

The scope of this group of puzzles is quite wide 
relations between numbers. Note that you must be careful with
divisions, as the propagators don't absorb it; transforming them into
multiplications can help. If you want fractions (e.g. somebody is 30
years and 6 months old), use multiples of the actual values, up to the
"granularity" you want  e.g. a factor of 12 for all months, or 365
for days.

The Abbot's Puzzle 

from "Amusements in Mathematics, Dudeney", number 110.
If 100 bushels of corn were distributed among 100 people in such
a manner that each man received three bushels, each woman two, and
each child half a bushel, how many men, women, and children were
there?
Dudeney added the condition that there are five times as many women
as men.
That way, the solution becomes unique (otherwise, there are seven
solutions).

Mother and Daughter 

from "Amusements in Mathematics, Dudeney", number 45.
"Mother, I wish you would give me a bicycle," said a girl of
twelve the other day. "I do not think you are old enough
yet, my dear," was the reply. "When I am only three times as old
as you are you shall have one." Now, the mother's age is fortyfive
years. When may the young lady expect to receive her present?

Curious Numbers 

from "Amusements in Mathematics, Dudeney", number 114.
The number 48 has this peculiarity, that if you add 1 to it the result
is a square number, and if you add 1 to its half, you also get
a square number. Now, there is no limit to the numbers that have
this peculiarity, and it is an interesting puzzle to find three more
of themthe smallest possible numbers. What are they?

Mamma's Age 

from "Amusements in Mathematics, Dudeney", number 40.
Tommy: "How old are you, mamma?"
Mamma: "Our three ages add up to exactly seventy years."
Tommy: "And how old are you, papa?"
Papa: "Just six times as old as you, my son."
Tommy: "Shall I ever be half as old as you, papa?"
Papa: "Yes, Tommy; and when that happens our three ages
will add up to exactly twice as much as today."
Can you find the age of Mamma?

Mrs Timpkin's Age 

from "Amusements in Mathematics, Dudeney", number 43.
When the Timpkinses married eighteen years ago, Timpkins was three
times as old as his wife, and today he is just twice as old as she.
How old is Mrs. Timpkin?

Contracting Costs 

from "Mathematical Puzzles of Sam Loyd, Volume 2", number 20.
A contractor planning the construction of a house found
that he would have to pay:
 $ 1,100 to the paper hanger and the painter,
 $ 1,700 to the painter and plumber,
 $ 1,100 to the plumber and electrician,
 $ 3,300 to the electrician and carpenter,
 $ 5,300 to the carpenter and mason,
 $ 3,200 to the mason and painter.
What does each man charge for his services?

Circling the Squares 

from "Amusements in Mathematics, Dudeney", number 43.
The puzzle is to place a different number in each of the ten squares
so that the sum of the squares of any two adjacent numbers shall be
equal to the sum of the squares of the two numbers diametrically
opposite to them. The four numbers placed, as examples, must stand as
they are. Fractions are not allowed, and no number need contain more
than two figures.

The Five Brigands 

from "Amusements in Mathematics, Dudeney", number 133.
The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and
Esteban, were counting their spoils after a raid, when it was found
that they had captured altogether exacly 200 doubloons. One of the
band pointed out that if Alfonso had twelve times as much, Benito
three times as much, Carlos the same amount, Diego half as much, and
Esteban onethird as much, they would still have altogether just 200
doubloons. How many doubloons had each?
There are a good many equally correct answers to this problem.
The puzzle is to discover exactly how many different answers there
are, it being understood that every man had something and there is to
be no fractional money.
