This creature, part man and part tree,
hates the termite as much as the flea.
His tracks do not match,
and his limbs may detach,
but he’s not a strange creature to see.
This is a place. The first letter is after A and before Z but nowhere in between. You can bring as many people as you want, as long as they’re all dead. Where is it?
The cemetery. The letter C is after A and before Z, and it’s not found in the word “between”. The cemetery won’t take live ones. Fortunately I don’t speak from experience on that.
You watch a group of words going to a party. A word either enters through one of two doors or is turned away by the guards. ‘HIM’ goes through door number one and ‘BUG’ goes through door number two. ‘HER’ is turned away. ‘MINT’ and ‘WEAVE’ go in through door one, ‘DOOR’ and ‘CORD’ take door two and ‘THIS’ and ‘That’ aren’t allowed in.
What determines whether a word can enter and which door they must use?
Door number one is for words composed entirely of capital letters written using only straight lines, such as A, E, F, H, and I. The entire set of letters allowed through door number one are AEFHIKLMNTVWXYZ. Door number two, as might be expected, is for words with capital letters that have a curve, including BCDGJOPQRSU. Any words composed of both straight and curved letters (or lowercase letters) are not allowed in. The word ‘THAT’ would have been sent through door number one, if the letters had been capitalized.
A similar problem can be found in L.A. Graham’s Ingenious Mathematical Problems and Methods with a range of 1 to 9, but the principle remains the same – the numbers with the smallest difference produce the largest product. You start out with the highest two digits, 7 and 6, then attach 5 and 4, putting the smaller of the two digits with the larger number, giving you 74 and 65. The next two highest digits are 3 and 2, giving you 742 and 653. Finally, you add the 1 to the lower number. Page 80 has the details of that solution.
A princess is as old as the prince will be when the princess is twice the age that the prince was when the princess’s age was half the sum of their present ages.
This one took a while to figure out and there are numerous valid ways of finding the answer.
Here is the solution I came up with
I created the following table from the riddle:
Current
Future
Past
Princess
x
2z
(x+y)/2
Prince
y
x
z
I then created three equations, since the difference in their age will always be the same.
d = the difference in ages
x – y = d
2z – x = d
x/2 + y/2 – z = d
I then created a matrix and solved it using row reduction.
x
y
z
1
-1
0
d
-1
0
2
d
.5
.5
-1
d
It reduced to:
x
y
z
1
0
0
4d
0
1
0
3d
0
0
1
5d/2
This means that you can pick any difference you want (an even one presumably because you want integer ages).
Princess age: 4d
Prince age: 3d
Ages that work
Princess
Prince
4
3
8
6
16
12
24
18
32
24
40
30
48
36
56
42
64
48
72
54
80
60
To see other solutions check out the comments from when I posted this on my blog.