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Triangles between squares
Prove that there are never more than two triangle numbers between two consecutive square numbers.
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Let \(T_a\) represent the \(a\)th triangle number. This means that \(T_a=\frac{1}{2}a(a+1)\).
Suppose that for some integer \(n\), \(n^2 \leq T_a <(n+1)^2\). This means that:
$$n^2 \leq T_a$$
$$n^2 \leq \frac{1}{2}a(a+1)$$
$$2n^2 \leq a^2+a$$
But for every positive integer \(a \leq a^2\), so:
$$2n^2 \leq 2a^2$$
$$n^2 \leq a^2$$
\(n\) and \(a\) are both positive integers, so:
$$n \leq a$$
Now consider \(T_{a+2}\):
$$T_{a+2}=\frac{1}{2}(a+2)(a+3)$$
$$=\frac{1}{2}(a^2+5a+6)$$
$$=\frac{1}{2}(a^2+a)+\frac{1}{2}(4a+6)$$
$$=\frac{1}{2}a(a+1)+2a+3$$
$$=T_a+2a+3$$
We know that \(a \geq n\) and \(T_a \geq n^2\), so:
$$T_a+2a+3 \geq n^2+2n+3$$
$$>n^2+2n+1 = (n+1)^2$$
And so \(T_{a+2}\) is not between \(n^2\) and \((n+1)^2\). So if a triangle number \(T_a\) is between \(n^2\) and \((n+1)^2\) then the next but one triangle number \(T_{a+2}\) cannot also be between \(n^2\) and \((n+1)^2\). So there cannot be more than two triangle numbers between \(n^2\) and \((n+1)^2\).
Extension
Given an integer \(n\), how many triangle numbers are there between \(n^2\) and \((n+1)^2\)?
Odd and even outputs
Let \(g:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) be a function.
This means that \(g\) takes two natural number inputs and gives one natural number output. For example if \(g\) is defined by \(g(n,m)=n+m\) then \(g(3,4)=7\) and \(g(10,2)=12\).
The function \(g(n,m)=n+m\) will give an even output if \(n\) and \(m\) are both odd or both even and an odd output if one is odd and the other is even. This could be summarised in the following table:
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | odd |
| e | odd | even | |
Using only \(+\) and \(\times\), can you construct functions \(g(n,m)\) which give the following output tables:
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| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | odd |
| e | odd | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | odd |
| e | odd | even | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | odd |
| e | even | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | odd |
| e | even | even | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | even |
| e | odd | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | even |
| e | odd | even | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | even |
| e | even | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | odd | even |
| e | even | even | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | odd |
| e | odd | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | odd |
| e | odd | even | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | odd |
| e | even | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | odd |
| e | even | even | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | even |
| e | odd | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | even |
| e | odd | even | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | even |
| e | even | odd | |
| \(n\) | |||
| odd | even | ||
| \(m\) | odd | even | even |
| e | even | even | |
Extension
Can you find functions \(h:\mathbb{N}\times\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) (call the inputs \(n\), \(m\) and \(l\)) to give the following outputs:
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etc
Tags: functions
Twenty-one
Scott and Virgil are playing a game. In the game the first player says 1, 2 or 3, then the next player can add 1, 2 or 3 to the number and so on. The player who is forced to say 21 or above loses. The first game went like so:
Scott: 3
Virgil: 4
Scott: 5
Virgil: 6
Scott: 9
Virgil: 12
Scott: 15
Virgil 17
Scott: 20
Virgil: 21
Virgil loses.
To give him a better chance of winning, Scott lets Virgil choose whether to go first or second in the next game. What should Virgil do?
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Virgil should go second. Whatever Scott adds, Virgil should then add to make four. For example, if Scott says 3, Virgil should say 1.
Using this strategy, Virgil will say 4, 8, 12, 16 then 20, forcing Scott to go above 21.
Extension
(i) If instead of 21, 22 cannot be said/beaten, how should Virgil win? How about 23? Or 24? How about \(n\)?
(ii) If instead of adding 1 to 3, 1 to 4 can be added, how should Virgil win? How about 1 to 5? Or 2 to 5? How about \(m\) to \(l\)?
(iii) Alan wants to join the game. Can Virgil win if there are three people? Can he win if there are \(k\) people?
Polya strikes out
Write the numbers 1, 2, 3, ... in a row. Strike out every third number beginning with the third. Write down the cumulative sums of what remains:
1, 2, 3, 4, 5, 6, 7, ...
1, 2, 3, 4, 5, 6, 7, ...
1, 2, 4, 5, 7, ...
1=1; 1+2=3; 1+2+4=7; 1+2+4+5=12; 1+2+4+5+7=19; ...
1, 3, 7, 12, 19, ...
Now strike out every second number beginning with the second. Write down the cumulative sums of what remains. What is the final sequence? Why do you get this sequence?
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1, 3, 7, 12, 19, ...
1, 7, 19, ...
1=1; 1+7=8; 1+7+19=27; ...
1, 8, 27, ...
The final sequence is the cube numbers. To show why, let \(n\) be an integer and follow through the process.
Cross out every third number:
1, 2, 3, 4, 5, 6, ..., 3n, \(3n+1\), \(3n+2\), ...
1, 2, 4, 5, ..., \(3n+1\), \(3n+2\), ...
Find the cumulative sums:
$$1=1$$
$$1+2=1+2=3$$
$$1+2+4=1+2+3+4-3=7$$
$$1+2+4+5=1+2+3+4+5-3=12$$
$$...$$
$$1+2+4+5+...+(3n+1)=\sum_{i=1}^{3n+1}-\sum_{i=1}^{n}3i$$
$$=\frac{1}{2}(3n+1)(3n+2)-\frac{3}{2}n(n+1)$$
$$=3n^2+3n+1$$
$$1+2+4+5+...+(3n+2)=3n^2+3n+1+(3n+2)$$
$$=3n^2+6n+3$$
$$...$$
1, 3, 7, 12, ..., \(3n^2+3n+1\), \(3n^2+6n+3\), ...
Cross out every second number, starting with the second:
1, 3, 7, 12, ..., \(3n^2+3n+1\), 3n2+6n+3, ...
1, 7, ..., \(3n^2+3n+1\), ...
Find the cumulative sums. The \(m\)th sum is:
$$\sum_{n=0}^{m}3n^2+3n+1$$
$$=3\sum_{n=0}^{m}n^2+3\sum_{n=0}^{m}n+\sum_{n=0}^{m}1$$
$$=\frac{3}{6}m(m+1)(2m+1)+\frac{3}{2}m(m+1)+m+1$$
$$=\frac{1}{2}(m+1)(m(2m+1)+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+m+3m+2)$$
$$=\frac{1}{2}(m+1)(2m^2+4m+2)$$
$$=(m+1)(m^2+2m+1)$$
$$=(m+1)(m+1)^2$$
$$=(m+1)^3$$
Hence the numbers obtained are the cube numbers.
Extension
What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every \(n\)th number starting at the \(m\)th?
Tags: numbers
Whist
Messrs. Banker, Dentist, Apothecary and Scrivener played whist last night. (whist is a four player card game where partners sit opposite each other.) Each of these gentlemen is the namesake of another's vocation.
Last night, the apothecary partnered Mr. Apothecary; Mr. Banker's partner was the scrivener; on Mr. Scrivener's right sat the dentist.
Who sat on the banker's left?
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Represent each player by a compass point. Let B, D, A and S represent Messrs. Banker, Dentist, Apothecary and Scrivener respectively and b, d, a and s represent the four jobs.
Mr. Banker (B) partners the scrivener (s). Let B sit at West. This means s sits at East. As no other player can also be B or s, the table looks like this:
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The dentist (d) sits on Mr. Scrivener's (S) right. East cannot be S, so North cannot be d. East cannot be d, so South cannot be S.
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By elimination, only North can be S. This means that d must sit to the right of North (at West):
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A and a are partners. This is only possible if A is South and a is North:
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Therefore, Mr. Banker the dentist sits to the left of the banker.
Extension
If each person is partnered with their job namesake, how many possible combinations of names and jobs are possible?
Exact change
Source: @AlexDBolton on Twitter
In the UK, the coins less than £1 are 1p, 2p, 5p, 10p, 20p and 50p. How many coins would I need to carry in my pocket so that I could make any value from 1p to 99p?
In the US, the coins less than $1 are 1¢, 5¢, 10¢, 25¢. How many coins would I need to carry in my pocket so that I could make any value from 1¢ to 99¢?
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In the UK, eight coins are needed: 1p, 1p, 2p, 5p, 10p, 20p, 20p, 50p.
In the US, ten coins are needed: 1¢, 1¢, 1¢, 1¢, 5¢, 10¢, 10¢, 25¢, 25¢, 25¢.
Extension
In a far away country, the unit of currency is the #, which is split into 100@ (# is like £ or $; @ is like p or ¢).
Let C be the number of coins less than #1. Let P be the number of coins needed to make any value between 1@ and 99@. Which coins should be the country mint to minimise the value of P+C?
Square cross
Source: Teach Further Maths Blog
![](javascript:showlimage("959058_orig.png"))
A figure in the shape of a cross is made from five 1 x 1 squares, as shown. The cross is inscribed in a large square whose sides are parallel to the dashed square, formed by four vertices of the cross.
What is the area of the large outer square?
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![](javascript:showlimage("cross+square.png"))
The smaller diagonal square is made up of a 1×1 square and four 1×2 right-angled triangles. Therefore its area is 5.
The line FM, and therefore the line DN, has gradient 2. The line JM, and therefore the line ON, has gradient -½. ON passes through (1,2) and DN passes through (2,0). Therefore, DN has equation \(y=2x-4\) and ON has equation \(y=\frac{5}{2}-\frac{1}{2}x\). These lines intersect at \((\frac{13}{5},\frac{6}{5})\), these are the co-ordinates of N. By the same method, the co-ordinates of P are \((-\frac{8}{5},-\frac{1}{5})\).
By Pythagoras' Theorem, The diagonal of the larger square is \(\frac{7\sqrt{10}}{5}\) and so the area of the larger square is 9.8.
Ten digit number
Source: Richard Wiseman's Blog
Can you create a 10-digit number, where the first digit is how many zeros in the number, the second digit is how many 1s in the number etc. until the tenth digit which is how many 9s in the number?
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6210001000 has 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights and 0 nines.
Extension
Are there any more numbers like this? Prove that you have them all.
Tags: numbers
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