Puzzles
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.
24 December
There are six 3-digit numbers with the property that the sum of their digits is equal to the product of their digits. Today's number is the largest of these numbers.
22 December
In bases 3 to 9, the number 112 is:
\(11011_3\),
\(1300_4\),
\(422_5\),
\(304_6\),
\(220_7\),
\(160_8\), and
\(134_9\).
In bases 3, 4, 6, 8 and 9, these representations contain no digit 2.
There are two 3-digit numbers that contain no 2 in their representations in all the bases between 3 and 9 (inclusive). Today's number is the smaller of these two numbers.
17 December
Eve picks a three digit number then reverses its digits to make a second number. The second number is larger than her original number.
Eve adds her two numbers together; the result is 584. What was Eve's original number?
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For the first and last digits of the result to be different, there must be a 1 carried from the middle column. Therefore the middle digit of Eve's number is 9.
Eve's number could be 292 or 193, but it cannot be the first as reversing this is not larger than her original number, so here number is 193.
1 December
If you write out the numbers from 1 to 1000 (inclusive), how many times will you write the digit 1?
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100 numbers will have 1 in the units column. Another 100 with have a 1 in the tens column; and other 100 will have a 1 in the hundreds column.
Finally, 1000 has a 1 in the thousands column.
In total, this makes 301 copies of the digit 1.
Not Roman numerals
The letters \(I\), \(V\) and \(X\) each represent a different digit from 1 to 9. If
$$VI\times X=VVV,$$
what are \(I\), \(V\) and \(X\)?
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For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.
If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.
Now that we know that \(VI\) is 37, we see that
$$37\times X=333,$$
and so \(X=9\) and the final solution is
$$37\times9=333.$$
22 December
In base 2, 1/24 is
0.0000101010101010101010101010...
In base 3, 1/24 is
0.0010101010101010101010101010...
In base 4, 1/24 is
0.0022222222222222222222222222...
In base 5, 1/24 is
0.0101010101010101010101010101...
In base 6, 1/24 is
0.013.
Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.
Today's number is the smallest base in which 1/10890 has a finite number of digits.
Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).
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If 1/10890 has a finite number of digits in base \(b\), then \(b^n/10890\) is an integer for some integer \(n\). In order for this to be
possible \(b\) must be a mutiple of each prime factor of 10890: the smallest such \(b\) will be when the exponent of each prime factor is 1.
The prime factorisation of 10890 is 2×33×5×112. Therefore the smallest base in which 1/10890 has a
finite number of digits is 2×3×5×11=330.
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.
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The number can contain no 1 (as all numbers are multiples of 1). It's first and last digits cannot be 2 (as it would then be even and so a multiple of 2).
The lowest three-digit palindrome containing no 0 or 1 and not ending in 2 is 323. This is not a multiple of 2 or 3 and so is the solution.
