Coin weightings with two irregular coins

wonderful

There are 8 coins of which 6 have the same weight of A g(A>10)] each. The other two are different. One weighs (A-2)g, and the other weighs (A+2)g

Please use the standard scale balance to identify the two differed coins using the minimum weightings?

Have A Great Day!

Rogerio

2 questions:
- Is A an integer?
- Do we know the A value a priori, or is A unknown?

wonderful

In principle, A can be any number greater than 10. For now, we can assume that A is an integer. A is known.

Have A Great Day!

Rogerio

I can do it with 6 weightings (maximum).
But, maybe, it could be done with only 5 weightings...I'm still working on it.

wonderful

I am glad that you are interested in solving this interesting question. You're heading in the righ direction. Can you do this with less than 5 weightings?

Have A Great Day!

Rogerio

Wow, it seems I can do it in 5 steps!
I'm gonna check it later...(too sleepy to continue)

Noone

1There are 8 coin's=true
2There are 6 coins that have the same weight of A g(A>10)] each.=true
3The other 2 coins are different, one weighs (A-2)g, and the other is (A+2)g=true
4 to answer the question you must identify the two differed coin's from the rest=true

So since 1,2,3,4 are true and then that would mean the answer is in the problem it's self...

Hence it identify's the two differed coin's, then turns it into a question at the end...

so one differed coin wight is (A-2)g and the other is (A+2)g, so the coin on the left has the wieght of (A+2)g and the one on the right has a wieght of (A-2)g, so i have identified wich coin's differed by how much... using logic and the information that was presented within the question/statments...

The value of (A) is not needed to know how much they differed by... because one is 4 less and one is 4 higher than (A).. which mean's (A) can stay an unknown factor within the problem.. because its not needed to find the correct answer..

So the one on the right min wieght is no less than -2g from (A) and the one on the left min wieght is no less of +2g from (A) so.. the differed scale is about +4g,-4g from each other's wieght's..i presented the answer in 3 difrent format's, and this last one is just how the question asked it to be presented in.

doodle

I think I am able to get it in 4 weighings (but not less). The initial steps are as follows:

Assuming that the coins are labelled 1 to 8, then the initial weighing is 12vs34. If 12=34, then the second weighing is 156vs278. Otherwise (whether 12>34 or 12<34), the second weighing is 137vs248.

Before I go on, tell me if I am on the right track.

Noone

Dose it really need a question mark at the end?

Schrodinger's Dog

I've seen the answer to this on Columbo. So I'll refrain from commenting. Although it's pretty obvious if you think about it.

Kittel Knight

What a trip!
What a waste of time!

Schrodinger's Dog

Yep the actual number of weighings is 1, it has to be. But the question was presented poorly. Effectively if you do the right thing you can always gain an answer from one weighing. It's a silly question. Columbo did it better in The Bye-bye Sky High I.Q. Murder Case.

How can you do it in one weighing? Well he doesn't specify how you use the weights too clearly.

wonderful

Thank you Rogerio and all who have participated in discussing this question. I agree the question could have been written clearer. However, Rogerio and Doodle understood the question perfectly. I really appreciate the truth that they spent time on the main purpose of this thread: finding the solution of the posted question.

Having said that, here is a more interesting question:

There are 13 coins. 11 of them weighs A g each. The other two are defected. One weighs (A+ 2) g and the other weighs (A-2) g.

Using a standard two arm balance, can you find out the two difffected coins using minimum number of weightings?

Have A Great Day!