Problematics |The fastest way to toast a sandwich
Given a toaster that can accommodate two slices but toast only one side of each at a given time, how do you toast three slices in the most efficient time?
Technology evolves in a way to phase out old items while ringing in the new, and I am not sure if people still use the kind of sandwich toaster that is central to this week’s main puzzle. There are modern types that toast both sides of a slice of bread simultaneously, and you can pack in several slices at the same time. The one being described here, on the other hand, can accommodate no more than two slices at a time, and requires you to toast only one side of any slice at any given time. Wikimedia Commons, thankfully, has an image of such a toaster, so that you can understand what I am describing.

You would obviously want both sides toasted, so you would let one side heat up for the time you need, then flip the slice and toast the other side. If you have two slices, you toast both together, first one side each, then flipping both slices to toast their reverse sides. With three slices, it becomes a little complicated if you want to manage your time effectively.
#Puzzle 108.1

You have three slices, and want to toast the six sides for 3 minutes each. You have just managed it in the most efficient way when three guests arrive.
“I was just having breakfast,” you greet them. “Would you like some toast?”
“Yes,” they chorus, “but just one slice for each of us, please!”
“How many minutes do you like your slice toasted for?”
“Two minutes each side,” says Guest #1. “Three minutes each side,” says Guest #2. “Four minutes each side,” says Guest #3.
What is the most efficient time you can manage when (a) toasting your own three slices, and (b) toasting the three slices for your guests? Remember, you had finished preparing your own slices before the guests arrived.
#Puzzle 108.2
A chessboard consists of 64 squares in the usual pattern, a side of each square measuring 1 inch. Now take 32 strips of cardboard, each 2 inches long and 1 inch wide. It is easy to arrange these 2 x 1 rectangles across the chessboard, each covering two squares, until the entire board is covered.
Now let’s mutilate the chessboard. Remove two squares at the corners, one diagonally opposite the other. That leaves you with 62 squares, so I ask you to discard one of your 32 cardboard strips. You now have 31 strips, each 2 x 1, or an equivalent of 62 chessboard squares. However, if you try, you will find that you cannot cover the entire 62-square board with these 31 strips.
Can you prove why it is impossible? (Hint: look at the colours. Source: Shakuntala Devi)
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 107.1

#Puzzle 107.2
Hi Kabir,
The ratio of the Good and the Ugly’s gold shares is 7:6 or 14:12; and the ratio of the shares of the Ugly and the Bad is 4:3 or 12:9. So the share ratio for all three is 14:12:9.
Let x be the common factor for gold. So, total gold = 14x + 12x + 9x = 735, or x = 21. So the gold shares are, Good = 14 × 21 = 294, Ugly = 12 × 21 = 252, Bad = 9 × 21 = 189
The ratio of the weights of the horses is the same, i.e. 14:12:9. Let the common factor here be y, so, total weight 14y + 12y + 9y = 1575kg, or y = 45kg. So the horses’ weights in kg are, Good = 14 × 45 = 630, Ugly = 12 × 45 = 540, Bad = 9 × 45 = 405.
— Shishir Gupta, Indore
Solved both puzzles: Swetha Senthilkumar (Coimbatore), Shishir Gupta (Indore), Dr Sunita Gupta (Delhi), Akshay Bakhai (Mumbai), Sanjay Gupta (Delhi), Sabornee Jana (Mumbai), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Anil Khanna (Ghaziabad), Ajay Ashok (Delhi), Aditya Krishnan (Coimbatore), Shri Ram Aggarwal (Delhi), YK Munjal (Delhi)
Solved #Puzzle 107.1: Rituparna Gupta (Indore)
Solved #Puzzle 107.2: Kanwarjit Singh (Chief Commissioner of Income-Tax, retired)