100 boxes of gold have a real box, how to use a scale to find out the real one?

Your problem of not playing cards according to the convention, it is generally difficult to ask why, the only way is to put 100 boxes in the scale at the same time, this time to get the weight of a hundred boxes, this time to remove a look at how much weight is left will know that time is to get true.

Yes. A box, each with its own.

balls, one of the box balls is at odds with the other balls.

Gram. Question: How do I find this box by weighing it once? Declaration:

There are no word traps;

There is nothing wrong with the box;

There is no difference in the appearance of the ball;

Other. The balls in the box are ** balls, the weight is exactly the same, but the weight is unknown;

Defective balls are all the same weight, but may be lighter or heavier than ** balls, error.

Gram. You can use a balance scale, or you can use a scale or rod scale. Analysis:

Arrange the boxes in a regular manner and label each box with a number.

easy to identify; Take out a different number of balls from each box and put them on the scale, and take a few balls from the box size.

Number of balls removed.

The number of the box. For example:

No. box, just take it.

A ball out. The number of balls taken out is, in order:

balls. If all the balls weigh the same all the same, then the total weight

Just after the exhaustion, the quotient is the weight of each ball.

In fact, the weight of the ball taken out of the defective ball box is biased and cannot be divided.

If the remainder is less than.

It means that the defective ball is heavier, and the remainder is a few, which is the number of defective balls in the box;

If the remainder is greater than.

Press closer. It means that the defective ball is lighter, and the remainder is ratio.

Worse, it's a few numbers.

Defective balls in a box. Example.

Suppose the ** balls are.

grams, defective balls are all.

grams (heavier.) grams) in.

number in the box. Ideas:

Let's assume for a moment that all ** balls are.

grams, then all the balls add up to:

Gram. And the defective ball is loaded.

box, it will be taken out.

Balls come out, each ball weighs.

grams, there will be more.

Gram, the total weight is.

Gram. I. The remainder is less than.

Explain that the defective ball is heavier, and you can conclude that it is.

No. box. Loading the defective ball. Example.

Suppose the ** balls are.

grams, defective balls are all.

grams (lighter.) grams) in.

number in the box. Ideas:

Let's assume for a moment that all ** balls are.

grams, then all the balls add up to:

Gram. And the defective ball is loaded.

box, it will be taken out.

Balls come out, and each ball is lighter than ** balls.

g, just. There will be less.

grams, the total weight will be.

Gram. I. The remainder is greater than.

Illustrates that the inferior ball is lighter and closer.

Compare. Less. It can be concluded that yes.

No. box contains a defective ball.

Open the chest and take out a different amount of gold nuggets from each box, such as 1 2 3 ...100 total of 5050 pieces. How many grams do you weigh? If there is an extra 50 grams, it means that the box of 50 pieces is real.

Find and compare the proportions of each other, and you will know the result.

Take one for the first belt and take two ,... for the second beltTake ten of the tenth, weigh them together, and see how many grams are less than 550, and a few grams less is that the first few bags are fake....

First of all, each bag is marked from 1 to 10, and the corresponding number of gold coins is taken from each bag according to the serial number, and weighed together. Obviously, because the normal gold coins are 10 grams each, no matter how many genuine coins are gone, the result is a multiple of 10, that is, the single digit is 0 and if there is a counterfeit coin, the single digit is 9 grams, if two counterfeit coins, because 2 * 9 = 18, then the single digit is 8, and the corresponding result can be obtained in the same way, because the number of gold coins is obtained according to the serial number of the bag, so the number of counterfeit coins corresponds to the serial number of the bag. If the single digit is 0, it means that 10 counterfeit coins have been taken, that is, bag No. 10 is counterfeit!

1.Number the boxes, from 1 to 10.

2.Take out one piece of gold from chest number 1, two pieces of 2, and so on.

3.Weigh it after taking it.

Under normal circumstances, it should be 550 taels (assuming a piece of gold is 10 taels), if it is less than 1 tael, it means that box 1 is fake gold, 2 taels and box 2, and so on.

Set aside all ten chests of gold, and take them in their likeness.

Dude, why do you have to do it twice, you can do it once? That's amazing.

Weigh only once? It's a little difficult, I won't.

Brain teasers!

The landlord asked the wrong place, here is the jewelry classification, the heads of jewelry experts are more old-fashioned, and they won't turn around! Hehe.

The boxes are numbered 1 10.

Take one piece for box No. 2, take 2 pieces for box No. 2, take 3 pieces for box 3, and so on ......3. Take out a total of 55 pieces of ** and weigh them together, which should be 55 taels under normal circumstances. If there is one less money, it means that it is the No. 1 box that is greedy, if it is less than two money, it means that it is the No. 2 box that is greedy, if it is less than three money, it means that it is the No. 3 box that is greedy, ......and so on

It's not a brain teaser.

Upstairs is very smart, if this answer is really up to you.

Is this a brain teaser?

Just 2 times and you're good to go.

First divide into three parts, 4 boxes per portion, weigh the two copies, 1: if the two boxes are not the same, then take out the light one, divide the 3 boxes into 3 parts, compare the two copies, if they are different, the light one is fake, if the same, the rest is fake;

2: If it is the same, then take out the remainder, divide the 3 boxes into 3 parts, compare the two copies, if they are different, the light one is fake, if the one is the same, the rest is fake;

1 time is fine.

Label each of the twelve boxes with numbers, e.g. ...... numbers 1 to 12Then take out the gold corresponding to the number from the box separately. Take one piece for the first box, and two for the second box......Take it and weigh ......

Assuming that the weight weighed at this time is 772 grams, if all 78 pieces of gold are true, it should be 78 times 10 grams ——— 78 grams. Therefore, there is a difference of 8 grams, that is, there is a problem with the box number 8

At the beginning, it was divided into two groups, each group of 6, called a group, if it was 60g, it was not here. Take out 3 scales from the 59g group, if it is 30 grams, it is not in this group, it is 29g in this group, use the 29g group to take one and weigh two, if it is 20g, it proves that the one taken away is fake, if it is 19g, take another one, see that the remaining one is 10g, and if it is 9g, it is 9g fake, if the rest is 9g, understand.

Didn't you say what to call it? If you are talking about the scales, you can find out three times if you think about it.

But I think of a very weird name.,Left and right sides are a piece.,If it's balanced, add another piece of each.,Until one side is cocked (light),Then the last piece on it is fake.,Can this be weighed once??

Upstairs are all pigs. The fastest one.

Each pot of gold is numbered, from 1, 2, 3 ,......10, and then each pot of gold is divided into 10 parts, which go to 1, 2, 3 ,......10 pieces up, it should be 5500 grams, but because there is a pot of gold is fake, so less than 5500 grams, at this time use 5500 to cut off the total mass, and then divide by 100, the number that comes out is the number of fake gold.

i refers to the number, a[x], which refers to the 1 10) of the mass of the x bucket of gold taken out of the x bucket of gold

i=[5500-(a[1]+a[2]+a[3]+…a[10])]/100

Give me points!!

Brother below, if you have the ability, you can think of a one that you don't need to separate the gold, but you can do it in one step!

It's a brain teaser problem, right? Only weigh once, no matter how you weigh it, you can distinguish the real from the fake, 900 grams is fake, and 1 kilogram is true--

You mean there's only 1 scale. It's OK with a 2-point method. Divide into 2 piles, 1 pile of 5 placed on both sides of the scale, the light one contains the false one.

Then take one of the fake piles out, and then weigh it, if it is equal, then the one drawn is fake. If the fake one is in one of the piles, then divide that pile. It's OK to divide it into 2 more times.

Each bucket of gold is divided into 10 equal parts (you can ?.. it1 10 barrels take 1 10 10 10 respectively, weigh it, subtract the actual weight with 5500, divide it by 10 to get the number of numbers

Dude, I've been wondering how you managed to divide all that gold into 10 parts!! Teach me, okay??

YMSZ2003 is the standard name, a very old intellectual question commonly known as: pirates weigh gold.

What do you mean? Weigh out 900 grams.

Mark the ten boxes as No. 1, No. 2, and No. 3... On the 10th, take out 1 gold bar in box 1, take out 2 gold bars in box 2, and so on, take out n gold bars in box n, and finally take out a total of 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 gold bars, weigh them with a scale, the ideal value should be 550 taels, but because there is a box of gold bars each less than one tael, the total mass must be less than 550 taels. Because 1 gold bar is taken out of box 1, if the gold bar in box 1 is less than two, the total mass must be less than 1 tael, which is 549 taels; If there are two gold bars in box No. 2, the total mass must be less than two taels, which is 548 taels. And so on, the total mass is n taels less than the ideal value, which means that the gold bars in box n are 1 tael less each.

Number 10 boxes from 1-10

Then take a piece from the No. 1 box, take two pieces from the No. 2 box, and take a few pieces from the number of boxes, so that a total of (1 + 10) x 10 2 = 55 pieces and then weigh the weight of the 55 pieces, because all of them are 1 tael is 55 taels, and now the weight of the scale is compared with 55 cars, and a few dollars less means which box has less **.

Take 1 in the 1st box, 2 in the 2nd box, 3 in the 3rd box, 4 in the 4th box, 5 in the 5th box, 6 in the 6th box, 7 in the 7th box, 8 in the 8th box, 9 in the 9th box, and 10 in the 10th box. Comparing the weight of these 55 sticks with the standard, a few dollars less, it means that there is a problem with a few sticks, that is, there is a problem with the corresponding first box.