Puzzles
18 December
If k = 21, then 28k ÷ (28 + k) is an integer.
What is the largest integer k such that 28k ÷ (28 + k) is an integer?
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If \(k\) is a multiple of 28, then let \(k=28a\) and see that:
$$\frac{28k}{28+k}=\frac{28a}{1+a}$$
This is an integer if either \(a=0\), \(a=1\), or \(1+a\) is a factor of 28. Hence the largest value of \(a\) is 27, leading to \(k\) = 756.
Similar working out can be done for the cases where the hcf of 28 and \(k\) is 14, 7, 2, or 1, and in each case a lower answer would be obtained.
What is the sum?
What is \(\displaystyle\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)?
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Start by rationalising the denominators:
$$\begin{array}{rl}
\frac{1}{\sqrt{n}+\sqrt{n+1}}
&=\frac{1}{\sqrt{n}+\sqrt{n+1}}\times\frac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}-\sqrt{n+1}}\\
&=\frac{\sqrt{n}-\sqrt{n+1}}{n-(n+1)}\\
&=\sqrt{n+1}-\sqrt{n}\end{array}$$
Therefore:
$$\begin{array}{rl}
\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}
&=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\\&...+(\sqrt{15}-\sqrt{14})+(\sqrt{16}-\sqrt{15})\\
&=-\sqrt{1}+\sqrt{16}\\
&=3
\end{array}
$$
19 December
1 = 1/2 + 1/4 + 1/8 + 1/16 + 1/16. This is the sum of 5 unit fractions (the numerators are 1).
In how many different ways can 1 be written as the sum of 5 unit fractions? (the same fractions in a different order are considered the same sum.)
Shooting hoops
You spend an afternoon practising throwing a basketball through a hoop.
One hour into the afternoon, you have scored less than 75% of your shots. At the end of the afternoon, you have score more than 75% of your shots.
Is there a point in the afternoon when you had scored exactly 75% of your shots?
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If there is not a point when you had scored exactly 75%, then there must be a shot which you score which took your average over 75%. In other words, there must be an $a$ and a $b$ so that:
$$\frac{a}{b}<\frac{3}{4}<\frac{a+1}{b+1}$$
Rearranging these inequalities gives:
$$4a<3b$$
$$3b+3<4a+4$$
Taking 3 from the bottom inequality gives:
$$4a<3b$$
$$3b<4a+1$$
Combining the inequalities gives:
$$4a<3b<4a+1$$
Therefore \(3b\) must be an integer which is between \(4a\) and \(4a+1\). But \(4a\) and \(4a+1\) are consecutive numbers so there is no other integer between them.
This means that this situation is impossible and there must have been a time when you had scored exactly 75% of your shots.
Extension
You spend another afternoon practising throwing a basketball through a hoop.
One hour into the afternoon, you have scored more than 75% of your shots. At the end of the afternoon, you have score less than 75% of your shots.
Is there a point in the afternoon when you had scored exactly 75% of your shots?
Odd sums
What is \(\frac{1+3}{5+7}\)?
What is \(\frac{1+3+5}{7+9+11}\)?
What is \(\frac{1+3+5+7}{9+11+13+15}\)?
What is \(\frac{1+3+5+7+9}{11+13+15+17+19}\)?
What is \(\frac{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ odd\ numbers}}{\mathrm{sum\ of\ the\ next\ }n\mathrm{\ odd\ numbers}}\)?
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They are all equal to one third.
The sum of the first \(n\) odd numbers is \(n^2\) (this can be proved by induction). This means that:
$$\frac{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ odd\ numbers}}{\mathrm{sum\ of\ the\ next\ }n\mathrm{\ odd\ numbers}}=\frac{n^2}{(2n)^2-n^2}\\
=\frac{n^2}{3n^2}=\frac{1}{3}$$
Extension
What is \(\frac{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ odd\ numbers}}{\mathrm{sum\ of\ the\ first\ }n\mathrm{\ even\ numbers}}\)?
