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2006 Paper 1 Q1
D: 1500.0 B: 1516.0
number theory perfect squares factorisation difference of two squares integer arithmetic systematic search

Find the integer, \(n\), that satisfies \(n^2 < 33\,127< (n+1)^2\). Find also a small integer \(m\) such that \((n+m)^2-33\,127\) is a perfect square. Hence express \(33\,127\) in the form \(pq\), where \(p\) and \(q\) are integers greater than \(1\). By considering the possible factorisations of \(33\, 127\), show that there are exactly two values of \(m\) for which \((n+m)^2 -33\,127\) is a perfect square, and find the other value.

Solution: \begin{align*} 180^2 &= 32400 \\ 181^2 &= 32761 \\ 182^2 &= 33124 \\ 183^2 &= 33489 \\ 184^2 &= 33856 \end{align*} Therefore \(182^2 < 33\,127 < (182+1)^2\). and \((182+2)^2 - 33\,127 = 729 = 27^2\). Therefore \(33\,127 = 184^2 - 27^2 = 211 \times 157\). (Note both of these numbers are prime). Suppose \((n+m)^2 - 33\,127 = k^2\) then \(33\,127 = (n+m)^2-k^2 = (n+m-k)(n+m+k)\). Since there are only two factorisations of \(33\,127\) into positive integer factors with one factor larger than the other, the other factorisation must be: \(n+m+k = 33\,127, n+m-k = 1 \Rightarrow k = \frac{33\, 126}{2} = 16563\), ie \(16564^2 - 33\,127 = 16563^2\)

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2003 Paper 1 Q7
D: 1484.0 B: 1516.0
proof inequality divisibility digit manipulation number theory cubic equations systematic search integer constraints

Let \(k\) be an integer satisfying \(0\le k \le 9\,\). Show that \(0\le 10k-k^2\le 25\) and that \(k(k-1)(k+1)\) is divisible by \(3\,\). For each \(3\)-digit number \(N\), where \(N\ge100\), let \(S\) be the sum of the hundreds digit, the square of the tens digit and the cube of the units digit. Find the numbers \(N\) such that \(S=N\). [Hint: write \(N=100a+10b+c\,\) where \(a\,\), \(b\,\) and \(c\) are the digits of \(N\,\).]

Solution: First note that \(10k - k^2 = 25 - (5-k)^2\) which is clearly bounded above by \(25\). The smallest it can be is when \(|5-k|\) is as large as possible, ie when \(k =0\) and we get a lower bound of \(0\). For \((k-1)k(k+1)\) notice this is the product of \(3\) consecutive integers, and therefore must be divisible by \(3\). (In fact, it's divisible by six, since \(\binom{k+1}{3}\) is the number of ways to choose \(3\) objects from \(k+1\). Let \(N = 100a + 10b + c\) where \(0 \leq a,b,c < 10\) and \(1 \leq a\). \(S = a + b^2 + c^3\) we want to find \begin{align*} && 100a +10b + c &= a + b^2 + c^3 \\ \Rightarrow && 0 &= \underbrace{99a}_{3 \mid 99 } + 10b - b^2 -\underbrace{c(c+1)(c-1)}_{3 \mid c(c+1)(c-1)} \\ \end{align*} Therefore \(3 \mid 10b - b^2 = b(10-b)\). Therefore \(3 \mid b\) or \(3 \mid 10-b\) so \(b = 0, 3, 6, 1, 4, 7\) We also have \(99a \geq 99\) and \(10b-b^2 \in [0, 25]\) so we need \(c^3-c \geq 99\), so \(c \geq 5\) Case \(c = 5\), Then \(c^3-c = 120\) so \(a = 1\) and \(10b-b^2 = 21\) so \(b= 3, 7\) \(N = 135, 175\) Case \(c = 6\), so \(c^3 - c = 210\) so \(a = 2\) and \(25-(5-k)^2 = 12\) so no solutions. Case \(c = 7\), so \(7^3 - 7 = 336\) so \(a = 3\) and \(25-(5-k)^2 = 39\) so no solutions. Case \(c = 8\) so \(8^3-8 = 504\) so \(a = 5\) and \(25-(5-k)^2 = 9\), so \(b = 1, 9\) and \(N = 518, 598\) Case \(c = 9\) so \(9^3 - 9 = 720\), so \(a = 7\) and \(25-(5-k)^2 = 27\) so no solutions. Therefore all the solutions are \(N = 135, 175, 518, 598\)

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1990 Paper 2 Q6
D: 1600.0 B: 1484.0
inequality proof positive integers fractions mediant Stern-Brocot systematic search quadratics and inequalities

Let \(a,b,c,d,p\) and \(q\) be positive integers. Prove that:

  1. if \(b > a\) and \(c > 1,\) then \(bc\geqslant2c\geqslant2+c\);
  2. if \(a < b\) and \(d < c\), then \(bc-ad\geqslant a+c\);
  3. if \({\displaystyle \frac{a}{b} < p < \frac{c}{d}}\), then \(\left(bc-ad\right)p\geqslant a+c\);
  4. if \({\displaystyle \frac{a}{b} < \frac{p}{q} < \frac{c}{d}},\) then \({\displaystyle p\geqslant\frac{a+c}{bc-ad}}\) and \({\displaystyle q\geqslant\frac{b+d}{bc-ad}}\).
Hence find all fractions with denominators less than 20 which lie between \(8/9\) and \(9/10\).

Solution:

  1. If \(b > a\) and \(c > 1\) then \(c \geq 2 \underbrace{\Rightarrow}_{\times c} bc \geq 2c = c+c \underbrace{\geq}_{c \geq 2} 2 + c\)
  2. If \(a < b\) and \(d < c\) then \(b \geq a+1\) and \(c \geq d+1\) so \begin{align*} bc - ad &\underbrace{\geq}_{b \geq a+1} (a+1)c - ad \\ &\underbrace{\geq}_{d \leq c-1} (a+1)c - a(c-1) \\ &= a+c \end{align*}
  3. If \(\displaystyle \frac{a}{b} < p < \frac{c}{d}\) then \(a < pb\) and \(pd < c\) so by the previous part \((pb)c - a(pd) \geq a + c \Leftrightarrow (bc-ad)p \geq a+c\).
  4. If \(\displaystyle \frac{a}{b} < \frac{p}{q} < \frac{c}{d}\) then \(\displaystyle \frac{qa}{b} < p < \frac{qc}{d}\) and so by the previous part we must have \((bc-ad)qp \geq q(a+c) \Rightarrow p \geq \frac{a+c}{bc-ad}\). Similarly we have \(\frac{d}{c} < \frac{q}{p} < \frac{b}{a}\) and so \(q \geq \frac{b+d}{bc-ad}\)
Suppose \(\frac{p}{q}\) is a fraction such that \(q \leq 20\) and \(\frac89 < \frac{p}q < \frac9{10}\) then: \begin{align*} q & \leq 20 \\ p & \geq \frac{17}{81-80} = 17 \\ q & \geq \frac{19}{81-80} = 19 \end{align*} Therefore the only fraction is \(\frac{17}{19}\) since \(\frac{18}{19} > \frac{18}{20} = \frac{9}{10}\)

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