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2021 Paper 2 Q3
D: 1500.0 B: 1500.0

In this question, \(x\), \(y\) and \(z\) are real numbers. Let \(\lfloor x \rfloor\) denote the largest integer that satisfies \(\lfloor x \rfloor \leqslant x\) and let \(\{x\}\) denote the fractional part of~\(x\), so that \(x = \lfloor x \rfloor + \{x\}\) and \(0 \leqslant \{x\} < 1\). For example, if \(x = 4.2\), then \(\lfloor x \rfloor = 4\) and \(\{x\} = 0.2\) and if \(x = -4.2\), then \(\lfloor x \rfloor = -5\) and \(\{x\} = 0.8\).

  1. Solve the simultaneous equations \begin{align*} \lfloor x \rfloor + \{y\} &= 4.9, \\ \{x\} + \lfloor y \rfloor &= -1.4. \end{align*}
  2. Given that \(x\), \(y\) and \(z\) satisfy the simultaneous equations \begin{align*} \lfloor x \rfloor + y + \{z\} &= 3.9, \\ \{x\} + \lfloor y \rfloor + z &= 5.3, \\ x + \{y\} + \lfloor z \rfloor &= 5, \end{align*} show that \(\{y\} + z = 3.2\) and solve the equations.
  3. Solve the simultaneous equations \begin{align*} \lfloor x \rfloor + 2y + \{z\} &= 3.9, \\ \{x\} + 2\lfloor y \rfloor + z &= 5.3, \\ x + 2\{y\} + \lfloor z \rfloor &= 5. \end{align*}

1998 Paper 1 Q1
D: 1516.0 B: 1500.0

How many integers between \(10\,000\) and \(100\,000\) (inclusive) contain exactly two different digits? (\(23\,332\) contains exactly two different digits but neither of \(33\,333\) and \(12\,331\) does.)


Solution: First consider \(5\) digit numbers containing at most \(2\) non-zero digits. Then there are \(\binom{9}{2}\) ways to choose the two digits, and \(2^{5}-2\) different ways to arrange them, removing the ones which are all the same. Considering all the pairs including zero, there are \(9\) ways to choose the non-zero (first) digit. There are \(2^4-1\) remaining digits where not all the numbers are the same. Finally we must not forget \(100\,000\). Therefore there are \(\binom{9}{2}(2^5-2) +9\cdot(2^4-1) + 1 = 1216\)