Question

There are 7 tumblers on a table, all standing upside down. You are allowed to turn any 2 tumblers simultaneously in one move. Is it possible to reach a situation when all the tumblers are right side up? (Hint: The parity of the number of upside down tumblers is invariant.)

Answer 1

Let u – No. of tumblers right side up

v – No. of tumblers up side down

Initial stage : u = 0, v = 7 (All tumblers upside down) 

Final stage output: u = 7, v = 0 (All tumblers right side up)

Possible Iterations:

(i) Turning both up side down tumblers to right side up u = u + 2, v = v – 2 [u is even]

(ii) Turning both right side up tumblers to upside down. u = u – 2, v = v + 2 [u is even]

(iii) Turning one right side up tumblers to upside down and other tumbler from upside down to right side up.

u = u + 1 – 1 = u, v = v + 1 – 1 = v [u is even] 

Initially u = 0 and continuous to be even in all the three cases. Therefore u is always even. Invariant: u is even (i. e. No. of right side up tumblers are always even)

But in the final stage (Goal), u = 7 and v = 0 i. e. u is odd.

Therefore it is not possible to reach a situation where all the tumblers are right side up.