Puzzles
6 December
When you add up the digits of a number, the result is called the digital sum.
How many different digital sums do the numbers from 1 to 1091 have?*
* There was a mistake in this question (it previously said 1092).
If you answered the typo'd question right, your answer should automatically correct itself to 9 less than it was...
5 December
Today's number is the number of ways that 35 can be written as the sum of distinct numbers, with none of the numbers in the sum being divisible by 9.
Clarification: By "numbers", I mean (strictly) positive integers. The sum of the same numbers in a different order is counted as the same sum: eg. 1+34 and 34+1 are not different sums.
The trivial sum consisting of just the number 35 counts as a sum.
4 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10. Today's number is the product of the digits in the red boxes.
| + | | ÷ | | = 2 |
| + | | ÷ | | - | |
| ÷ | | - | | = 5 |
| ÷ | | - | | × | |
| - | | × | | = 4 |
=
3 | | =
5 | | =
6 | |
3 December
What is the volume of the smallest cube inside which a rectangular-based pyramid of volume 266 will fit?
2 December
What is the maximum number of lines that can be formed by the intersection
of 30 planes?
1 December
One of the digits of today's number was removed to leave a two digit number.
This two digit number was added to today's number.
The result was 619.
Largest odd factors
Pick a number. Call it \(n\). Write down all the numbers from \(n+1\) to \(2n\) (inclusive). For example, if you picked 7, you would write:
$$8,9,10,11,12,13,14$$
Below each number, write down its largest odd factor. Add these factors up. What is the result? Why?
Show answer
Hide answer
Incredibly, the result will always be \(n^2\).
To see why, imagine writing every number, \(n+1\leq k\leq 2n\), in the form $$k=2^ab$$ where \(b\) is an odd number and also the \(k\)'s largest odd factor. The next largest number whose largest odd factor is \(b\) will be \(2^{a+1}b=2k\). But this will be larger than \(2n\), so outside the range. Therefore each number in the range has a different largest odd factor.
Each of the largest odd factors must be one of \(1, 3, 5, ..., 2n-1\), as they cannot be larger than \(2n\). But there are \(n\) odd numbers here and \(n\) numbers in the range, so each number \(1, 3, 5, ..., 2n-1\) is the highest odd factor of one of the numbers (as the highest odd factors are all different).
Therefore, the sum of the odd factors is the sum of the first \(n\) odd numbers, which is \(n^2\).
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.
Show answer & extension
Hide answer & extension
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?
