Cherry Count

Non-algebraic solution by working backwards: The mother saw $7$ cherries in the basket.

Before Charles took one extra cherry out, there were $8$ . Those $8$ cherries were $\frac{1}{6}$ of the number in the basket before he ate any. Thus there were $48$ cherries in the basket when he woke up.

Before Brian put two extra cherries in the basket, there were $46$ . This number is $\frac{2}{3}$ of the number in the basket before Brian took any. I.e., there were $69$ cherries in the basket when Brian came into the room.

Alex ate " $\frac{1}{2}$ the cherries, plus one". If we add the one cherry back, there were $70$ cherries in the basket after Alex took half of them. Thus the basket started with $140$ cherries.

Let there be $n$ cherry in the beginning. Then following the given conditions of the problem, we get $\dfrac{1}{6}\left (\dfrac{2}{3}\left (\dfrac{n}{2}-1\right )+2\right )-1=7\implies \dfrac{1}{6}\times \dfrac{n+4}{3}=8\implies n=8\times 6\times 3-4=\boxed {140}$ .

Let $N$ be the initial number of cherries, and let $x_1, x_2, x_3$ be the number of cherries eaten by the three brothers respectively. Then from the statement of the problem we can write the following linear equations.

$x_1 = \frac{1}{2} N + 1$

$x_2 = \frac{1}{3} (N - x_1) - 2$

$x_3 = \frac{5}{6} (N - x_1 - x_2) + 1$

$N - x_1 - x_2 - x_3 = 7$

Solving the above linear system of $4$ equations in $4$ unknowns, we find that

$x_1 = 71$ , $x_2 = 21$ , $x_3 = 41$ , $N = 140$

We can work backward to get the answer. Let the initial number of cherries be $n$ , what are left by Alex, Brian and Charles to be $n_a$ , $n_b$ , and $n_c$ respectively. Then we have:

$\begin{aligned} n_c & = 7 \\ n_b - \frac 56n_b - 1 & = 7 \\ \frac 16 n_b & = 8 \\ \implies n_b & = 48 \\ n_a - \frac 13 n_b + 2 & = 48 \\ \frac 23n_a & = 46 \\ \implies n_a & = 69 \\ n - \frac 12n - 1 & = 69 \\ \frac 12 n & = 70 \\ \implies n & = \boxed{140} \end{aligned}$

Solve Y=X-(X/2+1)

Z=Y-(Y/3)+2

7=Z-(5/6)Z-1

Answer

X=140