Puzzles
24 December
There are six ways to put two tokens in a 3 by 3 grid so that the diagonal from the top left to the bottom right is a line of symmetry:
Today's number is the number of ways of placing two tokens in a 29 by 29 grid so that the diagonal from the top left to the bottom right is a line of symmetry.
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Either both pieces must be on the diagonal, or one pieces is in the lower right half and the other is in the reflected position in the upper right half.
There are \(\left(\begin{array}{c}29\\2\end{array}\right)=406\) ways to pick two squares on the diagonal. There are 406 squares below the diagonal.
Therefore there are 406+406 = 812 ways to arrange the pieces.
23 December
198 is the smallest number that is equal to 11 times the sum of its digits.
Today's number is the smallest number that is equal to 48 times the sum of its digits.
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We're looking for a number \(n\) such that the digits of \(n\times48\) add up to \(n\). If we try numbers in order, we find that 9 works, so today's number is 432.
22 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.
| + | | + | | = 18 |
+ | | + | | + | |
| ÷ | | - | | = 1/2 |
+ | | + | | + | |
| + | | ÷ | | = 3/2 |
= 24 | | = 8 | | = 13 | |
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9 | + | 5 | + | 4 | = 18 |
+ | | + | | + | |
7 | ÷ | 2 | - | 3 | = 1/2 |
+ | | + | | + | |
8 | + | 1 | ÷ | 6 | = 3/2 |
= 24 | | = 8 | | = 13 | |
The largest number you can make with the digits in the red boxes is 984.
21 December
There are 3 ways to order the numbers 1 to 3 so that no number immediately follows the number one less that itself:
Today's number is the number of ways to order the numbers 1 to 6 so that no number immediately follows the number one less that itself.
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To make sequences of 4 numbers, we can insert 4s into three different places in the length 3 sequences given to obtain:
- 4, 3, 2, 1
- 3, 2, 4, 1
- 3, 2, 1, 4
- 4, 1, 3, 2
- 1, 4, 3, 2
- 1, 3, 2, 4
- 4, 2, 1, 3
- 2, 4, 1, 3
- 2, 1, 4, 3
There are some possibilities missing: those containing \(i, 4, i+1\). These can be found by taking the sequences of length 2, picking a number \(i\), adding 1 to every number larger than \(i\), then replacing \(i\) with \(i\ 4\ i+1\).
- 2, 1 → 3, 1 → 3, 1, 4, 2
- 2, 1 → 2, 1 → 2, 4, 3, 1
This gives a total of 3×3+2×1=11 sequences for 4 numbers.
To make sequences with 5 numbers, we can insert 5s into four different places in the length 4 sequences. This gives 4×11=44 sequences.
The missing sequences can then be found by taking the sequences of length 3, then doing the same process as above:
- 3, 2, 1 → 3, 2, 1 → 3, 5, 4, 2, 1
- 3, 2, 1 → 4, 2, 1 → 4, 2, 5, 3, 1
- 3, 2, 1 → 4, 3, 1 → 4, 3, 1, 5, 2
- 1, 3, 2 → 1, 4, 3 → 1, 5, 2, 4, 3
- 1, 3, 2 → 1, 3, 2 → 1, 3, 5, 4, 2
- 1, 3, 2 → 1, 4, 2 → 1, 4, 2, 5, 3
- 2, 1, 3 → 2, 1, 4 → 2, 5, 3, 1, 4
- 2, 1, 3 → 3, 1, 4 → 3, 1, 5, 2, 4
- 2, 1, 3 → 2, 1, 3 → 2, 1, 3, 5, 4
There are 3×3=9 of these, giving 44+9 = 53 total sequences of length 5.
To make sequences with 6 numbers, we can insert 6s into five different places in the length 5 sequences. This gives 5×53=265 sequences.
We can also make sequence by picking a number to replace in the length 4 sequences. This gives 4×11=44 more sequences.
Therefore there are 265+44 = 309 sequences in total.
20 December
18 can be written as the sum of 3 consecutive (strictly) positive integers: 5 + 6 + 7.
18 can also be written as the sum of 4 consecutive (strictly) positive integers: 3 + 4 + 5 + 6.
18 is in fact the smallest number that can be written as the sum of both 3 and 4 consecutive (strictly) positive integers.
Today's number is the smallest number that can be written as the sum of both 12 and 13 consecutive (strictly) positive integers.
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The sum of 13 consecutive integers is 13 times the middle number, so today's number is a multiple of 13.
The sum of 12 consecutive integers is 6 times the sum of the two middle numbers, so today's number is also a multiple of 6.
Therefore today's number is a multiple of 78.
78 and 156 do not work (the numbers would not all be strictly positive), so today's number is 234.
17 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums and product are correct.
Today's number is the product of the numbers in the red boxes.
| + | | + | | = 16 |
+ | | + | | + | |
| + | | + | | = 8 |
+ | | + | | + | |
| × | | × | | = 288 |
= 11 | | = 14 | | = 20 | |
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6 | + | 3 | + | 7 | = 16 |
+ | | + | | + | |
1 | + | 2 | + | 5 | = 8 |
+ | | + | | + | |
4 | × | 9 | × | 8 | = 288 |
= 11 | | = 14 | | = 20 | |
The product of the numbers in the red boxes is 189.
15 December
When talking to someone about this Advent calendar, you told them that the combination of XMAS and MATHS is GREAT.
They were American, so asked you if the combination of XMAS and MATH is great; you said SURE. You asked them their name; they said SAM.
Each of the letters E, X, M, A, T, H, S, R, U, and G stands for a different digit 0 to 9. The following sums are correct:
Today's number is SAM. To help you get started, the letter T represents 4.
14 December
The numbers 33, 404 and 311 contain duplicate digits. The numbers 120, 15 and 312 do not.
How many numbers between 10 and 999 (inclusive) contain no duplicate digits?
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There are 81 two-digit numbers with no duplicate digits: there are 9 choices of tens digit (1 to 9), and for each of these there are 9 remaining choices
for the units digit (0 to 9 but not the number already used).
There are 648 three-digit numbers with no duplicate digits: there are 9 choices of hundreds digit (1 to 9), and for each of these there are 9 remaining choices
for the tens digit (0 to 9 but not the number already used) and 8 choices for the units digit (0 to 9 but neight number already used).
648+81 = 729.