Puzzles
Multiple sums
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
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The multiples of 3 less than 1000 are 3,6,9,...,999; the multiples of 5 are 5,10,15,...,995. Multiples of 15 (15,30,...,990) will appear in both lists so we are trying to find (3+6+9+...+999)+(5+10+15+...+995)-(15+30+...+990). This is:
$$\sum_{i=1}^{333}3i+\sum_{j=1}^{199}5j-\sum_{k=1}^{66}15k$$
$$=3\sum_{i=1}^{333}i+5\sum_{j=1}^{199}j-15\sum_{k=1}^{66}k$$
$$=3\times\frac{333\times334}{2}+5\times\frac{199\times200}{2}-15\times\frac{66\times67}{2}$$
$$=166833+99500-33165$$
$$=233168$$
Extension
Find the sum of all the multiples of 3 or 5 below \(n\).
Complex squares
For which complex numbers, \(z\), are \(\mathrm{Re}(z^2)\) and \(\mathrm{Im}(z^2)\) both positive?
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Any complex number can be written in the form \(z=re^{i\theta}\).
This gives that \(z^2=r^2e^{2i\theta}\), which will have positive real and complex parts when \(0+2\pi n < 2\theta < \frac{\pi}{2}+2\pi n\).
This will occur when \(0 < \theta < \frac{\pi}{4}\) and \(\pi < \theta < \frac{5\pi}{4}\).
A complex number \(z\) falls in these regions when \(|\mathrm{Re}(z)|>|\mathrm{Im}(z)|\) and \(\mathrm{sign}(\mathrm{Re}(z))=\mathrm{sign}(\mathrm{Im}(z))\).
Extension
For which complex numbers, \(z\), are \(\mathrm{Re}(z^3)\) and \(\mathrm{Re}(z^3)\) both positive?
Adding bases
Let \(a_b\) denote \(a\) in base \(b\).
Find bases \(A\), \(B\) and \(C\) less than 10 such that \(12_A+34_B=56_C\).
Reverse bases again
Find three digits \(a\), \(b\) and \(c\) such that \(abc\) in base 10 is equal to \(cba\) in base 9?
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445 in base 10 is equal to 544 in base 7.
Extension
Find another pair of bases \(A\) and \(B\) so that there exist digits \(d\), \(e\) and \(f\) such that \(def\) in base \(A\) is equal to \(fed\) in base \(B\)?
Two
Find \(a\) such that \(a+(a+A)^{-1}=2\), where \(A=(a+A)^{-1}\).
ie. \(a + \frac{1}{a + \frac{1}{a + \frac{1}{a + \frac{1}{...}}}} = 2\).
Find \(b\) such that \(b+(b+B)^{\frac{1}{2}}=2\), where \(B=(b+B)^{\frac{1}{2}}\).
ie. \(b + \sqrt{b + \sqrt{b + \sqrt{b + \sqrt{...}}}} = 2\).
Find \(c\) such that \(c+(c+C)^{2}=2\), where \(C=(c+C)^{2}\).
In terms of \(k\), find \(d\) such that \(d+(d+D)^{k}=2\), where \(D=(d+D)^{k}\).
Reverse bases
Find two digits \(a\) and \(b\) such that \(ab\) in base 10 is equal to \(ba\) in base 4.
Find two digits \(c\) and \(d\) such that \(cd\) in base 10 is equal to \(dc\) in base 7.
Find two digits \(e\) and \(f\) such that \(ef\) in base 9 is equal to \(fe\) in base 5.
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If \(ab\) in base 10 is equal to \(ba\) in base 4, then \(10a+b=4b+a\).
So, \(9a=3b\).
\(a\) and \(b\) must both be less than 4, as they are digits used in base 4, so \(a=1\) and \(b=3\).
So 13 in base 10 is equal to 31 in base 4.
By the same method, we find that:
- 23 in base 10 is equal to 32 in base 7.
- 46 in base 10 is equal to 64 in base 7.
- 12 in base 9 is equal to 21 in base 5.
- 24 in base 9 is equal to 42 in base 5.
Extension
For which pairs of bases \(A\) and \(B\) can you find two digits \(g\) and \(h\) such that \(gh\) in base \(A\) is equal to \(hg\) in base \(B\)?
Ninety nine
In a 'ninety nine' shop, all items cost a number of pounds and 99 pence. Susanna spent £65.76. How many items did she buy?
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Every item bought will cause the pence in the total cost to fall by 1. So to spend £65.76, Susanna must have bought 24 items.
Extension
What is the smallest amount Susanna could spend for which we could not tell how many items she bought?
Triangle numbers
Let \(T_n\) be the \(n^\mathrm{th}\) triangle number. Find \(n\) such that: $$T_n+T_{n+1}+T_{n+2}+T_{n+3}=T_{n+4}+T_{n+5}$$
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\(T_{n} = \frac{1}{2}n(n+1)\), so:
$$T_{n}+T_{n+1} = \frac{1}{2}n(n+1) + \frac{1}{2}(n+1)(n+2)$$
$$= (n+1)^2$$
So, we are looking for \(n\) such that \((n+1)^2+(n+3)^2=(n+5)^2\). This is true when \(n=5\) (\(6^2+8^2=10^2\)).
Extension
Find \(n\) such that \(T_{n}+T_{n+1}+T_{n+1}+T_{n+2}=T_{n+2}+T_{n+3}\).
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