Derivatives Foundation puzzles, solved step by step

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All topicsMental maths and estimation9Random walks and Markov chains7Conditional probability and Bayes7Volatility and correlation7Option pricing intuition7Expected value and optimal stopping10Market making11Option payoffs and no-arbitrage10Probability and counting11Distributions and statistics8Games and logic8Betting and sizing5
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  1. 046Give the next term in each sequence: 2, 6, 12, 20, 30, ... ; 1, 1, 2, 6, 24, ... ; 3, 5, 9, 17, 33, ...Mental maths and estimationCoreProp trading firmsSell-side sales and trading

    Try it first

    What comes after 3, 5, 9, 17, 33?

    Show the worked solution

    42, 120 and 65. In the first, the differences 4, 6, 8, 10 rise by 2, so the next difference is 12 and the term is 42; the terms are n(n + 1). In the second, each term is the previous one times 1, 2, 3, 4, so the next multiplier is 5 and the term is 120; these are the factorials. In the third, the differences 2, 4, 8, 16 double, so add 32 to get 65; the terms are 2 to the n plus 1.

    What order should you test patterns in?

    A mechanic with an unknown rattle checks the cheap, common causes first and the exotic ones last. Sequence questions reward the same discipline: write the first differences, then the second differences, and only if neither settles try ratios and then rules that mix the two, such as double and subtract one. Each test takes a few seconds and most interview sequences give way to one of the first three. The order matters because the interviewer is timing you; the candidate who stares at the numbers hoping to recognise them is slower than the one who writes a row of differences under them without thinking.

    Test differences, then ratios, then a doubling rule: write the table under the numbers2, 6, 12, 20, 30constant second difference: n(n + 1)2612203042termsdifferences4681012second diff22221, 1, 2, 6, 24ratios climb by one: n!112624120termsratiosx1x2x3x4x53, 5, 9, 17, 33differences double: 2^n + 1359173365termsdifferences+2+4+8+16+32Each table needs at most two rows before the pattern is a constant; the lime box is the next term
    Writing a row under each sequence exposes the rule: the first has constant second differences of 2 and continues to 42, the second has ratios climbing by one and continues to 120, and the third has doubling differences and continues to 65.

    How does each of the three give way?

    For 2, 6, 12, 20, 30, the first differences are 4, 6, 8, 10 and the second differences are all 2. Constant second differences mean the sequence is a quadratic in n, so the next difference is 12 and the next term 42, and the closed form is n(n + 1): 1 x 2, 2 x 3, up to 6 x 7. For 1, 1, 2, 6, 24, the differences 0, 1, 4, 18 tell you nothing, so try ratios: 1, 2, 3, 4. The next ratio is 5 and the term is 120; the terms are 0!, 1!, 2!, 3!, 4!, and 5! is 120. For 3, 5, 9, 17, 33, the differences are 2, 4, 8, 16, a doubling, so the next difference is 32 and the term is 65. Spot the rule in one line too: each term is twice the previous minus 1, and every term is a power of two plus 1, so the next is 2 to the 6 plus 1.

    The relationship
    an=n(n+1)⇒a6=42,bn=(n−1)!⇒b6=5!=120,cn=2n+1⇒c6=65a_n = n(n+1) \Rightarrow a_6 = 42, \qquad b_n = (n-1)! \Rightarrow b_6 = 5! = 120, \qquad c_n = 2^n + 1 \Rightarrow c_6 = 65
    a nthe n-th term of the first sequence, a quadratic, so its second differences are constant
    b nthe factorials, each term the previous one times the next whole number
    c na power of two plus one, so its differences are powers of two
    What it says in wordsEach sequence has a one-line rule, and the difference or ratio row is how you find it in seconds.

    Is the answer really unique, and what should you say if pushed?

    Strictly, no. Any five numbers can be continued by any sixth, because a polynomial of degree five can be passed through all six points. For example n(n + 1) + (n - 1)(n - 2)(n - 3)(n - 4)(n - 5) matches 2, 6, 12, 20, 30 exactly and then gives 162. The expected answer is the simplest rule that fits, and the way to show you know that is to name the rule, not just the number. That habit carries over to the desk: a pattern in five data points is a hypothesis, and the trader who states the rule can test it on the sixth point, while the one who only extrapolates cannot tell when the pattern has broken. If the interviewer offers a sequence that resists all three tests, try alternating terms, or interleaved sequences, before guessing.

    Where candidates lose it

    The common loss is 66 for the third sequence: doubling 33 and forgetting that the rule doubles and then subtracts 1. Check the rule on an earlier pair, 17 to 33, before you say the answer.

    The second is hunting the factorials through differences, which give 0, 1, 4, 18 and lead nowhere. When differences grow faster than the terms, switch to ratios immediately.

    What the interviewer asks next

    • What comes next: 1, 4, 9, 16, 25, 36, and what are its second differences?
    • Next term: 1, 2, 6, 15, 31, ...
    • Next term: 2, 3, 5, 7, 11, 13, ...
    • Find a rule for 1, 3, 7, 15, 31 and give its closed form.
  2. 052A bus makes three stops. At each stop half the people on board get off, and then the number on board grows by a third. The bus reaches the end of the route with 16 people. How many were on board at the start?Mental maths and estimationCoreProp trading firms

    Try it first

    Before working it: each stop multiplies the number on board by what?

    Show the worked solution

    54 people. Each stop halves the load and then adds a third of what is left, so the load is multiplied by 1/2 x 4/3 = 2/3 at every stop. Undoing three stops means dividing 16 by 2/3 three times: 16 to 24 to 36 to 54. The forward check works: 54 to 27 to 36, then 18 to 24, then 12 to 16.

    Why work backwards rather than guess a start?

    If someone tells you a shirt was marked down by half, then marked up by a third, and now costs Rs 16, you do not guess the original tag; you undo the two moves in reverse order. Each stop is a pair of multiplications, and multiplications are undone by dividing in the opposite order, so the end number walks back to the start without any trial and error. The growth of a third comes last at each stop, so it is undone first: divide by 4/3, which is multiply by 3/4. Then undo the halving by doubling. From 16: times 3/4 is 12, times 2 is 24. Twice more gives 36 and then 54.

    Work backwards from 16: undo the growth, then undo the drop16243654end of routeat the startx 3/4 = 12then x 2x 3/4 = 18then x 2x 3/4 = 27then x 2Undo the growth by a third: divide by 4/3, which is x 3/4. Undo losing half: x 2.Forward check54starthalf off: 27+ a third: 3636stop 1half off: 18+ a third: 2424stop 2half off: 12+ a third: 1616stop 3Each stop multiplies the load by 1/2 x 4/3 = 2/3, so start = 16 x (3/2) cubed = 54
    Starting from 16 at the end of the route and undoing each stop, first multiplying by 3/4 then by 2, the count climbs 16 to 24 to 36 to 54, and running the three stops forward from 54 returns exactly 16, because each stop multiplies the load by 2/3.

    What is the one-line version an interviewer wants to hear?

    Collapse each stop to a single factor. Losing half is x 1/2 and gaining a third is x 4/3, so a stop is x 2/3, and three stops are x 8/27. The start is 16 x 27/8 = 54. Saying that in one breath shows you saw the structure rather than the arithmetic, which is what the question is for. The limitation is that it only works because every stop has the same two moves; a route where the fractions change needs the step-by-step walk back.

    The relationship
    N0=N3×(12×43)−3=16×(32)3=16×278=54N_0 = N_3 \times \left(\tfrac{1}{2}\times\tfrac{4}{3}\right)^{-3} = 16 \times \left(\tfrac{3}{2}\right)^3 = 16 \times \tfrac{27}{8} = 54
    N_0the number on board at the start
    N_3the number at the end of the route, 16
    (3/2)^3undoing the per-stop factor of 2/3 three times
    What it says in wordsThe start is the end count divided by the per-stop factor of two thirds, three times over.

    Where do the whole-number checks help you?

    The bus carries people, so every intermediate count must be a whole number, and that is a free check. If the backwards walk ever produces a fraction, you have undone the moves in the wrong order. Undoing the halving first from 16 gives 32, then times 3/4 gives 24, which happens to be whole here, but on the next stop it would give 48 then 36, and the order mistake would have cost you nothing visible until the end, where 72 x 3/4 = 54 again by luck. Order matters in general even when the numbers hide it, so say the order out loud.

    Where candidates lose it

    The common slip is treating the two moves at each stop as a net loss of a sixth, since a half minus a third is a sixth. The third is taken on the smaller number, so the stop is a factor of 2/3, not 5/6.

    The second slip is undoing the moves in the wrong order. Growth came last, so it is undone first. Say the per-stop factor, then walk back 16, 24, 36, 54, and finish with the forward check.

    What the interviewer asks next

    • If instead a third get off and then the number on board doubles at each stop, what is the per-stop factor?
    • How many stops would it take for a bus starting with 54 to get below 5 people?
    • The bus starts with 54 and the pattern continues. After how many stops is the number no longer a whole number?