Click an option to answer. A correct first attempt marks the question secure; a wrong one marks it needs work and opens the worked solution. Keys A-H answer the question nearest the top of the screen.
In a knockout tournament with \(37\) players, each match eliminates exactly one player and the tournament ends when one player remains. How many matches are played?
- A\(36\)
- B\(35\)
- C\(19\)
- D\(37\)
- E\(64\)
- F\(18\)
Worked solution
Each match eliminates one player and \(36\) must be eliminated, so \(36\) matches - whatever the draw looks like.
- At the end one player remains, so \(37 - 1 = 36\) players have been eliminated.Count the outcome, not the structure.
- Each match eliminates exactly one player.Given in the question - this is the invariant.
- So the number of matches equals the number of eliminations: \(36\).And this holds whatever byes or bracket shape are used, which is why the answer is unique.
- Trying to build the bracket and getting lost in byes because \(37\) is not a power of \(2\).
- Halving repeatedly (\(18, 19\)-type answers), which only works for tidy brackets.
The digits \(1\) to \(9\) are placed in a \(3 \times 3\) grid so that every row and every column has the same sum. What is that sum?
- A\(45\)
- B\(15\)
- C\(20\)
- D\(18\)
- E\(12\)
- F\(10\)
Worked solution
The three rows together contain every digit once, so \(3 \times\) (row sum) \(= 1 + 2 + \cdots + 9 = 45\), giving \(15\).
- The three rows between them contain each digit \(1\) to \(9\) exactly once.Every cell lies in exactly one row, so the rows partition the grid.
- So the three row sums add to \(1+2+\cdots+9 = 45\).Use \(\frac{9 \times 10}{2} = 45\).
- All three row sums are equal, so each is \(\frac{45}{3} = 15\).And the columns give the same total, so they must each be \(15\) too - consistent.
- Answering \(45\), the grand total rather than one row's sum.
- Trying to construct an actual magic square; the total settles it with no construction needed.
Each of \(100\) cards has a different integer from \(1\) to \(100\) written on it. What is the smallest number of cards that must be drawn (without looking) to be certain of having two cards whose numbers sum to \(101\)?
- A\(51\)
- B\(2\)
- C\(100\)
- D\(99\)
- E\(52\)
- F\(50\)
Worked solution
The \(50\) pairs \(\{1, 100\}, \{2, 99\}, \ldots, \{50, 51\}\) each sum to \(101\). Fifty cards (one from each pair) may avoid it; \(51\) cannot.
- Pair the numbers to sum to \(101\): \(\{1,100\}, \{2,99\}, \ldots, \{50,51\}\).Each of \(1\) to \(100\) lies in exactly one such pair, so there are \(50\) pairs.
- Drawing one card from each pair gives \(50\) cards with no pair summing to \(101\) - for example \(1\) to \(50\).So \(50\) is not enough; this constructs the worst case.
- Any \(51\) cards must include two from one pair, and those sum to \(101\).\(51\) cards into \(50\) pairs forces a repeat.
- Answering \(50\) - the largest number that can fail.
- Answering \(2\) by thinking about luck rather than certainty.
In a class every student studies at least one of Maths and Physics. \(70\%\) study Maths and \(60\%\) study Physics. A student is picked at random. What is the probability they study both?
- A\(0.42\)
- B\(0.2\)
- C\(0.1\)
- D\(0.3\)
- E\(0.5\)
- Fit cannot be determined
Worked solution
\(P(M \text{ or } P) = 1 = 0.7 + 0.6 - P(\text{both})\), so \(P(\text{both}) = 0.3\).
- 'At least one of Maths and Physics' means \(P(M \text{ or } P) = 1\).Nobody is outside the union, so the union has probability \(1\).
- Apply the addition rule: \(P(M \cup P) = P(M) + P(P) - P(M \cap P)\).Subtract the overlap, which the two percentages each count.
- Substitute: \(1 = 0.7 + 0.6 - P(\text{both})\), so \(P(\text{both}) = 1.3 - 1 = 0.3\).The excess of \(1.3\) over \(1\) is exactly the double-counted overlap.
- Multiplying \(0.7 \times 0.6 = 0.42\) by assuming independence - nothing here says the subjects are independent, and the union condition rules it out.
- Answering 'cannot be determined'; the 'at least one' condition is precisely what makes it determinable.
The sum of \(5\) consecutive odd numbers is \(S\). Which of the following is always a factor of \(S\)?
- A\(4\)
- B\(3\)
- C\(10\)
- D\(25\)
- E\(15\)
- F\(5\)
Worked solution
\((m - 4) + (m - 2) + m + (m + 2) + (m + 4) = 5m\) where \(m\) is the middle (odd) number; \(5m\) is odd, so \(10\) is never a factor.
- Let the middle number be \(m\) (odd). The five are \(m-4, m-2, m, m+2, m+4\).Consecutive odd numbers differ by \(2\); centring makes the offsets cancel in pairs.
- Add: the \(\pm2\) and \(\pm4\) cancel, leaving \(S = 5m\).So \(5\) is always a factor.
- Is \(10\) always a factor? \(m\) is odd, so \(5m\) is odd and never divisible by \(2\). No.For instance \(1+3+5+7+9 = 25\).
- Are \(3\), \(15\) or \(25\) always factors? \(25\) needs \(5 \mid m\), \(3\) needs \(3 \mid m\) - neither is guaranteed (\(m = 7\) gives \(S = 35\)).Only the factor \(5\) survives every case.
- Answering \(25\) from the single example \(1+3+5+7+9 = 25\) - that is \(5m\) with \(m=5\), a coincidence.
- Answering \(10\) by assuming an even total; five odd numbers sum to an odd number.
Seven consecutive integers are written down. Which of the following statements is/are true?
- ITheir sum is divisible by \(7\).
- IITheir product is divisible by \(7\).
- IIITheir sum is odd.
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
The sum is \(7m\) where \(m\) is the middle integer (I). Seven consecutive integers cover every remainder mod \(7\), so one is a multiple of \(7\) (II). \(7m\) is odd only when \(m\) is odd, so III need not hold.
- Let the middle be \(m\); the seven are \(m-3, \ldots, m+3\), and the offsets cancel, so the sum is \(7m\). I is true.Symmetric representation makes the cancellation immediate.
- II: seven consecutive integers take all seven remainders on division by \(7\), so exactly one is a multiple of \(7\); hence \(7\) divides the product. True.Among any \(k\) consecutive integers there is exactly one multiple of \(k\).
- III: the sum is \(7m\), which is odd only when \(m\) is odd. Take \(1,\ldots,7\): sum \(28\), even. Not always true.One counterexample settles it.
- So I and II.Match to the list.
- Accepting III after testing \(2,\ldots,8\) (sum \(35\), odd) and no other case.
- Confusing the sum and product claims - I is about \(7m\), II about a multiple of \(7\) among the factors.
A \(4 \times 4\) grid of squares is coloured so that every row and every column contains exactly two black squares. How many black squares are there in the grid?
- A\(6\)
- B\(16\)
- C\(10\)
- D\(8\)
- E\(4\)
- F\(12\)
Worked solution
Each of the \(4\) rows contains exactly \(2\) black squares: \(8\) in total.
- There are \(4\) rows and each contains exactly \(2\) black squares.Every black square lies in exactly one row, so rows partition them.
- Total \(= 4 \times 2 = 8\).No double counting.
- Check with the columns: \(4 \times 2 = 8\) as well - consistent, which is why such a colouring can exist.If the two counts disagreed, no such grid would exist.
- Adding rows and columns to get \(16\) - that counts every black square twice.
- Answering \(4\) by counting rows rather than squares.
A selection of \(n\) distinct integers is taken from \(1, 2, \ldots, 20\). What is the smallest \(n\) that guarantees two of the chosen integers differ by exactly \(10\)?
- A\(2\)
- B\(15\)
- C\(12\)
- D\(11\)
- E\(20\)
- F\(10\)
Worked solution
Pair up \(\{1, 11\}, \{2, 12\}, \ldots, \{10, 20\}\): ten pairs. Choosing \(11\) numbers forces two from the same pair (pigeonhole); choosing one from each pair gives \(10\) numbers with no such difference.
- Pair the numbers so that each pair differs by \(10\): \(\{1,11\}, \{2,12\}, \ldots, \{10,20\}\).Every number from \(1\) to \(20\) lies in exactly one pair, and two numbers differ by \(10\) precisely when they share a pair.
- There are \(10\) pairs, so \(10\) numbers can be chosen with no two in the same pair - for example \(1\) to \(10\).This shows \(10\) is not enough, which is the half of the argument people skip.
- With \(11\) numbers and only \(10\) pairs, two must come from the same pair - and those differ by \(10\).Pigeonhole: more items than boxes forces a repeat.
- So the smallest guaranteed \(n\) is \(11\).Boxes \(+ 1\).
- Answering \(10\) - the largest selection that can avoid the condition, not the smallest that forces it.
- Answering \(2\), confusing 'could happen' with 'guaranteed'.
- Forgetting to show \(10\) is achievable; without it you have not proved \(11\) is smallest.
Five friends each shake hands with some of the others (each pair shakes hands at most once). Which of the following is impossible?
- AEvery friend shakes hands with exactly two others.
- BEvery friend shakes hands with exactly four others.
- CThe numbers of handshakes made are \(1, 1, 2, 2, 2\).
- DThe numbers of handshakes made by the five friends are \(4, 4, 4, 4, 1\).
- EExactly one friend shakes no hands at all.
Worked solution
If four people each shook hands with all four others, everyone shook hands with everyone, so the fifth person also made \(4\) handshakes, not \(1\). (The total \(4+4+4+4+1 = 17\) is odd, which is a second reason.) The others can happen: 'exactly two each' is a pentagon of handshakes, 'exactly four each' is everyone with everyone, \(1,1,2,2,2\) is a chain, and one person shaking no hands just leaves them out.
- Take \(4,4,4,4,1\): four people each shook hands with all four others.With five people, the maximum possible is \(4\) handshakes each.
- If person \(A\) shook hands with everyone, that includes the fifth person. The same for the other three.Each of the four 'maximum' people contributes one handshake to the fifth person.
- So the fifth person made at least \(4\) handshakes, not \(1\). Impossible.The list contradicts itself.
- The others are all realisable: \(2,2,2,2,2\) is a pentagon; \(4,4,4,4,4\) is everyone with everyone; \(1,1,2,2,2\) is a chain \(A\)-\(B\)-\(C\)-\(D\)-\(E\); and \(0\) for one person just leaves them out.Constructing one arrangement is enough to show a list is possible.
- Rejecting 'everyone shakes four hands' as too many - it is exactly the complete case and is fine.
- Rejecting the list \(1,1,2,2,2\) without trying to build it; a chain works.
- Forgetting the parity check: the degree total must be even. \(4+4+4+4+1 = 17\) is odd, which is a second, instant reason it is impossible.
\(x\) and \(y\) are positive integers with \(x^2 - y^2 = 45\). How many possible pairs \((x, y)\) are there?
- A\(0\)
- B\(6\)
- C\(2\)
- D\(1\)
- E\(3\)
- F\(4\)
Worked solution
\((x - y)(x + y) = 45\) with \(0 < x - y < x + y\): \((1, 45), (3, 15), (5, 9)\) give \((x, y) = (23, 22), (9, 6), (7, 2)\). Three pairs.
- Factorise: \(x^2 - y^2 = (x-y)(x+y) = 45\).Difference of two squares converts the equation into a product of two integers.
- Both factors are positive with \(x - y < x + y\), so list factor pairs of \(45\) with the smaller first: \((1,45), (3,15), (5,9)\).\(x, y\) positive forces \(x + y > 0\), and then \(x - y > 0\) too since the product is positive.
- Solve each: \(x = \frac{\text{sum}}{2}\), \(y = \frac{\text{difference}}{2}\), giving \((23,22), (9,6), (7,2)\).Adding the two equations gives \(2x\); subtracting gives \(2y\). Both factors are odd, so both results are integers.
- All three have positive integer \(y\), so there are three pairs.Check one: \(9^2 - 6^2 = 81-36 = 45\) ✓.
- Including \((9,5)\) or other reversed pairs - \(x-y\) must be the smaller factor.
- Forgetting to check that \(x\) and \(y\) come out as integers; if \(45\) had a factor pair of mixed parity it would have to be discarded.
- Counting \(45 = 45 \times 1\) and \(1 \times 45\) as two different solutions.
Three statements are made about a positive integer \(N\):
- \(N\) is a perfect square.
- \(N\) is a multiple of \(6\).
- \(N\) is less than \(50\).
Exactly two of the statements are true. Which of the following could NOT be \(N\)?
- A\(144\)
- B\(12\)
- C\(36\)
- D\(48\)
- E\(16\)
- F\(25\)
Worked solution
\(36\) makes all three statements true, not exactly two. Each of the others makes exactly two true: \(16\) and \(25\) (square, \(< 50\)), \(48\) and \(12\) (multiple of \(6\), \(< 50\)), \(144\) (square, multiple of \(6\)).
- \(36\): perfect square (\(6^2\)) ✓, multiple of \(6\) ✓, less than \(50\) ✓ - three true.The condition demands exactly two, so \(36\) is ruled out.
- \(16\): square ✓, multiple of \(6\) ✗, \(<50\) ✓ - two. Allowed.Check each of the three statements, do not stop early.
- \(25\): square ✓, multiple of \(6\) ✗, \(<50\) ✓ - two. \(48\): square ✗, multiple of \(6\) ✓, \(<50\) ✓ - two. \(12\): the same pattern - two.All fine.
- \(144\): square (\(12^2\)) ✓, multiple of \(6\) ✓, \(<50\) ✗ - two. Allowed.So only \(36\) fails.
- Reading 'exactly two are true' as 'at least two', which would admit \(36\).
- Missing that \(144 = 12^2\) is both a square and a multiple of \(6\).
- Looking for the value making the fewest statements true instead of the most.
\(a\), \(b\) and \(c\) are real numbers with \(a < b < c\). Which of the following statements must be true?
- I\(a + b < 2c\)
- II\(ab < bc\)
- III\(\dfrac{a + c}{2} < c\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
I: \(a + b < c + c\). III: \(\frac{a + c}{2} < \frac{c + c}{2} = c\). II fails when \(b\) is negative: \(a = -3\), \(b = -2\), \(c = -1\) gives \(ab = 6 > bc = 2\).
- I: from \(a < c\) and \(b < c\), add: \(a + b < 2c\). True.Adding two inequalities of the same direction is always valid.
- III: \(a < c\) so \(a + c < 2c\), and dividing by \(2\) gives \(\frac{a+c}{2} < c\). True.Dividing by a positive constant preserves the direction. (The midpoint of two numbers lies below the larger.)
- II: \(ab < bc\) would follow from \(aMultiplying an inequality by a negative number reverses it.
- Counterexample: \(a=-3, b=-2, c=-1\) satisfies \(aAll-negative values are the natural place to test a multiplication claim.
- So I and III.Match to the list.
- Accepting II by 'multiplying \(a
- Testing only positive numbers, where II does hold.
A rectangle has integer side lengths and its area (in cm\(^2\)) is numerically equal to its perimeter (in cm). How many such rectangles are there (counting \(a \times b\) and \(b \times a\) once)?
- A\(4\)
- B\(3\)
- C\(0\)
- D\(1\)
- E\(2\)
- Finfinitely many
Worked solution
\(ab = 2a + 2b \Rightarrow (a - 2)(b - 2) = 4\): \((a - 2, b - 2) = (1, 4)\) or \((2, 2)\), giving \(3 \times 6\) and \(4 \times 4\).
- Set up: \(ab = 2a + 2b\) with \(a,b\) positive integers.Area equals perimeter, numerically.
- Rearrange: \(ab - 2a - 2b = 0\), then add \(4\) to both sides: \(ab - 2a - 2b + 4 = 4\).The added constant is the product of the two subtracted coefficients, chosen to make the left side factorise.
- Factorise: \((a-2)(b-2) = 4\).Check by expanding: \(ab - 2a - 2b + 4\) ✓.
- Positive factor pairs of \(4\) with \(a-2 \leq b-2\): \((1,4)\) and \((2,2)\), giving \(3\times 6\) and \(4 \times 4\).Negative factors would need \(a < 2\); \(a = 1\) gives \(b - 2 = -4\), i.e. \(b = -2\), not a length.
- So two rectangles.Check: \(3\times6\) has area \(18\) and perimeter \(18\) ✓; \(4\times4\) has \(16\) and \(16\) ✓.
- Counting \(3\times 6\) and \(6\times 3\) separately - the question says to count each rectangle once.
- Forgetting to add \(4\) and being unable to factorise.
- Including the factor pair \((4,1)\) as a third solution.
The product of three consecutive positive integers is \(N\), and the middle one of the three integers is \(m\). Which of the following statements must be true?
- I\(N\) is divisible by \(6\).
- II\(N\) is divisible by \(4\).
- III\(N + m\) is a perfect cube.
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
Among three consecutive integers there is a multiple of \(2\) and a multiple of \(3\) (I). II fails for \(1 \cdot 2 \cdot 3 = 6\). III: \(N + m = (m - 1)m(m + 1) + m = m^3 - m + m = m^3\).
- Write the three as \(m-1, m, m+1\), so \(N = (m-1)m(m+1)\).Centring on the middle term is what makes III come out.
- I: among any three consecutive integers there is at least one multiple of \(2\) and one multiple of \(3\), so \(6 \mid N\). True.Every second integer is even and every third is a multiple of \(3\).
- II: test the smallest case \(1 \times 2 \times 3 = 6\), which is not divisible by \(4\). False.Three consecutive integers contain only one or two evens; when only one, it may contribute a single factor of \(2\).
- III: \(N + m = (m^2-1)m + m = m^3 - m + m = m^3\) - a perfect cube. True.\((m-1)(m+1) = m^2-1\) is the difference of squares; the \(-m\) and \(+m\) cancel.
- So I and III.Match to the list.
- Accepting II by testing \(2\times3\times4 = 24\) only - one confirming case is not enough.
- Expanding \(N\) the long way instead of using \((m-1)(m+1) = m^2-1\).
How many positive integers less than \(1000\) are divisible by neither \(2\) nor \(5\)?
- A\(399\)
- B\(600\)
- C\(500\)
- D\(400\)
- E\(450\)
- F\(300\)
Worked solution
Of \(1\) to \(999\): \(499\) multiples of \(2\), \(199\) of \(5\), \(99\) of \(10\). Divisible by \(2\) or \(5\): \(499 + 199 - 99 = 599\). Neither: \(999 - 599 = 400\).
- The range is \(1\) to \(999\), so \(999\) integers.'Less than \(1000\)' excludes \(1000\) itself.
- Multiples of \(2\): \(\left\lfloor \frac{999}{2}\right\rfloor = 499\). Multiples of \(5\): \(199\). Multiples of both, i.e. of \(10\): \(99\).Divide and round down. \(10\) is the LCM of \(2\) and \(5\).
- Divisible by \(2\) or \(5\): \(499 + 199 - 99 = 599\).Subtract the overlap, which was counted twice.
- Neither: \(999 - 599 = 400\).Sanity check: in each block of \(10\) consecutive integers exactly \(4\) (ending in \(1,3,7,9\)) survive, and \(999 = 99\) full blocks plus \(991\)-\(999\), giving \(99 \times 4 + 4 = 400\) ✓.
- Using \(1000\) as the count and getting \(400\) by luck with wrong intermediate figures - or getting \(399\) by an off-by-one at the top.
- Forgetting to subtract the multiples of \(10\), giving \(999 - 698 = 301\).
\(p\) and \(q\) are prime numbers with \(p + q = 45\). Find \(pq\).
- A\(247\)
- B\(140\)
- C\(102\)
- D\(86\)
- E\(94\)
- F\(50\)
Worked solution
An odd sum of two primes needs one of them to be \(2\) (the only even prime): \(q = 43\), which is prime. \(pq = 86\).
- \(45\) is odd, so one prime is even and one odd.odd \(+\) odd \(=\) even, even \(+\) even \(=\) even; only odd \(+\) even gives an odd total.
- The only even prime is \(2\), so \(p = 2\) and \(q = 43\).\(2\) is the unique even prime - every other even number has \(2\) as a proper factor.
- Check \(43\) is prime: not divisible by \(2, 3, 5\) and \(7^2 = 49 > 43\), so yes.Test primes up to the square root only.
- \(pq = 2 \times 43 = 86\).The question asks for the product, not the primes.
- Trying odd pairs like \(2+43\) versus \(5+40\) at random; \(40\) is not prime and no odd pair can work at all.
- Answering \(43\) or \(45\) instead of the product.
A sequence of \(10\) integers is such that the sum of any three consecutive terms is positive and the sum of any five consecutive terms is negative. Which of the following statements must be true?
- IThe sum of all \(10\) terms is negative.
- IIThe sequence contains at least one negative term.
- IIIThe sequence contains at least one positive term.
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
Terms \(1\)-\(5\) and \(6\)-\(10\) are two blocks of five, each with negative sum (I). A five-block has negative sum, so some term is negative (II). A three-block has positive sum, so some term is positive (III).
- I: split the ten terms into terms \(1\)-\(5\) and terms \(6\)-\(10\) - two blocks of five consecutive terms.Each block has negative sum by hypothesis, and the two blocks together are the whole sequence.
- Negative \(+\) negative \(=\) negative, so the total is negative. True.The grouping is exact - no term is used twice or left out.
- II: terms \(1\)-\(5\) sum to a negative number, so at least one of them is negative (five non-negative terms could not). True.A sum is negative only if some summand is.
- III: terms \(1\)-\(3\) sum to a positive number, so at least one is positive. True.Same argument the other way round.
- So all three.Match to the list. (Such sequences do exist for \(10\) terms - the conditions become contradictory only for longer sequences.)
- Trying to construct an explicit sequence; the claims follow from the block structure alone.
- Thinking the sum of all ten is ambiguous because \(10\) is not a multiple of \(3\) - use the blocks of five, which do fit.
Alice, Ben and Cara each make one statement. Alice: 'Ben is lying.' Ben: 'Cara is lying.' Cara: 'Alice and Ben are both lying.' Exactly one of them is telling the truth. Who is it?
- ABen
- BAlice
- CCara
- DIt cannot be determined.
- EThe situation is impossible.
Worked solution
If Alice is truthful then Ben lies, so Cara is truthful: two truth-tellers, impossible. If Cara is truthful then Alice and Ben both lie; but Alice lying means Ben is truthful: contradiction. So Ben is the truth-teller: Cara lies (indeed Ben is not lying) and Alice lies (Ben is not lying), which is consistent.
- Suppose Alice is truthful. Then Ben lies, so 'Cara is lying' is false, so Cara is truthful. That makes two truth-tellers - contradiction.Exactly one truth-teller is the constraint; two breaks it.
- Suppose Cara is truthful. Then Alice and Ben both lie. But Alice lying means 'Ben is lying' is false, so Ben is truthful - contradiction.Cara's own statement collapses under its consequences.
- Suppose Ben is truthful. Then Cara lies, so Ben is not lying ✓ consistent. Alice says 'Ben is lying', which is false, so Alice lies ✓. Cara says both Alice and Ben lie - false, since Ben is truthful ✓.Every statement's truth value matches the assumed roles, with exactly one truth-teller.
- So Ben is the truth-teller.Two cases eliminated, one consistent - the answer is determined.
- Stopping at the first scenario that seems to work without checking every statement.
- Declaring the situation impossible after two failed cases instead of testing the third.
- Forgetting that a liar's statement must come out false, which is itself information.
A bag contains red and blue balls only. If one red ball is removed, one quarter of the remaining balls are red. If instead two blue balls are removed, one third of the remaining balls are red. How many balls are in the bag?
- A\(16\)
- B\(20\)
- C\(14\)
- D\(8\)
- E\(17\)
- F\(12\)
Worked solution
With \(r\) red and \(b\) blue: \(r - 1 = \frac14(r + b - 1)\) and \(r = \frac13(r + b - 2)\). The second gives \(b = 2r + 2\); substituting into the first, \(4r - 4 = 3r + 1\), so \(r = 5\), \(b = 12\): \(17\) balls.
- Let \(r\) be red and \(b\) blue. Removing one red: \(r - 1 = \frac14(r + b - 1)\).The total drops by \(1\) as well as the red count - both sides of the fraction change.
- Removing two blues: \(r = \frac13(r + b - 2)\).Here the red count is unchanged; only the total falls, by \(2\).
- From the second: \(3r = r + b - 2\), so \(b = 2r + 2\).Solve the simpler equation first and substitute.
- Substitute into the first: \(4(r-1) = r + (2r+2) - 1 = 3r + 1\), so \(4r - 4 = 3r + 1\) and \(r = 5\).Multiply through by \(4\) to clear the fraction before substituting.
- Then \(b = 12\) and the total is \(17\).Check: remove one red \(\to 4\) red of \(16\) \(= \frac14\) ✓; remove two blue \(\to 5\) red of \(15 = \frac13\) ✓.
- Forgetting that removing a ball also reduces the total, giving \(r - 1 = \frac14(r+b)\).
- Answering \(r\) or \(b\) rather than the total.
- Not checking the answer against both original statements.
\(n\) is a positive integer such that \(n^2\) has exactly \(3\) positive divisors. Which of the following must be true?
- A\(n = 3\)
- B\(n\) is prime
- C\(n\) is odd
- D\(n\) is a perfect square
- E\(n\) is even
Worked solution
A number with exactly \(3\) divisors is the square of a prime; so \(n^2 = p^2\) and \(n = p\) is prime.
- A number with exactly \(3\) divisors must be \(p^2\) for a prime \(p\): its divisors are \(1, p, p^2\).If a number had two distinct prime factors it would have at least \(4\) divisors; \(p^3\) has \(4\); \(p\) alone has \(2\).
- So \(n^2 = p^2\), and since both are positive, \(n = p\).Take the positive square root.
- Therefore \(n\) is prime.That is the only guaranteed property.
- Check the others: \(n = 2\) is even, \(n = 3\) is odd - so neither parity is forced; and \(n\) prime is never a perfect square.Both \(n=2\) and \(n=3\) satisfy the condition (\(4\) and \(9\) have three divisors each), which kills the parity options.
- Answering '\(n\) is a perfect square' by confusing \(n\) with \(n^2\).
- Answering \(n = 3\) from one example; \(n = 2, 5, 7\) all work too.
The positive integers \(a\), \(b\) and \(c\) satisfy \(a < b < c\) and \(a + b + c = 12\). Which of the following statements must be true?
- I\(a \leq 3\)
- II\(c \geq 5\)
- III\(b = 4\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
Since \(a < b < c\), \(3a < 12\) so \(a \leq 3\) (I); \(3c > 12\) so \(c \geq 5\) (II). III fails: \(1 + 2 + 9 = 12\) has \(b = 2\).
- I: since \(aReplacing \(b\) and \(c\) by the smallest value \(a\) gives a strict under-estimate of the sum.
- II: similarly \(3c > a+b+c = 12\), so \(c>4\), i.e. \(c \geq 5\). True.Replacing \(a\) and \(b\) by the largest value \(c\) over-estimates the sum.
- III: must \(b=4\)? Try \(1+2+9 = 12\) with \(1<2<9\) ✓ and \(b=2\). False.One valid triple with \(b \neq 4\) settles it.
- So I and II.Match to the list.
- Writing \(3a \leq 12\) instead of \(3a<12\); the strict ordering \(a
- Assuming \(b\) must be the mean of the three - true only for equally spaced values.
The integers \(a, b, c\) satisfy \(a + b + c = 0\) and \(abc = 12\). Which of the following statements must be true?
- IExactly one of \(a, b, c\) is positive.
- II\(a^2 + b^2 + c^2\) is even.
- IIIOne of \(a, b, c\) is divisible by \(3\).
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
\(abc > 0\) means either none or exactly two of them are negative; they cannot all be positive with sum \(0\), so exactly two are negative and one positive (I). \(a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc + ca) = -2(ab + bc + ca)\) is even (II). \(3\) is prime and divides \(abc\), so it divides one of them (III). (Example: \(-1, -3, 4\).)
- I: \(abc = 12 > 0\), so the number of negatives is even: \(0\) or \(2\). If none were negative, three positives could not sum to \(0\). So exactly two are negative and one positive. True.A positive product means an even count of negative factors; note \(0\) is impossible since \(abc \neq 0\).
- II: use \((a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)\). With \(a+b+c=0\), \(a^2+b^2+c^2 = -2(ab+bc+ca)\).The identity converts a sum of squares into twice an integer.
- That is \(2 \times\) an integer, hence even. True.\(ab+bc+ca\) is an integer because \(a,b,c\) are.
- III: \(3\) is prime and \(3 \mid 12 = abc\), so \(3\) divides one of the factors. True.Euclid's lemma: a prime dividing a product divides one of the factors. This would fail for a composite like \(4\).
- So all three. (Check with \(-1, -3, 4\): sum \(0\) ✓, product \(12\) ✓.)A concrete example confirms the constraints are satisfiable at all.
- Assuming exactly one negative because the product is positive - that would make the product negative.
- Trying to settle II by testing examples rather than using the identity.
- Thinking III needs a specific example; it follows from primality alone.
How many ordered pairs of positive integers \((m, n)\) satisfy \(\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{1}{4}\)?
- A\(4\)
- B\(3\)
- C\(2\)
- D\(6\)
- E\(8\)
- F\(5\)
Worked solution
\(4n + 4m = mn \Rightarrow (m - 4)(n - 4) = 16\). Positive factor pairs of \(16\) as ordered pairs: \((1,16), (2,8), (4,4), (8,2), (16,1)\): five. (Negative factor pairs would make \(m\) or \(n\) non-positive.)
- Multiply \(\frac1m + \frac1n = \frac14\) by \(4mn\): \(4n + 4m = mn\).Clearing denominators turns it into a polynomial equation in integers.
- Rearrange and factorise: \(mn - 4m - 4n + 16 = 16\), so \((m-4)(n-4) = 16\).Add \((-4)(-4) = 16\) to both sides to complete the rectangle.
- \(m, n\) positive: if \(m - 4\) were negative then \(m \in \{1,2,3\}\) and \(n-4 = \frac{16}{m-4}\) would make \(n\) negative. So both factors are positive.Check the negative branch explicitly rather than assuming; e.g. \(m = 3\) gives \(n - 4 = -16\), \(n = -12\) ✗.
- Positive factor pairs of \(16\): \((1,16),(2,8),(4,4),(8,2),(16,1)\) - five ordered pairs.'Ordered' means \((m,n)\) and \((n,m)\) count separately: \((5,20), (6,12), (8,8), (12,6), (20,5)\).
- Answering \(3\) by counting unordered pairs when the question says ordered.
- Forgetting \((4,4)\), the square factor pair.
- Including negative factor pairs without checking they give positive \(m, n\).
How this topic has been examined
Real TMUA questions on Chains of reasoning, newest first. Links open the official paper at the question and the online marking page.