Puzzles
15 December
Today's number is smallest three digit palindrome whose digits are all non-zero, and that is not divisible by any of its digits.
14 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 numbers in the red boxes.
- | + | = 10 | |||
÷ | + | ÷ | |||
÷ | + | = 3 | |||
+ | - | ÷ | |||
+ | × | = 33 | |||
= 7 | = 3 | = 3 |
13 December
There is a row of 1000 lockers numbered from 1 to 1000. Locker 1 is closed and locked and the rest are open.
A queue of people each do the following (until all the lockers are closed):
- Close and lock the lowest numbered locker with an open door.
- Walk along the rest of the queue of lockers and change the state (open them if they're closed and close them if they're open) of all the lockers that are multiples of the locker they locked.
Today's number is the number of lockers that are locked at the end of the process.
Note: closed and locked are different states.
12 December
There are 2600 different ways to pick three vertices of a regular 26-sided shape. Sometimes the three vertices you pick form a right angled triangle.
Today's number is the number of different ways to pick three vertices of a regular 26-sided shape so that the three vertices make a right angled triangle.
11 December
This puzzle is inspired by a puzzle Woody showed me at MathsJam.
Today's number is the number \(n\) such that $$\frac{216!\times215!\times214!\times...\times1!}{n!}$$ is a square number.
10 December
The equation \(x^2+1512x+414720=0\) has two integer solutions.
Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.
9 December
Today's number is the number of numbers between 10 and 1,000 that contain no 0, 1, 2 or 3.
8 December
Arrange the digits 1-9 in a 3×3 square so: each digit the first row is the number of letters in the (English) name of the previous digit, each digit in the second row is one less than the previous digit, each digit in the third row is a multiple of the previous digit; the second column is an 3-digit even number, and the third column contains one even digit.
The number in the first column is today's number.
each digit is the number of letters in the previous digit | |||
each digit is one less than previous | |||
each digit is multiple of previous | |||
today's number | even | 1 even digit |
Edit: There was a mistake in this puzzle: the original had two solutions. If you entered the wrong solution, it will automatically change to the correct one.