Fryer, Galois and Hypatia contests guide, 2027

Fryer-style question (ours). A school bake sale packs cookies
in boxes of 6 and boxes of 10. A customer buys whole boxes only.

(a) [2 marks, short answer] Show how to buy exactly 46 cookies.
(b) [2 marks, short answer] In how many ways can exactly 60 cookies
    be bought?
(c) [2 marks, full solution] Explain why an odd number of cookies
    can never be bought.
(d) [4 marks, full solution] Determine, with justification, the
    largest even number of cookies that cannot be bought.

(a) One box of 6 and four boxes of 10: 6 + 40 = 46.

(b) We need 6a + 10b = 60, that is 3a + 5b = 30, with a, b >= 0.
    b = 0 gives a = 10; b = 3 gives a = 5; b = 6 gives a = 0.
    Other values of b from 0 to 6 leave 30 - 5b not a multiple
    of 3. So there are 3 ways.

(c) 6a + 10b = 2(3a + 5b), which is even for every a and b.
    So an odd number of cookies is impossible.

(d) 14 cannot be bought. With no 10-box we need 6a = 14,
    impossible. With one 10-box we need 6a = 4, impossible.
    Two or more 10-boxes already give more than 14.

    Every even number from 16 up can be bought.
    16 = 6 + 10,   18 = 6 + 6 + 6,   20 = 10 + 10.
    Adding one more box of 6 to any of these gives the next
    three even numbers, 22, 24 and 26, and repeating this
    reaches every even number from 16 upward.

    Therefore the largest even number of cookies that cannot
    be bought is 14.

Part (c) is the key to part (d): once every purchase is known to be even, the search in (d) only has to look at even numbers. Part (b) quietly practises the same equation that (d) uses. The mark split shown is ours, chosen to resemble the cover's rules; CEMC sets its own split on each paper.

Galois-style question (ours). The parabola y = x^2 - 4x + k,
where k is a real number.

(a) [2 marks] If k = 3, what are the x-intercepts?
(b) [3 marks] For what value of k does the parabola touch
    the x-axis at exactly one point?
(c) [5 marks, full solution] The line y = 2x - 5 meets the
    parabola at two points whose x-coordinates differ by 4.
    Determine k and the two points.

(a) x^2 - 4x + 3 = (x - 1)(x - 3), so the intercepts are x = 1, x = 3.

(b) x^2 - 4x + k = (x - 2)^2 + (k - 4). This touches the x-axis
    exactly once when k - 4 = 0, so k = 4.

(c) At an intersection, x^2 - 4x + k = 2x - 5,
    so x^2 - 6x + (k + 5) = 0.
    By the quadratic formula,
        x = (6 +/- sqrt(36 - 4(k + 5))) / 2 = 3 +/- sqrt(4 - k).
    The two x-coordinates differ by 2 sqrt(4 - k).
    Setting 2 sqrt(4 - k) = 4 gives sqrt(4 - k) = 2, so k = 0.
    Then x = 5 or x = 1, and on the line y = 2x - 5 these give
    the points (5, 5) and (1, -3).
    Check on the parabola y = x^2 - 4x: 25 - 20 = 5 and 1 - 4 = -3.

    Therefore k = 0, and the points are (5, 5) and (1, -3).

Part (b) completes the square and part (c) uses the quadratic formula, which is completing the square in general form. A student who sees that link solves (c) calmly. The check on the parabola at the end costs ten seconds and catches arithmetic slips.

Hypatia-style question (ours). For a two-digit positive integer n,
R(n) is the number formed by reversing its digits,
so R(52) = 25 and R(40) = 4.

(a) [2 marks] What is 52 - R(52)?
(b) [3 marks, full solution] Prove that n - R(n) is a multiple
    of 9 for every two-digit n.
(c) [5 marks, full solution] Determine how many two-digit n
    make n - R(n) a positive perfect square.

(a) 52 - 25 = 27.

(b) Write n = 10a + b, where a is the tens digit (1 to 9) and
    b is the units digit (0 to 9). Then R(n) = 10b + a, and
        n - R(n) = 9a - 9b = 9(a - b),
    which is a multiple of 9.

(c) By (b), we need 9(a - b) to be a positive perfect square.
    If 9m = s^2 for a positive integer m, then 3 divides s, so
    s = 3t and m = t^2. Conversely 9t^2 = (3t)^2. So 9(a - b) is a
    perfect square exactly when a - b is a perfect square.
    Since 1 <= a - b <= 9, a - b is 1, 4 or 9.
      a - b = 1: a = 1 to 9, so 9 numbers (10, 21, ..., 98).
      a - b = 4: a = 4 to 9, so 6 numbers (40, 51, ..., 95).
      a - b = 9: only a = 9, b = 0, so 1 number (90).

    Therefore 9 + 6 + 1 = 16 two-digit numbers work.

Part (b) does the algebra that turns part (c) from a search through 90 numbers into a question about the digit difference a - b. The small lemma about 9m being a square is the line many students skip; without it the count is a guess.