Puzzles
11 December
Today's number is the number \(n\) such that $$\frac{216!\times215!\times214!\times...\times1!}{n!}$$ is a square number.
Square and cube endings
Source: UKMT 2011 Senior Kangaroo
How many positive two-digit numbers are there whose square and cube both end in the same digit?
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Only the units digit of the number will affect the last digit of the square and cube. This table shows how the last digits of the square and cube depend on the last digit of the number:
| Last digit of... |
| number | square | cube |
| 0 | 0 | 0 |
| 1 | 1 | 1 |
| 2 | 4 | 8 |
| 3 | 9 | 7 |
| 4 | 6 | 4 |
| 5 | 5 | 5 |
| 6 | 6 | 6 |
| 7 | 9 | 3 |
| 8 | 4 | 2 |
| 9 | 1 | 9 |
So numbers ending in 0, 1, 5 and 6 will have squares and cubes that end in the same digit. There are 4×9=36 two-digit numbers then end in one of these digits.
Extension
How many two-digit numbers are there in binary whose square and cube end in the same digit?
How many two-digit numbers are there in ternary whose square and cube end in the same digit?
How many two-digit numbers are there in base \(n\) whose square and cube end in the same digit?
What's the star?
In the Christmas tree below, the rectangle, baubles, and the star at the top each contain a number. The square baubles contain square numbers; the triangle baubles contain triangle numbers; and the cube bauble contains a cube number.
The numbers in the rectangles (and the star) are equal to the sum of the numbers below them. For example, if the following numbers are filled in:
then you can deduce the following:
What is the number in the star at the top of this tree?
You can download a printable pdf of this puzzle here. Show answer
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The two numbers between 14 are a cube number and a triangle number: these must be 8 and 6. Next you can see that 23 is the sum of 8, two times a triangle number and a square number: the triangle and square numbers must be 3 and 9.
Next, call the square number at the bottom left \(a\), the square number at the top left \(b\), and the triangle number at the top right \(c\). Adding upwards, we find that \(a+b+45=106\) and \(a+141+c=198\); and so \(a+b=61\) and \(a+c=57\). The only two square numbers that add to 61 are 25 and 36. Therefore \(c\) must be 21 or 32, but must be 21 as 32 is not a triangle number. And so \(a\) is 36 and \(b\) is 25.
Putting all these numbers into the tree gives the top number as
433. This is fitting because 433 is in fact a
star number:
Square pairs
Source: Maths Jam
Can you order the integers 1 to 16 so that every pair of adjacent numbers adds to a square number?
For which other numbers \(n\) is it possible to order the integers 1 to \(n\) in such a way?
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Yes: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16.
It is clearly possible for the numbers 1 to 15 (remove the 16 from the end of the sequence above) and 1 to 17 (add 17 to the start of the sequence above).
The OEIS sequence
A090461 gives other numbers for which this is possible. It starts 15, 16, 17, 23, then includes every number from 25 onwards. It is conjectured, but not proven, that it is possible for every number above 25.
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?
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.
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![](javascript:showlimage("puzzletree.png"))
![](javascript:showlimage("433star.png"))