Puzzles
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.
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.
13 December
There is a row of 1000 lockers numbered from 1 to 1000. Locker 1 is closed and locked and the rest are open.
A queue of people each do the following (until all the lockers are closed):
- Close and lock the lowest numbered locker with an open door.
- Walk along the rest of the queue of lockers and change the state (open them if they're closed and close them if they're open) of all the lockers
that are multiples of the locker they locked.
Today's number is the number of lockers that are locked at the end of the process.
Note: closed and locked are different states.
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The squarefree numbered lockers will be locked. (Squarefree number are numbers that are not divisible by any square number apart from 1.) There are 608 squarefree numbers between 1 and 1000.
7 December
There is a row of 1000 closed lockers numbered from 1 to 1000 (inclusive). Near the lockers, there is a bucket containing the numbers 1 to 1000 (inclusive) written on scraps of paper.
1000 people then each do the following:
- Pick a number from the bucket (and don't put it back).
- Walk along the row of lockers and change the state
(open them if they're closed and close them if they're open)
of all the lockers that are multiples of the number they picked (including the number they picked).
Today's number is the number of lockers that will be closed at the end of this process.
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The square numbered lockers will be open at the end of the process. There are 969 non-square numbers between 1 and 1000, so this many lockers will be closed.
6 December
Write down the numbers from 12 to 22 (including 12 and 22). Under each number, write down its largest odd factor*.
Today's number is the sum of all these odd factors.
* If a number is odd, then its largest odd factor is the number itself.
Digitless factor
Ted thinks of a three-digit number. He removes one of its digits to make a two-digit number.
Ted notices that his three-digit number is exactly 37 times his two-digit number. What was Ted's three-digit number?
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Ted's number was 925: \(925\div25=37\).
If Ted had removed the final digit of his number, then he would be looking for a solution of \(ABC = 37\times AB\). But \(ABC\)
is between 10 and 11 times \(AB\) (it is \(10\times AB + C\)) and so cannot be 37 times \(AB\). So Ted cannot have removed the final digit.
Therefore, Ted must have removed one of the first two digits: so two- and three- digit numbers have the same final digit (\(C\)).
The final digit of the three-digit number (\(C\)) will be the final digit of \(7\times C\) (7 times the final digit of the two digit number).
This is only possible if the final digit is \(C\) is 0 or 5.
This only leaves four possible solutions—10, 15, 20 and 25—as \(30\times37>1000\). Of these only \(925=37\times25\) works.
Extension
How many three-digit numbers are there that are a multiple of one of the two-digit numbers you can make by removing a digit?
24 December
Today's number is the smallest number with exactly 28 factors (including 1 and the number itself as factors).
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If \(p\) and \(q\) are prime numbers, then the number \(p^a\times q^b\) will have \((a+1)(b+1)\) factors. This is because all its factors are
of the form \(p^\alpha\times q^\beta\), with \(\alpha=0,1,...,a\) and \(\beta=0,1,...,b\). The same idea can be used on numbers with three or more prime factors; in general the number \(p_1^{a_1}\times...\times p_n^{a_n}\) has \((a_1+1)\times...\times(a_n+1)\) factors.
28 can be written as: 28, 14×2, 7×4, or 7×2×2. Therefore the following numbers have 28 factors:
$$2^{27},\quad
2^{13}\times3^1,\quad
2^{6}\times3^3,\quad
2^{6}\times3^1\times5^1
$$
and any other number with 28 factors will have larger prime factors, so will be larger.
These numbers are 134217728, 24576, 1728 and 960. Therefore the smallest number with 28 factors is 960.
