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Puzzles

Advent 2016 murder mystery

2016's Advent calendar ended with a murder mystery, with each of the murderer, motive, weapon and location being a digit from 1 to 9. Here are the clues:
10
None of the digits of 171 is the location.
3
None of the digits of 798 is the motive.
7
One of the digits of 691 is the location.
16
None of the digits of 543 is the location.
5
One of the digits of 414 is the murderer.
20
The first digit of 287 is the number of false red clues.
8
Clues on days that are factors of 768 are all true.
22
The murderer is the square root of one of the digits of 191.
11
One of the digits of 811 is the weapon.
19
The highest common factor of the weapon and 128 is 1.
13
None of the digits of 512 is the murderer.
18
One of the digits of 799 is the motive.
17
None of the digits of 179 is the motive.
6
None of the digits of 819 is the location.
24
One of the digits of 319 is total number of false clues.
23
One of the digits of 771 is the murderer.
2
The weapon is not one of the digits of 435.
14
The final digit of 415 is the number of true blue clues.
4
The weapon is a factor of 140.
12
The number of false clues before today is the first digit of 419.
9
One of the digits of 447 is the motive.
1
None of the digits of 563 is the motive.
21
One of the digits of 816 is the murderer.
15
One of the digits of 387 is the motive.

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24 December

Today's number is 191 more than one of the other answers and 100 less than another of the answers.
Tags: numbers

23 December

Today's number is the number of three digit numbers that are not three more than a multiple of 7.
Tags: numbers

22 December

Today's number is a palindrome. Today's number is also the number of palindromes between 111 and 11111 (including 111 and 11111).

21 December

Today's number is a multiple of three. The average (mean) of all the answers that are multiples of three is a multiple of 193.
Tags: averages, mean

20 December

Earlier this year, I wrote a blog post about different ways to prove Pythagoras' theorem. Today's puzzle uses Pythagoras' theorem.
Start with a line of length 2. Draw a line of length 17 perpendicular to it. Connect the ends to make a right-angled triangle. The length of the hypotenuse of this triangle will be a non-integer.
Draw a line of length 17 perpendicular to the hypotenuse and make another right-angled triangle. Again the new hypotenuse will have a non-integer length. Repeat this until you get a hypotenuse of integer length. What is the length of this hypotenuse?

19 December

The sum of all the numbers in the eighth row of Pascal's triangle.
Clarification: I am starting the counting of rows from 1, not 0. So (1) is the 1st row, (1 1) is the 2nd row, (1 2 1) is the 3rd row, etc.

18 December

The smallest number whose sum of digits is 25.

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