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придётся немного поработать с «подбором»:
пусть сначала было k коробок, потом n, затем m.
тогда: 6k = 9n + 6,
а также
6k = 7m + 3.
или:
9n + 6 = 7m + 3.
выразим отсюда: n = (7m – 3)/9.
но n (равно как и k и m) должно быть целым. подбираем варианты:
m = 3 => n = 2; (m увеличиваем в каждом шаге на 9)
m = 12 => n = 9; k = 1,5n + 1 = 14,5.
m = 21 => n = 16; k = 24 + 1 = 25.
m = 30 => n = 23; k = 34,5.
m = 39 => n = 30; k = 45 + 1 = 46.
при k = 25 имеем: 6k = 150, это < 200.
при k = 46 получаем: 6k = 276.
то число подарков «подходит» под условие .
проверяем: 306 = 9•30 + 6 =276; 306 = 7•39 +3 = 276.
итак, число подарков было
Получается надо 6 графиков;
Из условия имеем такие функции:
1)![y=1x^2](/tpl/images/1498/6779/7ee94.png)
2)![y=-1x^2](/tpl/images/1498/6779/75b70.png)
3)![y=0,125x^2](/tpl/images/1498/6779/0307f.png)
4)![y=-0,125x^2](/tpl/images/1498/6779/bd797.png)
5)![y=4x^2](/tpl/images/1498/6779/33bd5.png)
6)![y=-4x^2](/tpl/images/1498/6779/69a34.png)
Фото графиков прикрепил.
P.s Графиков 6, а фото 5. Просто к функциям ( 3 и 4 ) одинаковые графики получаются.