According to a, a must be singular. Because only two people will be different, if there is an odd number left.

According to B, how to calculate this situation? From what A said, we know that B must be singular, because only in this way can A and C be the same.

If b is 1 1, then a and c are 1/2.

If b is 9, then AC-3/2- 1/4.

If it is seven, then AC-5/2-3/4- 1/6.

If it is five, then AC- 1/8-3/6-5/4-7/2.

If there are three, then AC-110-3/9-5/6-7/4-9/3.

If it is 1, then AC-112-3/10-5/8-7/6-9/4-1/2.

C has a hidden condition here. How to guess the combination when C knows his own number and AB is singular?

First of all, make sure that the opposite side is singular after taking c, so that the only number that can exclude the combination with even number is 10.

C takes 8, the remaining 5, and the combination of the two, the singular number of AB is known, and it is determined when B is known, but ABC is different. Then b should be 5 and a is 1.

So the password should be 85 1.

Thanks for correcting me. I made a mistake earlier, and it has been revised.