Math problem

You have ten identical open-topped boxes with 10 visually identical coins in each.

In nine of the boxes each of the 10 coins has a mass of 10g.
In one box the 10 coins have masses of only 9g each.

How can you find which box is the odd one out?You have a normal mass measurer with a single pan and a scale. With just one weighing you can identify the box with the lighter coins.

What will you weigh?

Reply 1

brianeverit

Original post by Zenarthra

You take 1 coin from the first box, 2 from the second, 3 from the third and so on.
and record the total weight

(edited 9 years ago)

Reply 2

Zenarthra

Original post by brianeverit

You take 1 coin from the first box, 2 from the second, 3 from the third and so on.
and record the total weight

I dont understand O_O?

Reply 3

the_chemist

When you say "one weighing" do you mean you can only take one measurement? Is this even possible? Cuz it seems pretty impossible

Reply 4

Kenan and Kel

Hope you get lucky first time round. If you don't, make balancing scales with the equipment left and use it to find the odd box against the original box measured. Boom!

Reply 5

brianeverit

Original post by Zenarthra

I dont understand O_O?

If the coins all weighed 10 gm then the total of the weigjhing would be 550 gm.
Suppose instead it was (550 -k) gm then the box trhat you took k coins from is the one with the lighter coins.

Reply 6

Zenarthra

Original post by brianeverit

If the coins all weighed 10 gm then the total of the weigjhing would be 550 gm.
Suppose instead it was (550 -k) gm then the box trhat you took k coins from is the one with the lighter coins.

Oh -.-
That was so simple *face palm

THanks!

Reply 7

Zenarthra

Original post by the_chemist

When you say "one weighing" do you mean you can only take one measurement? Is this even possible? Cuz it seems pretty impossible

I guess so xD

Reply 8

Zenarthra

Original post by Kenan and Kel

Hope you get lucky first time round. If you don't, make balancing scales with the equipment left and use it to find the odd box against the original box measured. Boom!

LOOL

Reply 9

the_chemist

Original post by Zenarthra

Oh -.-
That was so simple *face palm

THanks!

I thought you could only take one measurement? Isn't his method taking ten?

EDIT: Or at least, more than one?

Reply 10

MathMeister

Isn't it because you can find out how many 9g coins you took out by taking increasing amounts of coins. Therefore, you know what box had the 9g coins.

Reply 11

MathMeister

Isn't it because you can find out how many 9g coins you took out by taking increasing amounts of coins. Therefore, you know what box had the 9g coins. You can find out how many 9g coins are taken out because if the 9g was in a different box, a different amount of 9g coins would be taken out and when you measure it would measure a different amount. You can calculate the weight and see how many 9g coins there are and thus what box it was in.

Reply 12

RoyalBlue7

If you don't mind it - where did you get this problem? :smile:

Reply 13

BabyMaths

Original post by the_chemist

I thought you could only take one measurement? Isn't his method taking ten?

EDIT: Or at least, more than one?

No. It's only one weighing.

Take 1 from the first, 2 from the second etc and weight them all together.

If they were all 10g then the result would be 10(1+2+3+...+10)=550

The actual result will be less by 1, 2, 3...or 10 depending which box held the light coins.