Puzzles
3 December
If you write out the numbers from 1 to 1000 (inclusive), how many times will you write the digit 0?
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The numbers 10, 20, ..., 1000 end with a 0. There are 100 of these numbers.
The numbers 100, 101, ..., 109, 200, ..., 209, 300, ..., 309, ..., 908, 909, 1000 contain a 0 in their tens column. There are \(10\times9+1=91\) of these numbers.
The number 1000 has a 0 in its hundreds column.
In total, there are 100+91+1=192 zeros.
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.
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.
10 December
Today's number is the smallest multiple of 24 whose digits add up to 24.
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There are one three-digit even numbers whose digits add to 24: 798, 888 and 996. Of these, only 888 is a multiple of 24.
8 December
The residents of Octingham have 8 fingers. Instead of counting in base ten, they count in base eight: the digits of their numbers represent ones, eights, sixty-fours, two-hundred-and-fifty-sixes, etc
instead of ones, tens, hundreds, thousands, etc.
For example, a residents of Octingham would say 12, 22 and 52 instead of our usual numbers 10, 18 and 42.
Today's number is what a resident of Octingham would call 11 squared (where the 11 is also written using the Octingham number system).
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The Octingham resident's 11 is equal to our number 9. 9 squared is 81. 81 in base eight is 121.
Interestingly, this is the same answer as "just" doing 11 squred in base ten.
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.
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.
