of taking logarithms here.
4
digits $= ⌊ 20.33 ⌋ + 1 = 21$
Why: $1 0^{20} \leq 10 8^{10} < 1 0^{21}$ , and every number in that range has $21$ digits.
Number of digits from a common logarithm

Common mistakes

Option A

Filler — no route reaches $19$ .

Option B

Stopped at the characteristic $20$ , forgetting that the digit count is one more than it.

Option D

Rounded $20.33$ up to $21$ and then added $1$ as well.

Skills used

Number of digits from a common logarithm Prime factorisation

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

and

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

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

14Arithmetic • Number System • Prime Numbers

ModerateIdeal time: about 1.5 min

The sum of three prime numbers is $100$ . If one of them exceeds another by $36$ , then one of the numbers is

17

29

43

DNone of the above

Show answer & solution

Option D

None of the above

Approach

An even total forces one of the three primes to be $2$ ; the other two then sum to $98$ and differ by $36$ .

Write an odd number as

2 m + 1

Step-by-step solution

1
Three odd numbers add to an odd number, but the total is $100$ , so one prime must be the only even prime, $2$ .
Why: Parity settles which prime is special before any equation is written.
Write an odd number as $2 m + 1$
2
The other two satisfy $p + q = 100 - 2 = 98$ .
Why: Removing the known prime leaves a two-unknown problem.
3
The stated gap of $36$ must be between these two, so $q - p = 36$ .
Why: If the gap were measured from $2$ , the other prime would be $38 = 2 \times 19$ , which is not prime.
Turn a worded condition into a linear equation in one unknown
4
$q = \frac{98 + 36}{2} = 67$ and $p = \frac{98 - 36}{2} = 31$ ; both are prime.
Why: Adding and subtracting the two equations gives each in one line, and primality then has to be checked rather than assumed.
Prime factorisation
5
The three primes are $2$ , $31$ and $67$ — none of which is $17$ , $29$ or $43$ .
Why: The derived numbers must be matched against the printed options, and here none of them appears.

Common mistakes

Option A

Paired $17$ with $53$ (a gap of $36$ ) and stopped there, without checking that the remaining $100 - 70 = 30$ is prime.

Option B

Paired $29$ with $65$ , overlooking that $65 = 5 \times 13$ is not prime.

Option C

Paired $7$ with $43$ and left $50$ as the third number, which is not prime.

Skills used

Write an odd number as

2 m + 1

Turn a worded condition into a linear equation in one unknown Prime factorisation

15Arithmetic • Number System • Rational Numbers

ModerateIdeal time: about 1.5 min

If a, b and c are positive integers such that $a + \frac{1}{b + \frac{1}{c + \frac{1}{2}}} = \frac{16}{23}$ , then what is the mean of a, b and c ?

1

2

1.33

2.33

Show answer & solution

Option B

2

Approach

Collapse the continued fraction from the innermost level outwards, match it against the given value to read off $a$ , $b$ , $c$ , then average them.

Evaluate a nested fraction from the inside out

Step-by-step solution

1
$c + \frac{1}{2} = \frac{2 c + 1}{2}$ , so $\frac{1}{c + \frac{1}{2}} = \frac{2}{2 c + 1}$
Why: The innermost level must be a single fraction before the level outside it means anything.
Evaluate a nested fraction from the inside out
2
$b + \frac{2}{2 c + 1} = \frac{b ( 2 c + 1 ) + 2}{2 c + 1}$ , so the fraction sitting after $a$ is $\frac{2 c + 1}{b ( 2 c + 1 ) + 2}$
Why: Each level is combined over a common denominator and then inverted; keeping everything as fractions keeps the numbers small.
3
With $a = 1$ the remaining fraction is $\frac{23}{16} - 1 = \frac{7}{16}$ . Then $2 c + 1 = 7$ gives $c = 3$ , and $7 b + 2 = 16$ gives $b = 2$ .
Why: $2 c + 1$ is odd, so it must match the odd numerator $7$ directly rather than any multiple of it; $a$ is forced to $1$ because the whole value lies between $1$ and $2$ .
Clear the denominators of an equation
4
mean $= \frac{a + b + c}{3} = \frac{1 + 2 + 3}{3} = 2$
Why: The question asks for the average of the three integers, not for the integers themselves.
Mean of a frequency distribution

Common mistakes

Option A

Filler — no route reaches a mean of $1$ .

Option C

Mis-solved the innermost level and took $c = 1$ , giving $\frac{1 + 2 + 1}{3} = \frac{4}{3} \approx 1.33$ .

Option D

Read $2 c + 1 = 7$ as $2 c - 1 = 7$ and took $c = 4$ , giving $\frac{1 + 2 + 4}{3} = \frac{7}{3} \approx 2.33$ .

Skills used

Evaluate a nested fraction from the inside out Clear the denominators of an equation Mean of a frequency distribution

16Algebra • Sets • Operations

ModerateIdeal time: about 1.5 min

Consider the following for the next three (03) items :

In a certain town of population size $1, 00, 000$ three types of newspapers (I, II and III) are available. The percentages of the people in the town who read these papers are as follows :

Newspaper	Proportion of readers
I	$10%$
II	$30%$
III	$5%$
Both I and II	$8%$
Both II and III	$4%$
Both I and III	$2%$
All the three (I, II and III)	$1%$

What is the number of people who read only one newspaper ?

20, 000

25, 000

30, 000

35, 000

Show answer & solution

Option A

20, 000

Approach

Work in per cent as if it were a count, build the three 'exactly one' Venn regions, add them, and convert to people at the end.

Exactly one, exactly two and at least two of three sets

Step-by-step solution

1
$n (I) = 10$ , $n (II) = 30$ , $n (III) = 5$ , $n (I \cap II) = 8$ , $n (II \cap III) = 4$ , $n (I \cap III) = 2$ , $n (I \cap II \cap III) = 1$ , all in per cent
Why: Percentages of a fixed total behave exactly like counts out of $100$ . Each pair figure already includes the people who read all three, since the table does not say 'but not the third'.
$x %$ of a quantity
2
only I $= 10 - 8 - 2 + 1 = 1$
Why: Strip both overlaps that involve I; the all-three group was removed by each of them, so one copy must be added back.
Exactly one, exactly two and at least two of three sets
3
only II $= 30 - 8 - 4 + 1 = 19$ and only III $= 5 - 4 - 2 + 1 = 0$
Why: The same correction applies to each circle; nobody in this town reads only paper III.
4
exactly one $= 1 + 19 + 0 = 20$ per cent $= \frac{20}{100} \times 1, 00, 000 = 20, 000$
Why: The three regions are disjoint, so they add; convert to people only once, at the end.

Shortcut

With $P = 10 + 30 + 5 = 45$ , $Q = 8 + 4 + 2 = 14$ and $R = 1$ , 'exactly one' is $P - 2 Q + 3 R = 45 - 28 + 3 = 20$ per cent, giving $20, 000$ directly.

Common mistakes

Option B

Filler — no route reaches $25, 000$ .

Option C

Took the $30%$ row for paper II as the answer; that counts everyone who reads II, including those who also read another paper.

Option D

Filler — no route reaches $35, 000$ .

Skills used

Exactly one, exactly two and at least two of three sets

x %

of a quantity

17Algebra • Sets • Operations

EasyIdeal time: about 1.5 min

Same directions as Q16

What is the number of people who read at least two newspapers ?

12, 000

13, 000

14, 000

15, 000

Show answer & solution

Option A

12, 000

Approach

Add the three pairwise overlaps, then remove the double-counting of the group that reads all three.

Exactly one, exactly two and at least two of three sets

Step-by-step solution

1
$Q = n (I \cap II) + n (II \cap III) + n (I \cap III) = 8 + 4 + 2 = 14$ per cent
Why: Each pair figure includes the readers of all three papers, so this sum counts that group once for each of the three pairs.
Exactly one, exactly two and at least two of three sets
2
at least two $= Q - 2 R = 14 - 2 (1) = 12$ per cent
Why: Everyone in exactly two papers is counted once by $Q$ and is already correct; the all-three group is counted three times and should be counted once, so two of its copies are removed.
3
$\frac{12}{100} \times 1, 00, 000 = 12, 000$
Why: Convert the percentage to people only at the end.
$x %$ of a quantity

Common mistakes

Option B

Removed the all-three group once instead of twice, giving $14 - 1 = 13$ per cent.

Option C

Added the three pair figures and stopped, counting the $1%$ who read all three once for each pair.

Option D

Added the all-three group instead of subtracting, giving $14 + 1 = 15$ per cent.

Skills used

Exactly one, exactly two and at least two of three sets

x %

of a quantity

18Algebra • Sets • Operations

EasyIdeal time: about 1 min

Same directions as Q16

What is the number of people who do not read any of these three newspapers ?

62, 000

64, 000

66, 000

68, 000

Show answer & solution

Option D

68, 000

Approach

Find the union of the three readerships by inclusion-exclusion, then take its complement in the town.

Inclusion-exclusion for three sets

Step-by-step solution

1
$n (I \cup II \cup III) = 10 + 30 + 5 - 8 - 4 - 2 + 1 = 32$ per cent
Why: Each pairwise overlap is subtracted once; that removes the all-three group three times after it was added three times, so one copy is restored.
Inclusion-exclusion for three sets
2
read none $= 100 - 32 = 68$ per cent
Why: 'None of these three' is the complement of the union within the town.
3
$\frac{68}{100} \times 1, 00, 000 = 68, 000$
Why: Convert once, at the end.
$x %$ of a quantity

Common mistakes

Option A

Filler — no route reaches $62, 000$ .

Option B

Filler — no route reaches $64, 000$ .

Option C

Used the 'at least two' figure $12%$ in place of the sum of the pairwise overlaps, giving a union of $45 - 12 + 1 = 34$ per cent.

Skills used

Inclusion-exclusion for three sets

x %

of a quantity

19Arithmetic • Number System • Unit Digit

EasyIdeal time: about 45 s

What is the unit place digit in the expansion of $7^{73}$ ?

1

3

7

9

Show answer & solution

Option C

7

Approach

The unit digits of powers of $7$ run in a cycle of four; locate $73$ within that cycle.

Unit digit of a power from its cycle

Step-by-step solution

1
$7^{1} = 7$ , $7^{2} = 49$ , $7^{3} = 343$ , $7^{4} = 2401$ — unit digits $7, 9, 3, 1$ , then repeating
Why: The cycle closes as soon as the unit digit $1$ reappears, so its length here is $4$ .
Unit digit of a power from its cycle
2
$73 = 4 \times 18 + 1$ , so the remainder is $1$
Why: Only the position within the cycle matters, and that is the remainder on division by the cycle length.
3
Position $1$ in the cycle $7, 9, 3, 1$ is $7$
Why: Counting starts at $7^{1}$ , so a remainder of $1$ points at the first entry.

Common mistakes

Option A

Sent the remainder $1$ to the end of the cycle, or worked with $7^{72}$ instead of $7^{73}$ .

Option B

Took $73 \equiv 3 (mod 4)$ and read the third entry of the cycle.

Option D

Counted the cycle from $7^{0} = 1$ , so the remainder $1$ landed on $9$ .

Skills used

Unit digit of a power from its cycle

20Arithmetic • Number System • Perfect Squares

ModerateIdeal time: about 1.5 min

Suppose n is a positive integer such that $(n^{2} + 48)$ is a perfect square. What is the number of such n ?

AOneBTwoCThreeDFour

Show answer & solution

Option C

Three

Approach

Set $n^{2} + 48 = m^{2}$ , factorise the difference of squares, and keep only the factor pairs of $48$ with matching parity.

Difference of two squares

Step-by-step solution

1
$m^{2} - n^{2} = 48 \Rightarrow (m - n) (m + n) = 48$
Why: Factorising the difference of squares turns one equation in two unknowns into a short list of factor pairs — the move that makes the count finite.
Difference of two squares
2
$m - n$ and $m + n$ add to $2 m$ , so they share a parity; two odd factors would give an odd product, so both must be even.
Why: This parity check is what discards half the factor pairs before any arithmetic is done.
Write an odd number as $2 m + 1$
3
Write $m - n = 2 s$ , $m + n = 2 t$ with $s t = 12$ and $t > s > 0$ : $(s, t) = (1, 12), (2, 6), (3, 4)$
Why: Dividing out the $2$ s reduces the search to the factor pairs of $12$ , and $t > s$ keeps $n$ positive.
Prime factorisation
4
$n = t - s$ gives $n = 11, 4, 1$ — three values
Why: Each admissible pair yields exactly one positive $n$ , so the count is three.

Common mistakes

Option A

Found a single solution, usually $n = 4$ , and stopped searching.

Option B

Missed one factor pair — most often $(3, 4)$ , which supplies the small solution $n = 1$ .

Option D

Counted a factor pair of opposite parity, such as $1 \times 48$ , which gives non-integer $m$ and $n$ .

Skills used

Difference of two squares Write an odd number as

2 m + 1

Prime factorisation