Puzzles
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7 December
There are 15 dominos that can be made using the numbers 0 to 4 (inclusive):
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The sum of all the numbers on all these dominos is 60.
Today's number is the sum of all the numbers on all the dominos that can be made using the numbers 5 to 10 (inclusive).
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Each number will appear 7 times: one time paired with the numbers six numbers 5 to 10, plus an extra appearance on the tile containing the same number twice.
The total of all the numbers is therefore 7×(5+6+...+10)=7×45=315.
6 December
There are 12 ways of placing 2 tokens on a 2×4 grid so that no two tokens are next to each other horizonally, vertically or diagonally:
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Today's number is the number of ways of placing 5 tokens on a 2×10 grid so that no two tokens are next to each other horizonally, vertically or diagonally.
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First, consider placing 5 tiles in a 1×9 grid. There is only one way to do this:
O | O | O | O | O |
To get the number of ways of placing 5 tiles in a 1×10 grid, imagine adding an extra blank square to either the start or end of the grid or between two of the counters.
There are 6 places this tile could be inserted leading to 6 arrangements of 5 tiles in a 1×10 grid.
For 5 tiles in a 2×10 grid, you can first pick the columns the tiles go in (as a tile being in a column means nothing can be placed the columns either side, the number of ways to pick
columns is the same and the number of wats to arrange 5 tokens in a 1×10 grid). For each of these column choices, there are two locations for each tile (top or bottom).
This leads to a total number of arrangements of 6×25=192.
Tags: numbers, combinatorics
5 December
Carol rolled a large handful of six-sided dice. The total of all the numbers Carol got was 521. After some calculating, Carol worked out that the probability that of her total being 521
was the same as the probability that her total being 200. How many dice did Carol roll?
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The totals that are equally likely add up to 7 times the number of dice.
This can be seen by using the fact that the opposite sides of a dice add up to 7: for each way of making a given total with \(n\) dice, there is a way of making \(7n\) minus that total
by looking at the dice from below. Therefore \(T\) and \(7n-T\) are equally likely. (This also holds true (but is harder to explain) if you rearrange the faces of the dice so that the opposite
faces no longer add to 7.)
Therefore today's number is \((521+200)/7\), which is 103.
4 December
Today's number is a three digit number which is equal to the sum of the cubes of its digits. One less than today's number also has this property.
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If the final digit of the number is 0, then some carrying takes place when 1 is subtracted. Otherwise, no carrying happens.
If no carrying happens, call the three digits of today's number \(A\), \(B\), and \(C\). We know that \(A^3+B^3+C^3\) is one more than \(A^3 + B^3 + (C-1)^3\).
This implies that \(C^3=(C-1)^3+1\), which is only possible if \(C\) is 1.
Therefore either the final digit of today's number is 0 or the final digt of one less that today's number is 0. In both cases, we need to find a number
with the desired property whose final digit is 0: we are looking for digit \(A\) and \(B\) such that \(A^3+B^3\) is a multiple of 10.
Looking at all the cube numbers, there are a few combinations that add up to multiple of 10:
$$0^3+0^3=0$$
$$1^3+9^3=730$$
$$2^3+8^3=520$$
$$3^3+7^3=370$$
$$4^3+6^3=280$$
$$5^3+5^3=250$$
The only one of these that has the required property is 370. By checking 369 and finding it doesn't have the property, we see that the two numbers must be
370 and 371.
3 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 largest number you can make with the digits in the red boxes.
÷ | - | = 3 | |||
+ | + | ÷ | |||
÷ | × | = 1 | |||
× | - | + | |||
- | × | = 20 | |||
= 91 | = 6 | = 8 |
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9 | ÷ | 1 | - | 6 | = 3 |
+ | + | ÷ | |||
4 | ÷ | 8 | × | 2 | = 1 |
× | - | + | |||
7 | - | 3 | × | 5 | = 20 |
= 91 | = 6 | = 8 |
The largest number you can make with the digits in the red boxes is 321.
2 December
Carol draws a square with area 62. She then draws the smallest possible circle that this square is contained inside.
Next, she draws the smallest possible square that her circle is contained inside. What is the area of her second square?
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By drawing an appropriate diagram, it can be seen that the small square has half the area of the large square.
Therefore the area of the large square is 124.
1 December
It is possible to write 325 different numbers using the digits 1, 2, 3, 4, and 5 at most once each (and using no other digits).
How many of these numbers are odd?
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There are 3 one-digit numbers using these digits (1, 3 and 5).
To make two-digit odd numbers, there are 3 choices for the units digit, and 4 remaining choices for the tens digit.
To make three-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, and 3 remaining choices for the hundreds digit.
To make four-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, 3 remaining choices for the hundreds digit, and 2 remaining choices for the thousands digit.
To make five-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, 3 remaining choices for the hundreds digit, 2 remaining choices for the thousands digit and 1 remaining choice for the ten-thousands digit.
In total, this gives 3 + 3×4× + 3×4×3 + 3×4×3×2 + 3×4×3×2×1 = 195 odd numbers.
Advent 2019 logic puzzle
It's nearly Christmas and something terrible has happened: while out on a test flight, Santa's sled was damaged and Santa, Rudolph and Blitzen fell to the ground over the Advent Isles.
You need to find Santa and his reindeer before Christmas is ruined for everyone.
You have gathered one inhabitant of the four largest Advent Isles—Rum, Land, Moon and County—and they are going to give you a series of clues about where Santa and his reindeer landed.
However, one or more of the islanders you have gathered may have been involved in damaging Santa's sled and causing it to crash: any islander involved in this will lie to you to attempt to stop
you from finding Santa and his reindeer.
Once you are ready to search for Santa, Rudolph and Blitzen, you can find the map by following this link.
Each of the clues will be about Santa's, Rudolph's or Blitzen's positions in Advent Standard Coordinates (ASC): ASC are given by six two-digit numbers with dots inbetween, for example
12.52.12.13.84.55.
For this example coordinate, the islanders will refer to
(the first) 12 as the first coordinate,
52 as the second coordinate,
(the second) 12 as the third coordinate,
13 as the fourth coordinate,
84 as the fifth coordinate, and
55 as the sixth coordinate.
Here are the clues:
3
Rum says: "The product of all the digits in Blitzen's six coords is 432."21
Moon says: "Blitzen's fifth coord is 23."9
Moon says: "Blitzen's third coord is 23."1
Land says: "Santa's third coord ends in 3, 0 or 1."2
Land says: "Santa's third coord ends in 2, 0 or 3."4
Rum says: "Santa's second coord ends in 3, 4 or 1."12
Rum says: "Rudolph's second and sixth coords are both 64."10
Rum says: "All six of Rudolph's coords are factors of 256."18
Moon says: "Santa's fourth and fifth coords are both 79."24
County says: "Santa's third coord ends in 3, 2 or 1."22
Land says: "Santa's sixth coord is not 43."7
Rum says: "Santa's sixth coord is 43."23
County says: "One of the digits of Santa's third coord is 7."25
✔14
Land says: "Santa's third coord is 12."5
Rum says: "Santa's first coord is 36."15
Rum says: "Blitzen's first coord is 23."17
Rum says: "The first digit of Santa's third coord is 1."8
County says: "Santa's third coord shares a factor (≠1) with 270."6
County says: "Santa's second coord is 21."16
Land says: "Blitzen's second coord is 21."20
Moon says: "All six of Rudolph's coords are multiples of 8."11
Moon says: "The sum of Rudolph's six coords is 192."13
Moon says: "Santa's second coord is 21 or 11."19
Moon says: "Blitzen's fourth and sixth coords are both 11."To find a point's ASC coordinates, split a map of the islands into a 9×9 grid, then number the rows and columns 1 to 9: the first two digits of ASC give the vertical then horizontal position of a square in this grid.
The next two digits then give a smaller square when this square is then itself split into a 9×9 grid, and so on. An example is show below.
You can view the map here.
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Santa was at 36.11.19.79.79.43.
Rudolph was at 16.64.16.16.16.64.
Blitzen was at 23.12.23.11.23.11.
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