In detail

The eccentric teacher

The eccentric teacher

We are searching data for your request:

Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.

Background Vector created by brgfx

Here is a remarkable problem of ages that I am sure will entertain young people and will open, at the same time, a new line of reasoning to some smarties who have made statistical calculation their specialty.

It seems that an ingenious or eccentric teacher - since both cases can be treated -, eager to gather a certain number of older students in a class he was training, offered to give a prize every day to the side of boys or girls whose ages would add more.

Well, on the first day only one boy and one girl attended, and as the boy's age doubled that of the girl, the prize went to him.

The next day, the girl took her sister to school. It was discovered that their combined ages were twice that of the boy, so that both girls shared the prize.

When the school opened the next day, however, the boy had recruited one of his brothers. It was discovered that the combined ages of both double the ages of the two girls, so the boys took all the honors that day and divided the prize.

The fight began to warm up then between the Jones and Brown families, so on the fourth day the two girls appeared accompanied by their older sister, so that day the combined ages of the three girls competed against those of the boys. Of course they won this time, since their ages together doubled that of the two boys.

The battle continued until the class was filled, but it is not necessary for our problem to go further. We want to know the age of that first boy, knowing that the last girl joined the class on the day of his twenty-first birthday.

It is a simple but beautiful puzzle, which requires more ingenuity than mathematical knowledge, and easily deciphered by means of methods typical of all riddles.


The first girl was only 638 days old, and the boy double, that is, 1,276 days.

The next day the youngest girl would have 639 days, and the new recruit 1,915 days, totaling 2,544 days, which would double the age of the first boy who, with one more day, would have 1,277.

The next day the boy, of 1,278 days of age, brings his older brother, who is 3834 days, so that their combined ages total 5,112 days, just twice the age of the girls, who at that time would be 640 and 1916, that is, 2,566 days.

We arrive at 7,670 days as follows. The young woman has reached her twenty-first birthday, so 21 times 365 of 7,665, plus 4 days for four leap years and 1 extra day that is her twenty-first birthday.

Those who assumed that the boy's age was 3 and a half years overlooked the fact of increasing the age of the students day by day.