20 November 2020

Puzzles in a Singapore class

Yesterday, my blog reminded me of this incident from seven years back when my friend from Singapore – Jyotsna – had pointed out that puzzles from my blog were being discussed in the classes of Singapore. And she asked me to post more.

So, here goes a fairly interesting puzzle:

Imagine a pharmaceutical company in a race to find a cure for coronavirus. It has come up with 1000 chemical formulas as potential anti-virus vaccinations. Those chemical formulas are sitting as solutions in 1000 different beakers in their lab.

However, only 1 of them is effective. In fact, it is supereffective. Even a small trace of it injected in a rat’s body will extinguish the virus within 7 days. The other 999 are of no use. They also know that you can mix those chemical formulas up and the resultant concoction will retain all the properties of the original formulas. Put simply, in a combination of solutions, if you have even a little of the effective formula, the whole combination is supereffective. Else, any combination of those other 999 solutions is totally useless.

One catch is that the pharma company has exactly 7 days to win the race to a vaccination. Which means it gets only one time to inject rats with the various combinations of the chemicals and see what happens by the 7th day.

As if that was not enough, there is way too much demand for rats – apparently lots of pharma companies are trying to do experiments for the vaccination.

Finally, they have to find out the exact solution with the effective chemical formula – there is enough manufacturing capability to make only one of those solutions in a scaled manner.

What is the minimum number of rats the company has to procure to conclusively prove which one of the chemical formula is the ultimate solution against the virus? How would they do the experiment?

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13 June 2020

Really hard puzzle. Can you help me?

I have recently started doing Kenken puzzles (the 6×6 version). Last night, a question came to my mind and ever since I can neither solve the problem nor get it off my mind.

The usual Kenken is a 4×4 grid where you have to fill in every row and every column with the digits 1,2,3 and 4. One rule is that you have to have each one of those digits in every row and every column. (meaning therefore, no repeats in a row or column either – like Sudoku).

There are other rules for each individual Kenken problem. However, assume there is no other rule. You can fill in every row and every column with the digits 1,2,3 and 4 – with the only rule being you have to use each one of them once in every row and every column… how many such combinations are possible?

(I am attaching a picture here of an actual Kenken puzzle – but that

30 May 2020

The girl does make me proud at times… (puzzle)

Usually, it is her sharp wit and acerbic comebacks. And at times it is her math capabilities. (It does take a nerd to recognize another).

Recently, my nephews had sent me a puzzle. Using 1, 1 and 1, you have to make 6. Similarly, using 2, 2 and 2… and then 3,3, and 3… and so on. You can use the normal mathematical operators – but no other digits. For example 2+2+2 = 6; They had given me till 10,10,10.

8 is a little tricky. It would be 8 – sqrt (sqrt (8+8)). Note that sqrt symbol does not use a digit.

During the walk last evening, Niki and tried 11,11,11 and then 12,12,12 and so. Once we reached 19,19,19… she said something that helped us find a completely generic way of getting to 6 given any number (repeated thrice).

For example, can you get 6 using 73, 73 and 73?

Note that you cannot split 73 into 7 and 3.

The mathematical operators allowed (any number of times) are plus, minus, multiply, divide, parentheses, square root, log and factorial. Give it a shot!

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18 May 2020

This week’s puzzle

This week is another area problem. Take a rectangle ABCD (as shown here). Draw two lines from the adjacent vertices A and D to two random points E and F on the opposite side such that they intersect within the rectangle. See picture below. Nothing is drawn to scale. The area of the two triangles are 4 and 16. One of the quadrilaterals has an area of 10. What is the area of the other quadrilateral?

Again, this is geometry from elementary school with a twist to it.

Note that the original version of the problem that I had posted had the quadrilateral area as 1 instead of 10. Harmindar Matharu successfully proved that such a diagram is not possible. I should have been more careful while picking some values for those areas…

5 May 2020

A new square puzzle

Take any square. Take any random point within the square. Draw straight lines from that point to the midpoints of every side of the square. You will get four quadrilaterals. Three of them have areas of 40 units, 60 units and 80 units. What is the size of the total square?

This is an interesting problem and requires a little visualization and some basic memory of geometry.

14 March 2020

Which airport was I in?

We just took off from the airport and as the plane started gaining height, I noticed an interesting landscape on the ground. There is a man made water body and it has been carefully dug up to leave a landmass looking like an airplane!

Which airport was I in?

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24 February 2020

Followup puzzle

This is a followup puzzle to the previous one to understand why the one-black-dot case is not possible. Some find the answer to be controversial. To me the logic is bulletproof though.

So, it is the same problem as before except that the lady skips the portion where she asks the three men to raise their hands if they see at least one white dot. Instead, after removing the blindfolds, she straightaway asks “What color dot do you have on your forehead”.

After quite some thinking, the smartest person came up with the correct answer.

What was the answer and how did he figure it out?

For completeness sake, I am restating the whole problem here…

Three smart men were trying to find out who is the smartest of them all. Or was at least most logical and quickest thinker. To resolve that, they went to a woman who had the reputation of being the smartest of them all. She devised a quick problem for them. She blindfolded them and painted a dot on each of their foreheads.
She then told them “I have painted a dot on each of your foreheads. The dots are either white or black in color. I am going to soon open your blindfolds. You will be able to see other’s dots but not your own. Then I am going to ask you to guess the color of your own dot”.
Saying so, she opened their blindfolds.
“Okay, now guess what color is on your forehead”.
The three men started pondering and soon one of them – who was indeed a little smarter than the others – spoke up.
“Correct”, said the lady.
Question: What was the answer and how did he figure it out?