Puzzles
Square factorials
Multiply together the first 100 factorials:
$$1!\times2!\times3!\times...\times100!$$
Find a number, \(n\), such that dividing this product by \(n!\) produces a square number.
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First, look at how many times each number will appear in the product.
$$1!\times2!\times3!\times...\times100!
=1^{100}\times2^{99}\times3^{98}\times...\times100^1$$
Now split the odd and even numbers.
$$=\left[1^{100}\times3^{98}\times...\times99^2\right]\times\left[2^{99}\times4^{97}\times...\times100^1\right]$$
As all the powers in the first bracket are even, the first bracket is a square number.
$$=\left[1^{50}\times3^{49}\times...\times99^1\right]^2\times\left[2^{99}\times4^{97}\times...\times100^1\right]$$
Next, take a factor of two out of each number in the second bracket.
$$=\left[1^{50}\times3^{49}\times...\times99^1\right]^2\times\left[(2\times1)^{99}\times(2\times2)^{97}\times...\times(2\times50)^1\right]$$
$$=\left[1^{50}\times3^{49}\times...\times99^1\right]^2\times2^{99+97+...+1}\left[1^{99}\times2^{97}\times...\times50^1\right]$$
$$=\left[1^{50}\times3^{49}\times...\times99^1\right]^2\times2^{2500}\left[1^{99}\times2^{97}\times...\times50^1\right]$$
The only odd powers involved are now in the last bracket. Dividing by \(50!\) would make each of these powers even, hence the overall number would be square.
$$\frac{1!\times2!\times3!\times...\times100!
}{50!}=\left[1^{50}\times3^{49}\times...\times99^1\right]^2\times2^{2500}\left[1^{98}\times2^{96}\times...\times50^0\right]$$
$$=\left[1^{50}\times3^{49}\times...\times99^1\times2^{1250}\times1^{49}\times2^{48}\times...\times50^0\right]^2$$
Extension
For which numbers \(m\) is it possible to find a number \(n\) such that $$\frac{1!\times2!\times...\times m!}{n!}$$ is a square number?
An arm and a leg
If 60% of people have lost an eye, 75% an ear, 80% an arm and 85% a leg, what is the least percentage of people that have lost all four?
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40% of people still have both eyes, 25% both ears, 20% both arms and 15% both legs. If none of these overlap, they add to 100%. Therefore at least 0% of people have lost all four.
Blackboard sums II
The numbers 1 to 20 are written on a blackboard. Each turn, you may erase two adjacent numbers, \(a\) and \(b\) (\(a\) is to the left of \(b\)) and write the difference \(a-b\)
in their place. You continue until only one number remains.
What is the largest number you can make?
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Erasing 2 and 3; then the result and 4; then the result and 5 and so on will lead to having \(1\) and \(2-\sum_{i=3}^{20}i\) on the board.
Erasing these will give \(1-2+\sum_{i=3}^{20}i=206\).
Extension
The numbers 1 to 20 are written on a blackboard. Each turn, you may erase two adjacent numbers, \(a\) and \(b\) (\(a\) is to the left of \(b\)) and write the quotient \(a/b\)
in their place. You continue until only one number remains.
What is the largest number you can make?
Blackboard sums
The numbers 1 to 20 are written on a blackboard. Each turn, you may erase two numbers, \(a\) and \(b\) and write the sum \(a+b\) in their place. You continue until only one number remains.
What is the largest number you can make?
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After each turn, the total of the numbers written on the board remains the same. Therefore the sum of the numbers is always the same: 210
Extension
The numbers 1 to 20 are written on a blackboard. Each turn, you may erase two numbers, \(a\) and \(b\) and write the product \(a\times b\) in their place. You continue until only one number remains.
What is the largest number you can make?
Combining multiples
In each of these questions, positive integers should be taken to include 0.
1. What is the largest number that cannot be written in the form \(3a+5b\), where \(a\) and \(b\) are positive integers?
2. What is the largest number that cannot be written in the form \(3a+7b\), where \(a\) and \(b\) are positive integers?
3. What is the largest number that cannot be written in the form \(10a+11b\), where \(a\) and \(b\) are positive integers?
4. Given \(n\) and \(m\), what is the largest number that cannot be written in the form \(na+mb\), where \(a\) and \(b\) are positive integers?
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1. 7
2. 11
3. 90
First, if \(n\) and \(m\) share a common factor other than 1, there will be no largest number: any number that is not a multiple of the common factor will be impossible to make.
If \(n\) and \(m\) are coprime, then considered the remainders when the multiples of \(n\) are divided by \(m\). For example, if \(n=3\) and \(m=7\):
| Multiple | Remainder |
| 0 | 0 |
| 3 | 3 |
| 6 | 6 |
| 9 | 2 |
| 12 | 5 |
| 15 | 1 |
| 18 | 4 |
| 21 | 0 |
Once a number with a given remainder is reached, then all other numbers with that remainder can be reached by repeatedly adding \(m\). Once \(mn\) is reached, the remainders column will repeat itself. Before \(mn\), all remainders will appear (this can be shown by showing that there are \(m\) rows which much all have different remainders). Hence above \(mn\) all numbers can be made.
In the 3,7 example, the last remainder to be hit is 4. The highest number that cannot be made will be the highest number with remainder 4 that is less than 18 (when remainder 4 is hit).
In general, the last remainder will be hit at \(mn-n\). The number before this with the same remainder will be \(mn-n-m\). This will be the highest number that cannot be made.
Extension
Given \(n\), \(m\) and \(k\), what is the largest number that cannot be written in the form \(na+mb+kc\), where \(a\), \(b\) and \(c\) are positive integers?
Cross diagonal cover problem
Draw with an \(m\times n\) rectangle, split into unit squares. Starting in the top left corner, move at 45° across
the rectangle. When you reach the side, bounce off. Continue until you reach another corner of the rectangle:
How many squares will be coloured in when the process ends?
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$$\mathrm{lcm}(m-1,n-1)+1 - \frac12\left(\frac{\mathrm{lcm}(m-1,n-1)}{m-1}-1\right)\left(\frac{\mathrm{lcm}(m-1,n-1)}{n-1}-1\right)$$
Lots of ones
Is any of the numbers 11, 111, 1111, 11111, ... a square number?
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No. If one of them were a square number, then its square root must end in 1 or 9 (as this is the only way to make the final digit a one). So the square root is of the form \(10n\pm1\).
$$111...1=(10n\pm1)^2$$$$=100n^2\pm20n+1$$
$$=10(10n^2\pm2n)+1$$
If \(10(10n^2\pm2n)+1\) is of the form 111...1, then \(10n^2\pm2n\) is also of the form 111...1 (as it has just had the final 1 taken off). But \(10n^2\pm2n\) is even and 111...1 is odd, so this is not possible.
Subsum
1) In a set of three integers, will there always be two integers whose sum is even?
2) How many integers must there be in a set so that there will always be three integers in the set whose sum is a multiple of 3?
3) How many integers must there be in a set so that there will always be four integers in the set whose sum is even?
4) How many integers must there be in a set so that there will always be three integers in the set whose sum is even?
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1) Yes, there must be either at least two even integers or at least two odd integers. The sum of two even integers is even. The sum of two odd integers is even.
2) Five.
3) Five.
4) An infinite number: in a list of odd integers, there will never be three integers which add up to an even number.
This puzzle leads naturally into the following extension:
Extension
How many integers must there be in a set so that there will always be \(a\) integers in the set whose sum is a multiple of \(b\)?
For which values of \(a\) and \(b\) will the answer to this be finite?
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