Problematics | Small car vs big car

Someone has proved that a small car occupies the same area as a big car. Can you prove them wrong?

Many tricks can make convincing arguments that appear to prove obvious fallacies. Proving that 1 = 0 or 1 = 2 is very common, and the “proofs” come in various forms. There is as much fun in detecting where the mathematics is wrong as in observing how smartly the “flaw” has been introduced.

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Below is an elaborately devised ”proof” of yet another fallacy. I found it fun, so I adapted it from the translated writings of Russian author Boris A Kordemsky.

#Puzzle 183.1

At a mall, a small car had found a nice slot and was about to park there when the owner of a bigger car stopped the other owner. For simplicity, we shall call them Mr Big and Mr Small, both male and Indian, with no other details given.

Mr Big’s argument was that the slot was large enough for a big car but not enough for two small cars, so some space would be wasted if Mr Small parked there. Mr Small, on the other hand, argued that both cars would occupy the same space. Mr Big was unconvinced, so Mr Small offered to prove it mathematically. Mr Big, who had never been strong in that subject, reluctantly agreed.

“Suppose my car occupies an area of s,” said Mr Small, “while your car occupies an area of b, Mr Big. For simplicity, let us assume that the two areas add up to an even number, say 2n, i.e. s + b = 2n.”

“I see,” said Mr Big, having made no sense of it. Mr Small then proceeded with his equation:

s + b = 2n

=> s – 2n = –b (equation 1)

=> s = –b + 2n (equation 2)

Multiplying equations 1 and 2,

s² – 2ns = b² – 2nb

Adding n² to both sides

s² – 2ns + n² = b² – 2nb + n²

=> (s – n)² = (b – n)²

Taking the square root of each side,

s – n = b – n,

=> s = b

“Therefore, you see, your car and mine each occupy an equal area,” explained Mr Small.

Mr Big, too puzzled to argue, withdrew.

Since b is obviously bigger than s, something must be wrong in Mr Small’s proof that they are equal. Where is the flaw?

#Puzzle 183.2

At a party being celebrated mainly by the adults of three families, their children (seven in all) are made to sit on a bench and told not to make a noise. The children, four girls and three boys, grumblingly take their seats at random along the length of the bench.

What is the probability that a boy is sitting on each end?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 182.1

Hello Kabir,

Let the original three-digit number be x, formed with the digits a, b and c. The number formed with the opposite spots will have the digits (7 – a), (7 – b) and (7 – c), so adding the two three-digit numbers will give us 777. In other words, the opposite three-digit number is (777 – x). By combining the two into a six-digit number, we get

1000x + (777 – x) = 999x + 777

Dividing the above by 111 gives (9x + 7), which is announced to the magician. The magician subtracts 7 to get 9x, then divides by 9 to get x, which is the original three-digit number.

— Vinod Mahajan, New Delhi

#Puzzle 182.2

Hi Kabir,

The pyramid puzzle took a bit longer than the one with dice.

The square pyramid will have in each layer 1, 4, 9, 16 oranges and so on. The total number of oranges in such a pyramid will be: for 1 layer = 1; for 2 layers = 1 + 4 = 5; for 3 layers = 5 + 9 = 14; for 4 layers = 14 + 16 = 30; and for 5 layers = 30 + 25 = 55.

The triangular pyramid will have 1, (2 + 1), (3 + 2 + 1), (4 + 3 + 2 + 1) oranges etc in successive layers, that is 1, 3, 6,10 etc. The total oranges in such a pyramid will be: for 1 layer =1; for 2 layers = 1 + 3 = 4; for 3 layers = 4 + 6 = 10; for 4 layers = 10 + 10 = 20; for 5 layers = 20 + 15 = 35; and for 6 layers = 35 + 21 =56.

From the condition in the puzzle, we have to add one orange to a square pyramid to get a triangular pyramid. So, there were 55 oranges in 5 layers in the square pyramid, and 56 oranges in 6 layers in the triangular pyramid.

— Dr Nitin Trasi, Sydney

***

Dear Kabir,

There are two options within plausible limits. A square pyramid of 5 layers (55 oranges) can be given an extra orange to form a triangular pyramid of 6 layers (56 oranges). Also, a square pyramid of 9 layers (285 oranges) can be given an extra orange to form a triangular pyramid of 11 layers (286 oranges).

— Y K Munjal, Delhi

While (285, 286) is a mathematically correct combination, it is debatable whether a child would plausibly want to form pyramids with so many oranges (even if she got hold of so many, which again is debatable). The only answer I had in mind was (55, 56). Still, thanks to every reader who has suggested the (285, 286) alternative.

Solved both puzzles: Vinod Mahajan (Delhi), Dr Nitin Trasi (Sydney), YK Munjal (Delhi), Dr Sunita Gupta (Delhi), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Ajay Ashok (Delhi), Sabornee Jana (Mumbai)

Solved Puzzle 182.1: Anil Khanna (Ghaziabad), Shishir Gupta (Indore)

Solved Puzzle 182.2: Dr Vivek Jain (Baroda)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com