Puzzles
Fair dice
Timothy and Urban are playing a game with two six-sided dice. The dice are unusual: Rather than bearing a number, each face is painted either red or blue.
The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they're different. Their chances of winning are equal.
The first die has 5 red faces and 1 blue face. What are the colours on the second die?
Show answer & extension
Hide answer & extension
Let \(A\) and \(B\) be the outcomes of the two dice. Let \(p\) be the probability that the second die lands on red. The probability of the dice being the same is:
$$\frac{1}{2}=\mathbb{P}(A=r)\mathbb{P}(B=r)+\mathbb{P}(A=b)\mathbb{P}(B=b)\\
=\frac{5}{6}p+\frac{1}{6}(1-p)\\
=\frac{1}{6}+\frac{4}{6}p
$$
This means that:
$$\frac{4}{6}p=\frac{1}{2}-\frac{1}{6}\\
=\frac{1}{3}\\
p=\frac{\frac{1}{3}}{\frac{4}{6}}=\frac{1}{2}$$
Extension
If the first die has \(n\) red faces and \(6-n\) blue faces, what colours are on the second die?
Half digits
Can you use each of the digits 1 to 9 to make a fraction which is equal to a half?
Pizza
Twelve friends want to share a pizza. One of the friends is very fussy and will not eat the centre of the pizza.
Is it possible to split a (circular) pizza into twelve identical pieces such that there is at least one piece which does not touch the centre?
Frogs
Two frogs and two toads are standing on five lily pads.
The frogs and toads need to pass each other. They can only move by jumping one or two lily pads forward. In jumping two pads forwards they can jump over other frogs or toads.
How many jumps need to be made to get the frogs and toads past each other?
Show answer & extension
Hide answer & extension
Representing the frogs as \(F\), the toads as \(T\) and the spaces as \(\), the solution is as follows:
$$
F\ F\ \_\ T\ T\\
F\ \_\ F\ T\ T\\
F\ T\ F\ \_\ T\\
F\ T\ F\ T\ \_\\
F\ T\ \_\ T\ F\\
\_\ T\ F\ T\ F\\
T\ \_\ F\ T\ F\\
T\ T\ F\ \_\ F\\
T\ T\ \_\ F\ F\\
$$
Eight moves are required.
Extension
If there are three frogs on each side, how many moves are needed?
If there are three frogs on one side and two on the other, how many moves are needed?
If there are \(n\) frogs on one side and \(m\) on the other, how many moves are needed?
The blue-eyed sisters
If you happen to meet two of the Jones sister (two sisters chosen at random from all the Jones sisters), it is exactly an even-money bet that both will be blue-eyed. What is your best guess of the total number of Jones sisters?
Show answer & extension
Hide answer & extension
If there are \(n\) sisters and \(k\) of these sisters have blue eyes, then the probability of the two sisters having blue eyes is:
$$\frac{\left(\begin{array}{c}k\\2\end{array}\right)}{\left(\begin{array}{c}n\\2\end{array}\right)}
\\
=\frac{k(k-1)}{n(n-1)}=\frac{1}{2}
$$
This means that:
$$2k(k-1)=n(n-1)$$
The smallest integer solution of this is when there are 4 sisters, 3 of whom have blue eyes.
The next smallest integer solution is when there are 21 sisters, 15 of whom have blue eyes. There are unlikely to be as many as 21 sisters, so 4 sisters are the most likely.
Extension
What is the next integer solution after 21 sisters?
1089
Take a three digit number. Reverse the digits then take the smaller number from the larger number.
Next add the answer to its reverse.
For example, if 175 is chosen:
$$571-175=396$$
$$396+693=1089$$
What numbers is it possible to obtain as an answer, and when will each be obtained?
Show answer & extension
Hide answer & extension
Call the digits of the starting number \(A\), \(B\) and \(C\).
If the number is a palindrome (if \(A=C\)) then the answer to the subtraction will be 0, so the answer is 0.
If the first and last digits differ one, we can consider the case \(A=C+1\). The same analysis will apply to \(C=A+1\).
$$100C+10B+A-(100A+10B+C)=99C-99A$$
$$=99A+99-99A$$
$$=99$$
Then, \(99+99=198\) so the answer is 198
If the first and last digits differ by at least two, we can consider the case \(A=C+n\), where \(2\leq n\leq 9\). The same analysis will apply to \(C=A+n\).
$$100C+10B+A-(100A+10B+C)=99C-99A$$
$$=99(A+n)-99A$$
$$=99n$$
This number will be 198, 297, 396, 495, 594, 693, 792, 891 or 990. In each of these numbers, the middle digit is 9 and the two other digits add up to 9, so they can be written \(100n+90+(9-n)\). If we add this to its reverse:
$$100n+90+(9-n)+100(9-n)+90+n$$
$$=100n+99-n+900-100n+90+n$$
$$=900+99+90+100n-100n+n-n$$
$$=1089$$
So the three possible answers are 0, 99 and 1089 which will occur when the first and list digits of the number differ by 0, 1 or more respectively.
Extension
Which answers are possible if the numbers are written in different bases?
Which answers are possible if four digit numbers are taken? Or five digit numbers?
Integrals
$$\int_0^1 1 dx = 1$$
Find \(a_1\) such that:
$$\int_0^{a_1} x dx = 1$$
Find \(a_2\) such that:
$$\int_0^{a_2} x^2 dx = 1$$
Find \(a_n\) such that (for \(n>0\)):
$$\int_0^{a_n} x^n dx = 1$$
Show answer & extension
Hide answer & extension
$$1=\int_0^{a_1} x dx=\frac{a_1^2}{2}$$
So, \(a_1=\sqrt{2}\).
$$1=\int_0^{a_2} x^2 dx=\frac{a_2^3}{3}$$
So, \(a_2=3^{\frac{1}{3}}\).
$$1=\int_0^{a_n} x^n dx=\frac{a_n^{n+1}}{n+1}$$
So, \(a_n={(n+1)}^{\frac{1}{n+1}}\).
Extension
Find \(b_n\) such that (for \(n>1\)):
$$\int_{b_n}^{\infty} x^{-n} dx = 1$$
Tetrahedral die
When a tetrahedral die is rolled, it will land with a point at the top: there is no upwards face on which the value of the roll can be printed. This is usually solved by printing three numbers on each face and the number which is at the bottom of the face is the value of the roll.
Is it possible to make a tetrahedral die with one number on each face such that the value of the roll can be calculated by adding up the three visible numbers? (the values of the four rolls must be 1, 2, 3 and 4)
Show answer & extension
Hide answer & extension
Let \(a\), \(b\), \(c\) and \(d\) be the numbers on the four faces. The following simultaneous equations must hold:
$$a+b+c=1$$
$$a+b+d=2$$
$$a+c+d=3$$
$$b+c+d=4$$
These can be solved to find that the numbers on the faces must be \(-\frac{2}{3}\), \(\frac{1}{3}\), \(\frac{4}{3}\) and \(\frac{7}{3}\).
Extension
Is it possible to make a six-sided die with one number on each face such that the value of the roll can be calculated by adding up the five visible numbers?
Is it possible to make an \(n\)-sided die with one number on each face such that the value of the roll can be calculated by adding up the \((n-1)\) visible numbers?
Is it possible to make a die with one integer on each face such that the value of the roll can be calculated by adding up the visible numbers?
![](javascript:showlimage("pizza-answer.png"))
![](javascript:showlimage("frogs-and-toads.png"))