problem 77

[HardAttack]

10 little bags.
10 balls inside each bags.

bags and balls inside bags are pretty much same in shape / size / color, i mean there is no visible idedifier to call which ball belongs to which bag.
every ball in 9 of 10 bags to weigh 10 grams each.
but, 1 of 10 bags, has the balls inside 9 grams each.

you are given a scales, and you are given only one measurement try/chance.

your objective is to find out which bag has the 9 gram balls.

good luck.

[WingCmdr Ginkapo]

Weigh at once,

1 ball from bag 1
2 balls from bag 2
3 balls from bag 3
Etc etc

Realise your scales arent that accurate and this was a complete waste of time.

[Symmetry]

Sym brand logic problems- Accept no substitutes.

[mrswdk]

I'd just guess, because YOLO.

[Army of GOD]

use a string

/TG's shitty solutions

[TA1LGUNN3R]

Weigh at once,

1 ball from bag 1
2 balls from bag 2
3 balls from bag 3
Etc etc

Realise your scales arent that accurate and this was a complete waste of time.

Why would you pull individual balls out? What if there's a hundred balls in each bag? Couldn't you simply weigh all ten bags at once, then pull off each bag sequentially? The one bag that subtracts the lesser amount would be the lighter bag.

use a string

/TG's shitty solutions

Now you're using your noggin!

-TG

[Army of GOD]

Weigh at once,

1 ball from bag 1
2 balls from bag 2
3 balls from bag 3
Etc etc

Realise your scales arent that accurate and this was a complete waste of time.

Why would you pull individual balls out? What if there's a hundred balls in each bag? Couldn't you simply weigh all ten bags at once, then pull off each bag sequentially? The one bag that subtracts the lesser amount would be the lighter bag.

we don't have to make our solution solve for all x, we just need it for x=10

besides, your solution would end up in x number of weighings any way

[TA1LGUNN3R]

x number of weighings? How is that different than pulling out individual balls?

-TG

[WingCmdr Ginkapo]

Why would you pull individual balls out? What if there's a hundred balls in each bag? Couldn't you simply weigh all ten bags at once, then pull off each bag sequentially? The one bag that subtracts the lesser amount would be the lighter bag.

No because that would be called weighing it multiple times. My method increases the prep work, but as the problem requires, minimises the use of the weighing scales, aka the limiting factor.

If you assume that this operation is being undertaken in a white room with no one else in it and just the bags and scales then sure it makes no difference how long you take. However, if you go with the actual application of the problem, a busy room in which 50 people all want access to the scales, you would be the complete jerk who hogs the scales for ages.

Are you a complete jerk TG? Is that the guy you want to be?

[Army of GOD]

x number of weighings? How is that different than pulling out individual balls?

-TG

pretend you only have a quarter and to see the value of the measurement, you have to insert the quarter. in our case, we only care about the measurement once (when 1 ball from the first bag, 2 from the second...9 from the ninth, 10 from the tenth) are on the scale. In your case, we have to take 10 measurements (one for all the bags on, one for all the bags but the first...one for just the ninth and tenth, one for just the tenth)

[mrswdk]

Why would you pull individual balls out? What if there's a hundred balls in each bag? Couldn't you simply weigh all ten bags at once, then pull off each bag sequentially? The one bag that subtracts the lesser amount would be the lighter bag.

No because that would be called weighing it multiple times. My method increases the prep work, but as the problem requires, minimises the use of the weighing scales, aka the limiting factor.

If you assume that this operation is being undertaken in a white room with no one else in it and just the bags and scales then sure it makes no difference how long you take. However, if you go with the actual application of the problem, a busy room in which 50 people all want access to the scales, you would be the complete jerk who hogs the scales for ages.

Are you a complete jerk TG? Is that the guy you want to be?

What if all those people get cake while they wait?

[TA1LGUNN3R]

x number of weighings? How is that different than pulling out individual balls?

-TG

pretend you only have a quarter and to see the value of the measurement, you have to insert the quarter. in our case, we only care about the measurement once (when 1 ball from the first bag, 2 from the second...9 from the ninth, 10 from the tenth) are on the scale. In your case, we have to take 10 measurements (one for all the bags on, one for all the bags but the first...one for just the ninth and tenth, one for just the tenth)

Ah, i see know, the weight of the 55 separated balls would be a function of whichever bag had the lighter balls. 1a+2b+...10j, a-j being 10grams except one which would be 9grams.

I misinterpreted wing's solution originally as weighing all ten bags and pulling out balls from the pile.

Why would you pull individual balls out? What if there's a hundred balls in each bag? Couldn't you simply weigh all ten bags at once, then pull off each bag sequentially? The one bag that subtracts the lesser amount would be the lighter bag.

No because that would be called weighing it multiple times. My method increases the prep work, but as the problem requires, minimises the use of the weighing scales, aka the limiting factor.

If you assume that this operation is being undertaken in a white room with no one else in it and just the bags and scales then sure it makes no difference how long you take. However, if you go with the actual application of the problem, a busy room in which 50 people all want access to the scales, you would be the complete jerk who hogs the scales for ages.

Are you a complete jerk TG? Is that the guy you want to be?

I am a bit. But I think in real life I could do my method faster than yours. There are digital balances now.

-TG