Question

На памятнике древнегреческому математику деофанту (3в)имеется надпись:Прохожий! под этим камнем покоится прах диофант умершего в старости шестую часть его жизни заняло детство двенадцатую отрочество седьмую юность затем он жинился и через 5 лет у него родился сын который прожил вдвое много отца и года до самой своей кончины диофант оплачивал сына сколько лет жил диофант?

Answer 1

Решение.
1способ.

1) 1/6+1/12+1/7+1/2=75/84 (ч)
2) 1- 75/84=9/84 (ч)
3) 5+4=9 (л)
4) 9: 9/84 = 84(г)
Ответ:84 года.
2 способ.

Пусть х - лет Диофанту.
По условию задачи составляем уравнение:
1/6х+1/12х+1/7х+5+1/2х+4=х;
75/84х+9=х;
х -75/84х =9;
9/84х=9;
х=84(г).
Ответ:84 года.

Answer 2

The Inscription on the Monument to Diophantus

The inscription on the monument to the ancient Greek mathematician Diophantus states that he lived for a certain number of years and that his life can be divided into different stages. Let's analyze the inscription to determine how long Diophantus lived.

The inscription states that Diophantus spent one-sixth of his life in childhood, one-twelfth in adolescence, and one-seventh in youth. After that, he got married, and five years later, he had a son. The son lived twice as long as his father and died at the same age as his father.

To find out how long Diophantus lived, we can represent his age as a variable, let's say x. Based on the information given in the inscription, we can create the following equation:

x = (1/6)x + (1/12)x + (1/7)x + 5 + 2x

Simplifying the equation, we get:

1 = 1/6 + 1/12 + 1/7 + 2 + 2

To solve this equation, we can find a common denominator and simplify:

1 = (14 + 7 + 12 + 84x + 12x) / 84

84 = 14 + 7 + 12 + 84x + 12x

84 = 33 + 96x

96x = 51

x = 51/96

Simplifying the fraction, we get:

x = 17/32

Therefore, Diophantus lived for 17/32 of a unit of time, which could represent years, months, or any other unit mentioned in the inscription.

Please note that the inscription does not provide specific information about the unit of time used.