Puzzles
24 December
Today's number is the smallest number with exactly 28 factors (including 1 and the number itself as factors).
Show answer
Hide answer
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.
21 December
The factors of 6 (excluding 6 itself) are 1, 2 and 3. \(1+2+3=6\), so 6 is a perfect number.
Today's number is the only three digit perfect number.
20 December
What is the largest number that cannot be written in the form \(10a+27b\), where \(a\) and \(b\) are nonnegative integers (ie \(a\) and \(b\) can be 0, 1, 2, 3, ...)?
Show answer & extension
Hide answer & extension
Any number can be written as \(10a+27b\) with integer \(a\) and \(b\), since \(1=3\times27-8\times10\).
So the problem may be thought of as asking when one of \(a\) and \(b\) must be negative.
Given one way of writing a number, you can get the others by shifting by 14*29. For example,
$$219 = 10\times30 - 27\times3$$ $$= 10 + 10\times29 - 27\times3 $$ $$= 1\times10 + 27\times11$$
So the question now becomes: When does this adjustment fail to eliminate negative numbers?
This is when you are at what Pedro calls "limit coefficients":
$$10\times(-1) + 27\times13 = 10\times28 + 27\times(-1) = 233$$
So the answer is 233.
Extension
Let \(n\) and \(m\) be integers. What is the largest number that cannot be written in the form \(na+mb\), where \(a\) and \(b\) are nonnegative integers?
19 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the products are correct. Today's number is the smallest number that can be made using the digits in the red boxes.
| × | | × | | = 90 |
| × | | × | | × | |
| × | | × | | = 84 |
| × | | × | | × | |
| × | | × | | = 48 |
=
64 | | =
90 | | =
63 | |
18 December
Today's number is the maximum number of pieces that a (circular) pancake can be cut into with 17 straight cuts.
17 December
Arrange the digits 1-9 in a 3×3 square so that every row makes a three-digit square number, the first column makes a multiple of 7 and the second column makes a multiple of 4.
The number in the third column is today's number.
| | | square |
| | | square |
| | | square |
| multiple of 7 | multiple of 4 | today's number |
15 December
The string ABBAABBBBB is 10 characters long, contains only A and B, and contains at least three As.
Today's number is the number of different 10 character strings of As and Bs that have at least three As.
Show answer
Hide answer
The number of 10 character strings containing at least 3 As is equal to
the number of 10 character strings containing exactly 3 As
plus
the number of 10 character strings containing exactly 4 As
plus ... plus
the number of 10 character strings containing exactly 10 As.
The number of 10 character strings containing exactly \(a\) As is \(\left(\begin{array}{c}10\\a\end{array}\right)\) ("10 choose \(a\)").
Therefore the number we are looking for is
$$
\left(\begin{array}{c}10\\3\end{array}\right)
+
\left(\begin{array}{c}10\\4\end{array}\right)
+
\left(\begin{array}{c}10\\5\end{array}\right)
+
\left(\begin{array}{c}10\\6\end{array}\right)
+
\left(\begin{array}{c}10\\7\end{array}\right)
+
\left(\begin{array}{c}10\\8\end{array}\right)
+
\left(\begin{array}{c}10\\9\end{array}\right)
+
\left(\begin{array}{c}10\\10\end{array}\right)
$$
You can work this out by either using the formula \(\left(\begin{array}{c}n\\r\end{array}\right)=\frac{n!}{r!(n-r)!}\), or by remembering that
these numbers all appear in the 10th row of
Pascal's triangle.
The answer is 968.
