Question

A woman has some peanuts. She sees a friend at the door and eats a peanut before her friend enters and then takes half of the remaining peanuts. She sees another friend at the door and eats a peanut then takes one half of the remaining peanuts. Then she sees another friend at the door and eats one peanut then removes half of the remaining peanuts, this leaves her with only one peanut. How many peanuts did the woman originally have?

Answer 1

She starts with x peanuts.

After her first friend takes their share of nuts, she has (x-1)/2 peanuts. Call that y.

After the second friend takes their share of nuts, she has (y-1)/2 peanuts. Call that z.

After the third friend takes their share of nuts, she has (z-1)/2 peanuts. We know that's 1.

Substitute variables until you get one long equation with only x. Solve for x.

Answer 2

his method looks good.

Answer 3

to get you started...we start at the end and work our way back. At the end of the process she has one peanut.

(z-1)/2=1
z=3

So she had 3 peanuts before her final friend arrived...

Answer 4

So,
(y-1)/2=3
y=7

She had 7 peanuts when the second friend arrived....

Answer 5

and,
(x-1)/2=7
x=15
she had 15 peanuts originally.

Answer 6

Okay, so you're doing it for her.

I may as well post the formulas then, although I was going to wait for her to figure it out.

\[\frac{z-1}{2} = 1\]
\[\frac{\frac{y-1}{2}-1}{2} = 1\]
\[\frac{\frac{\frac{x-1}{2}-1}{2}-1}{2} = 1\]

Answer 7

@geoffb my bad

Answer 8

That made more sense. I understand better. Thank you very much.

Answer 9

@eseidl Not to worry.