Gardner’s book goes on to a more advanced version. Give instructions as before, but this time you want all three coins to be heads up. Again, what is the minimum number of moves required?

#Puzzle 23.2:

You have three glasses, one of them upside down. Flipping any two glasses at a time, one move is enough to get all three glasses upside down. Can you make three such moves and still get all three glasses upside down?

#Puzzle 22.1:

The puzzle about cats and dogs, in hindsight, could have been made more difficult by withholding some information. Kudos to Gopal Menon for a solution that is better than my own:

You have provided excess information for this puzzle. The total cost of the three dogs need not be given. It can be determined.

Let the first letter in the name of each animal denote its cost. Then, a = d + 56 and c = d – 56.

Let x, y and z be three distinct numbers {0.5, 1, 2}, not necessarily in that order. Then, the costs of the cats are xa, yc and zd.

a + c + d + xa + yc + zd = 100000

=> (d + 56) + (d – 56) + d + x(d + 56) + y(d – 56) + zd = 100000.

=> (3 + x + y +z)d + 56(x – y) = 100000

=> 6.5d + 56(x – y) = 100000, since x + y + z = 3.5.

=> d = [200000 – 112(x – y)] ÷ 13.

(x – y) can be 0.5, 1, 1.5, (–0.5), (–1) or (–1.5). Now, for these values, check whether [200000 – 112(x – y)] is divisible by 13.

Only x – y = 1 satisfies this requirement. From this, we get d = 15376 and, a = 15432, c = 15320.

x – y = 1 implies that x = 2, y = 1, z = 0.5.

Using the equation for the total cost, we get, b = 30864, m = 7688 and s = 15320.

We can then determine the combinations in the enclosures.

— Gopal Menon, Mumbai.

#Puzzle 22.2:

Hello Sir,

The longest word formed by anagramming the letters of DICTIONARY would be INDICATORY (10 letters).

— Yojit Manral, Faridabad

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