10 identical coins are lying on a table having head ‘H’ face as the upper face. In one attempt, exactly four coins can be turned upside down. What is the minimum total number of attempts in which tail ‘T’ face of all the 10 coins can be brought to be the upper face?
4
5
6
7
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2012
The correct option provided is B (5). However, a mathematical analysis indicates that the minimum number of attempts is 3. Assuming the provided options are correct and the question is solvable within them, there might be an unstated constraint or nuance not captured by simple state transition analysis. Based on standard interpretations of this type of coin-flipping puzzle, 3 attempts are sufficient and necessary. Let’s explain the 3-attempt solution, acknowledging the discrepancy with the provided options.
Let H be the number of heads and T be the number of tails. Initially, state is (H=10, T=0). Target state is (H=0, T=10). In each attempt, exactly 4 coins are flipped.
Let $h$ be the number of heads flipped and $t$ be the number of tails flipped in an attempt, where $h+t=4$. The new state is $(H’ = H-h+t, T’ = T-t+h)$.
Alternatively, consider the number of coins flipped an odd number of times ($o$) and an even number of times ($e$). Initially $(o=0, e=10)$. Target $(o=10, e=0)$. Flipping $x$ coins from the odd group and $y$ from the even group ($x+y=4$) results in a new state $(o’ = o-x+y, e’ = e-y+x)$.
Using the (H,T) state:
Attempt 1: From (10,0). Only H coins are available. Must flip 4 H and 0 T (h=4, t=0). New state: (10-4+0, 0-0+4) = (6,4).
Attempt 2: From (6,4). Possible flips (h,t) with h+t=4, h<=6, t<=4: (4,0), (3,1), (2,2), (1,3), (0,4).
- (4,0) -> (6-4+0, 4-0+4) = (2,8)
– (3,1) -> (6-3+1, 4-1+3) = (4,6)
– (2,2) -> (6-2+2, 4-2+2) = (6,4)
– (1,3) -> (6-1+3, 4-3+1) = (8,2)
– (0,4) -> (6-0+4, 4-4+0) = (10,0)
To make progress towards (0,10), we can go to (2,8) or (4,6).
Attempt 3:
– From (4,6). Possible flips (h,t) with h+t=4, h<=4, t<=6: (4,0), (3,1), (2,2), (1,3), (0,4).
- (4,0) -> (4-4+0, 6-0+4) = (0,10). Goal reached. This requires flipping 4 H and 0 T from (4,6), which is possible.
Thus, a path of 3 attempts exists: (10,0) -> (6,4) -> (4,6) -> (0,10).
Using the $(o,e)$ state (number of coins flipped odd/even times):
Initial state: (0,10) (all coins flipped 0 times, even). Target state: (10,0) (all coins flipped an odd number of times).
Attempt 1: From (0,10). Flip 4 coins. All must be from the ‘even’ group (y=4, x=0). New state: (0-0+4, 10-4+0) = (4,6).
Attempt 2: From (4,6). Flip 4 coins (x from odd, y from even; x+y=4, x<=4, y<=6). Possible (x,y): (0,4), (1,3), (2,2), (3,1), (4,0).
- (0,4) -> (4-0+4, 6-4+0) = (8,2)
– (1,3) -> (4-1+3, 6-3+1) = (6,4)
– (2,2) -> (4-2+2, 6-2+2) = (4,6)
– (3,1) -> (4-3+1, 6-1+3) = (2,8)
– (4,0) -> (4-4+0, 6-0+4) = (0,10)
Attempt 3: From (6,4). Flip 4 coins (x from odd, y from even; x+y=4, x<=6, y<=4). Possible (x,y): (0,4), (1,3), (2,2), (3,1), (4,0).
- (0,4) -> (6-0+4, 4-4+0) = (10,0). Goal reached. Requires picking 0 coins from the 6 odd coins and 4 coins from the 4 even coins, possible.
This also shows a 3-attempt path: (0,10) -> (4,6) -> (6,4) -> (10,0).
Both state models confirm that 3 attempts are sufficient. Minimum steps cannot be less than 3 as shown by exploring states after 1 and 2 steps. Since 3 is not among the options (4, 5, 6, 7), the question or options likely contain an error. If forced to choose from the provided options and accepting the provided answer might be 5 (Option B), it suggests a constraint was missed or a more complex sequence is required. However, standard puzzle analysis leads to 3.