Puzzles
Cooked turkey
An old invoice showed that seventy-two turkeys had been purchased for "—67.9—". The first and last digits were illegible.
How much did one turkey cost?
Show answer & extension
Hide answer & extension
Let the first digit be \(a\) and the final digit be \(b\). The cost of the 72 turkeys (in pence) is \(10000a+6790+b\). This must be divisible by 72.
$$10000a+6790+b=(138\times 72+64)a+(94\times 72+22)+b$$
$$=(138a+94)\times 72+64a+22+b$$
\(10000a+6790+b\) is divisible by 72, so \(64a+22+b\) is divisible by 72. This means that \(b\) must be even (as 64,22,72 are all even). Let \(b=2c\).
Dividing by two, we find that \(32a+11+c\) is divisible by 36. \(c\) must be odd, so that \(32a+11+c\) is even. Let \(c=2d+1\) and so \(b=4d+2\).
Dividing by two again, we find that \(16a+6+d\) is divisible by 18. \(d\) must be even, so that \(16a+6+d\) is even. Let \(d=2e\) and so \(b=8e+2\). But \(b\) is a single digit number, so \(e=0\), \(b=2\) and \(16a+6\) is divisible by 18.
Dividing by two yet again, we find that \(8a+3\) is divisible by 9. \(a\) must be divisible by 3. Let \(a=3f\).
Dividing by three, we find that \(8f+1\) is divisible by 3. \(8f+1=6f+2f+1\) so \(2f+1\) is divisible by 3. This will be true when \(f=1,4,7,10,...\). \(a\) must be a single digit, so \(f=1\) and \(a=3\). And so the price of the 72 turkeys is £367.92, and one turkey will cost £5.11.
Extension
Which numbers could 72 be replaced with in the original problem so that the problem still has a unique solution?
Coming and going
In my house are a number of rooms. (A hall separated from the rest of the house by one or more doors counts as a room.) Each room has an even number of doors, including doors that lead outside. Is the total number of outside doors even or odd?
Show answer & extension
Hide answer & extension
Add up the number of doors leaving each room; call the sum
\(S\).
As
the number in each room is even, \(S\) will be even. Each interior door
has
been counted twice (as they can be seen in two rooms) and each exterior
door has been counted once. Let \(I\) be the number of interior doors
and
\(E\)
be the number of exterior doors. We have:
$$S=2I+E$$
$$E=S-2I$$
\(S\) and \(2I\) are even, so \(E\) must be even.
Extension
If the number of doors in each room is odd, is the number of
exterior doors odd or even?
Wool circles
\(n\) people stand in a circle. The first person takes a ball of
wool, holds the end and passes the ball to his right, missing a
people. Each person who receives the wool holds it and passes the
ball on to their right, missing \(a\) people. Once the ball returns to
the first person, a different coloured ball of wool is given to
someone who isn't holding anything and the process is repeated. This is
done until everyone is holding wool.
For example, if \(n=10\) and \(a=3\):
In this example, two different coloured balls of wool are needed.
In terms of \(n\) and \(a\), how many different coloured balls of
wool
are
needed?
Show answer & extension
Hide answer & extension
Starting with the person who starts with the wool and
going anti-clockwise, number the people \(0,1,2,3,4,...\). As the
wool is passed, it will be held by people with numbers:
$$0,a+1,2(a+1),3(a+1),...,k(a+1),...$$
The first person will have the wool again when
$$k(a+1)\equiv 0 \mathrm{\ \ mod\ } n$$
or
$$k(a+1)=ln.$$
This will first occur when (hcf is highest common factor):
$$l=\frac{a+1}{\mathrm{hcf}(a+1,n)}\mathrm{\ \ and\ \ }k=\frac{n}{\mathrm{hcf}(a+1,n)}$$
\(k\) is also the number of people who are holding the wool. So
the number of different coloured balls needed is:
$$\frac{n}{\left(\frac{n}{\mathrm{hcf}(a+1,n)}\right)}$$
$$=\mathrm{hcf}(a+1,n)$$
Extension
The ball is passed around the circle of \(n\) people again. This
time,
the number of people missed alternates between \(a\) and \(b\). How many
different coloured balls of wool are now needed?
Sum equals product
\(3\) and \(1.5\) are a special pair of numbers, as \(3+1.5=4.5\)
and
\(3\times 1.5=4.5\) so \(3+1.5=3\times 1.5\).
Given a number \(a\), can you find a number \(b\) such that
\(a+b=a\times b\)?
Show answer & extension
Hide answer & extension
If \(a+b=a\times b\), then:
$$ab-b=a$$
$$b(a-1)=a$$
$$b=\frac{a}{a-1}$$
This will work for any \(a\not=1\) (\(a=1\) will not work as this
will
mean
division by zero).
Extension
(i) Given a number \(a\), can you find a number \(b\) such that
\(b-a=\frac{b}{a}\)?
(ii) Given a number \(a\), can you find a number \(b\) such that
\(b-a=\frac{a}{b}\)?
(iii) Given a number \(a\), can you find a number \(b\) such that
\(a-b=\frac{b}{a}\)?
(iv) Given a number \(a\), can you find a number \(b\) such that
\(a-b=\frac{a}{b}\)?
Multiples of three
If the digits of a number add up to a multiple of three, then the number is a multiple of three. Therefore if a two digit number, \(AB\) (first digit \(A\), second digit \(B\); not \(A\times B\)), is a multiple of three, then \(A0B\) is also a multiple of three.
If \(AB\div 3=n\), then what is \(A0B\div 3\)?
Seven digits
"I'm thinking of a number. I've squared it. I've squared the square. And I've multiplied the second square by the original number. So I now have a number of seven digits whose final digit is a 7," said Dr. Dingo to his daughter.
Can you work out Dr. Dingo's number?
Show answer & extension
Hide answer & extension
Let's call Dr. Dingo's number \(n\). If the number is squared twice then multiplied by \(n\), we get \(n^5\).
For all integers \(n\), the final digit of \(n^5\) is the same as the final digit of \(n\). In other words:
$$n^5\equiv n \mod 10$$
Therefore, the final digit of Dr. Dingo's number is 7.
$$7^5=16807$$
$$17^5=1419857$$
$$27^5=14348907$$
So, in order for the answer to have seven digits, Dr. Dingo's number was 17.
Extension
For which integers \(m\) does there exist an integer \(n\) such that for all integers \(x\):
$$x^n\equiv x \mod m$$
Square numbers
Towards the end of his life, Lewis Carroll recorded in his diary that he had discovered that double the sum of two square numbers could always be written as the sum of two square numbers. For example
$$2(3^2 +4^2 )=1^2 +7^2$$
$$2(5^2 +8^2 )=3^2 +13^2$$
Prove that this can be done for any two square numbers.
Show answer & extension
Hide answer & extension
Let \(a^2\) and \(b^2\) be the two square numbers.
$$2(a^2 +b^2 ) = 2a^2 +2b^2$$
$$= a^2 + 2ab + b^2 + a^2 - 2ab + b^2$$
$$= (a+b)^2 +(a-b)^2$$
Extension
Prove that 3 times the sum of 3 squares is also the sum of 4 squares.
N
Consider three-digit integers \(N\) such that:
(a) \(N\) is not exactly divisible by 2, 3 or 5.
(b) No digit of \(N\) is exactly divisible by 2, 3 or 5.
How many such integers \(N\) are there?
Show answer & extension
Hide answer & extension
(b) implies that the digits of \(N\) are all 1 or 7, so \(N\) can only be 111, 117, 171, 177, 711, 717, 771 or 777. These are all divisible by 3, so no such integers \(N\) exist.
Extension
Consider 21-digit integers \(N\) such that:
(a) \(N\) is not exactly divisible by 2, 3 or 5.
(b) No digit of \(N\) is exactly divisible by 2, 3 or 5.
How many such integers \(N\) are there?
![](javascript:showlimage("woolcircles.png"))