Puzzles
19 December
120 is the smallest number with exactly 16 factors (including 1 and 120 itself).
What is the second smallest number with exactly 16 factors (including 1 and the number itself)?
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If \(p_1^{a_1}\times p_2^{a_2}\times\dots\times p_n^{a_n}\) is the prime factorisation of a number, then the number has \((a_1+1)(a_2+1)\dots(a_n+1)\) factors.
The prime factorisation of 120 is \(2^3\times3\times5\). The next smallest number with 16 factors must be one of:
- \(2\times3^3\times5=270\)
- \(2^3\times3\times7=168\)
- \(2^3\times3^3=216\)
The smallest of these is 168.
2 December
What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 7, and 8?
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8×7×5×3=840. This is also a multiple of 4, 6, 2, and 1. If anything is removed from the product it will no longer be a multiple of all the numbers.
6 December
When 12345 is divided by today's number, the remainder is 205. When 6789 is divided by today's number, the remainder is 112.
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The first division tells that 12345 is 205 more than a mutliple of today's number, and so today's numwber is a factor of 12140.
The second division tells that 6789 is 112 more than a mutliple of today's number, and so today's number is a factor of 6677.
The only numbers that are factors of both 12240 and 6677 are 1 and 607.
2 December
The number \(7n\) has 37 factors (including 1 and the number itself). How many factors does \(8n\) have?
There was a typo in this puzzle. It originally read "38 factors" when it was meant to say "37 factors".
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If \(p_1\), \(p_2\), etc are distinct prime numbers, then the factors of the number \(N=p_1^{a_1}p_2^{a_2}...\) all have the form \(p_1^{b_1}p_2^{b_2}...\),
where \(b_1\) is 0 or 1 or ... or \(a_1\) (and similar conditions for \(a_2\), \(a_3\), etc). There are therefore \(a_1+1\) possible values of \(b_1\), \(a_2+1\) possible values
of \(b_2), and so on. Therefore, \(N\) has \((a_1+1)(a_2+1)(...)\) factors.
From this we see that the only way to have a number with 37 factors is if it is a prime number to the power of 36, and so \(n=7^{35}\).
\(8n\) is therefore equal to \(2^3\times7^{35}\). This has 4×36=144 factors.
Extension
Which other numbers can 37 be replaced with leaving a puzzle that still has a unique solution?
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.
15 December
There are 5 ways to make 30 by multiplying positive integers (including the trivial way):
Today's number is the number of ways of making 30030 by multiplying.
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
$$37\times9=333.$$
20 December
Today's number is the sum of all the numbers less than 40 that are not factors of 40.
