Problematics | The IPL equation

14 days ago | 5 Views

Your mathematics teacher at school will have told you that you can only solve a set of simultaneous equations if there are as many equations as there are variables, but that is because they don’t teach you Diophantine equations at school. As puzzlers know, these are equations that are usually one fewer in number than the variables, and the most common kind involves one equation with two variables. You can still solve them because the questions involve certain restrictions, such as only integer solutions being acceptable.

I don’t know if anyone has ever tried introducing Diophantine equations to the IPL. If they have, hats off to them. If not, here is the IPL’s first Diophantine problem.

#Puzzle 86.1

The batting side is struggling with three wickets having fallen for very little. The number five batsman then saves the day by scoring the bulk of the runs in the remainder of the innings. The wickets continue to fall, though. The fourth falls when No. 5 has scored only 1/7th of his eventual total. The fifth one falls after No. 5 has scored another 1/5th of his personal total. When the sixth wicket falls, No. 5 has added just five runs more.

The real partnership begins now. Number five continues to score while number eight lends a supporting hand, giving his senior partner most of the strike. As it turns out, the number of runs scored by No. 5 during the partnership is exactly twice the runs scored by No. 8.

Then No. 8 falls and disaster strikes again. The bowler goes on to complete a hat-trick with No.9 and No 10, then No. 11 survives a tense couple of deliveries, and No. 5 breathes a sigh of relief as he gets back on strike in the subsequent over.

He refuses a single on each of the first four deliveries. The fifth ball goes for four. On the last delivery, he attempts a quick single and runs his partner out.

None of the dismissed batters has reached 50. What is No. 5’s score, and what is No. 8’s?

#Puzzle 86.2

A book of puzzles tells me with authority that only four English words end in the letters DOUS. How many can you find?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 85.1

Hi Kabir,

The parties and their old and current members can be summarised as shown in the table.

— Anil Khanna, Ghaziabad

#Puzzle 85.2

Hi,

As 6 cigarette butts make one new cigarette, 216 cigarette butts can make 36 cigarettes. From the butts of these 36 cigarettes, six more cigarettes can be made, and out of these six butts, one more cigarette can be made. So the total number of cigarettes the miser can smoke is 36 + 6 + 1 = 43.

— Ajay Ashok, Mumbai

***

Hi Kabir,

Consider that instead of making the cigarettes in batches, he makes one cigarette and smokes it, then makes another one and smokes it and so on. He uses 6 butts to make the first cigarette. For making every subsequent cigarette he uses the butt left from the previous cigarette plus 5 additional butts.

So, the total number of cigarettes is 1 + (216 - 6)/5 = 43.

Of course, there will be one butt left at the end.

— Professor Anshul Kumar, Delhi

Solved both puzzles: Anil Khanna (Ghaziabad), Ajay Ashok (Mumbai), Professor Anshul Kumar (Delhi), Yadvendra Somra (Sonipat), Shruti M Sethi (Ludhiana), Akshay Bakhai (Mumbai), Dr Sunita Gupta (Delhi), Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Shishir Gupta (Indore)

Solved #Puzzle 85.1: Vishnuvardhan (University of Technology Sydney), Sanjay S (Coimbatore)

Solved #Puzzle 85.2: Dr Vivek Jain (Baroda), Jaikumar Inder Bhatia & Disha Bhatia (Ulhasnagar, Thane)