7th grade algebra problem

[jyl]

In case you want to test your math skills against the kids in my daughter's algebra class.

You have seven friends and an unknown quantity of chocolates. You give the first friend half of your chocolates, plus one. You give the second friend half of your remaining chocolates, plus one. You give the third friend half of your remaining chocolates, plus one. And so on. After you have given chocolates to all seven friends, you have one chocolate left. How many chocolates did you start with?

And how did you solve the problem?

[gprsh924]

can you give out half chocolates

[kstar]

I'm only on 4th grade math.

[Rearden]

I'm thinking you start with 382 chocolates.

[masraum]

sounds like a progression problem.

[afterburn 549]

Sounds like the same weird way they still teach math to those of us that were smart enough but not patient enough to put up with the BS.

[TerryH]

seems to me, you'd start at the end and work backwards. but I've wrong before.... most of the time.

[kstar]

1+1 x2

4+1 x2

10+1 x2

22+1x2

46+1x2

94+1x2

190+1x2

382 ???

What Rearden said.

[Arizona_928]

11 1/2

[ikarcuaso]

382, yup, backed into it

[strupgolf]

I think I understand what you have stated and you did say,but I have to say that you might not think what has been shown is what it is and has what it has to be of any relevance to any and all of us who think the way we do and show any regard to any and all of this position to what it is. Any more questions?

[trekkor]

Now for a seventh grade question:

"How will I ever use this"?


KT

[jyl]

You guys are funny. Yes, 382 and you work backwards from 1. I think it was a good problem, teaches the kids that you can't simply grind the most obvious algebraic approach. Creativity counts. That said, I think it is too hard for her class. I don't think any of the kids will solve the problem without at least a hint from their parents. Disadvantage for the kids whose parents aren't into math. Her teacher is odd, every now and then she nails the kids with a homework problem or a test that is much harder than than the large majority of what they've been doing. All the "A" students get "C"s on those, unless their parents help out. I'm not quite sure what the point of it is.

[masraum]

a limit with some sort of geometric or arithmetic progression, but I'm having a hard time wrapping my head around the answer (how you get the number, not the actual number). Maybe I'm over thinking it a bit. I'm thinking in terms of Calculus, not algebra. This isn't finding the volume of the difference of two curves that have been rotated about an axis or the dot product of vectors.

[jyl]

I emailed my daughter a hint ( I'm in a hotel) and we will see if she solved it.

[jyl]

Let x be the chocolates left after you give to a friend and let y be the chocolates before you give to that friend.

y - ( y/2 + 1 ) = x

Rewrite to y = f ( x ) form. Then start with x = 1 and work backwards.

[HardDrive]

Wow am I fecking dumb.

[bonkers]

382?

To work backwards, for each give y = x*2 +1

So the regression would be:
1
3
7
15
31
63
127
255

[HardDrive]

For the record, I got 339.

[artplumber]

It's not that hard to brute force this problem.

you're left with 1=1/128x-127/64, ergo x =382