Problematics | Don Vito’s grandchildren
When the don dies, which grandson lays wreaths of which colour? And how do the Good, the Bad and the Ugly share their wealth unequally?
It has been quite a while since we last had an Einstein puzzle (aka zebra puzzle) in Problematics. The reason is that I have been setting such puzzles more frequently than earlier, but they have been appearing in our online Games section every Sunday. You can check them here.
This has left me with little time to create an Einstein puzzle exclusively for Problematics, but we cannot allow this to go on. While the online puzzles will carry on as usual, we will resume our Einstein puzzles in Problematics too, once every few weeks. Here comes one of them.
Puzzle #107.1
You know that Don Vito’s children are called Connie, Fredo, Michael and Sonny, in alphabetical order. What not everyone knows, however, is that a film on the family impressed the four siblings so much that they named their own sons after the male actors. In alphabetical order, these grandsons are Brando, Caan, Cazale and Pacino, each of whom has now grown up and plays a different role in the mob.
When Don Vito dies, the grandsons lay wreaths beside his coffin. Placed side by side, their flowers are cyan, magenta, yellow and black, although not necessarily in that order. Your clues:
1. The hitman’s wreath is on the far right.
2. Cazale’s wreath is on the far left.
3. The yellow wreath is second from right.
4. Michael’s son is an arms dealer.
5. The counterfeiter’s wreath is at one of the ends.
6. The wreath from Sonny’s son is immediately right of the arms dealer’s wreath.
7. The black flowers are on the far right.
8. The mob accountant lays his flowers adjacent to Brando’s wreath.
9. Connie’s son places his flowers adjacent to the wreath with magenta flowers.
10. Fredo’s son lays his wreath on the far right.
11. Sonny’s son places his wreath immediately right of the wreath placed by Caan.
12. The wreaths from Michael’s son’s and the mob accountant are side by side.
Please send me a table placing the wreaths from left to right, with the corresponding grandson’s name, whose son he is, his job in the mob, and the colour of the flowers he chose for Grandpa.
Puzzle #107.2
Let’s take a well-known story and add a puzzling plot twist. The Good, the Bad and the Ugly have just collected a treasure buried in the cemetery but, rather than shoot one another, let’s say they decide to share the wealth. The treasure consists of 735 gold coins, which is a multiple of 3, but the Good cons the other two into agreeing to unequal shares.
“Our horses are not equally strong. My horse, being the heaviest, will be able to carry more gold than yours,” the Good says. The Bad and the Ugly nod in agreement. “Let’s share the coins in the proportion of the weight of our horses. They are 1,575kg together, and each one of them is a decent weight, so everyone gets a sizeable share even if it’s unequal,” the Good adds, and smiles to himself as the other two nod again.
After the division, the Ugly cross-checks in a manner he has learnt from previous escapades with the Good. He first compares his share with the Good’s: “Six for me, seven for you. Six for me, seven for you…” and so on. It’s a neat division.
Bundling his coins together again, the Ugly now compares them with the coins with the Bad: “Three for you, four for me. Three for you, four for me…” and so on. This division too is neat.
Who gets how much, and whose horse weighs how much?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 106.1
Hi Kabir,
The matches when 32 seeded players are remaining are shown in the outer columns of the illustration. The progressive rounds are shown converging to the final in the middle column.
— Sabornee Jana, Mumbai
#Puzzle 106.2
Hi Kabir,
With the encyclopedia volumes placed from left to right, the first page of Volume I would be on its right extreme and the last page of Volume X would be on its left extreme, as shown in the illustration. The hole bored by the worm extends from the first page of Volume I to the last page of Volume X. Therefore, it includes one page each of volumes I and X and all 1,000 pages of volumes II to IX. The total number of pages affected = 2 + 8*1000 = 8002.
— Professor Anshul Kumar, Delhi
While the above is indeed the solution I had in mind, and the way a number of readers have solved the puzzle, there is a twist to this tale, too. Here it comes:
Hello Kabir,
The answer is 8002 pages. The worm must have started between pages #1 & #2 of Vol I. This assumes that the term “page” actually refers to a sheet with two printed pages (back to back), and the actual pagination is assumed to be from page #1 to page #2000 in each volume. If, however, the numbering is actually from 1 to 1000, then each volume will have only 500 sheets. This will create a fallacy since the worm cannot bore through 1 page alone. By default it will bore through 2 pages each time it goes through a single sheet.
— Sanjay Gupta, Delhi
Point noted, Sanjay Gupta. The puzzle needs to be more carefully worded the next time I present it anywhere.
Solved both puzzles: Sabornee Jana (Mumbai), Professor Anshul Kumar (Delhi), Sanjay Gupta (Delhi), Aditya Krishnan (Coimbatore)
Solved #Puzzle 107.1: Akshay Bakhai (Mumbai), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Shishir Gupta (Indore)
Solved #Puzzle 107.2: Dr Vivek Jain (Baroda), Ajay Ashok (Delhi)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com