Problematics | 6 men from 6 cities: who gets the job?
This one belongs to a family that we call Einstein puzzles. I have adapted my 1993 original just a little bit, for 2023 readers.
The following puzzle comes from nearly 30 years ago: September 26, 1993, when I set it to readers in my hometown. When I found it among my old newspaper clippings, I was pleased to note that I could still solve it (there have been times when I have struggled to solve my own old puzzles).
This one belongs to a family that we call Einstein puzzles. I have adapted my 1993 original just a little bit, for 2023 readers.
* The six men are from six different cities: Bhopal, Chandigarh, Fatehpur, Guwahati, Haldia and Thane.
* Each plays a different sport: badminton, cricket, football, golf, hockey and table tennis.
* The one who got the job plays a sport whose spelling begins with the same letter as his home city (for everyone else, it is two different letters).
* The man from Haldia sat between the table tennis player and the man from Fatehpur.
* The cricketer and the man from Chandigarh sat at opposite ends of the row.
* The man from Thane sat between the badminton player and the man from Guwahati.
* The hockey player did not sit next to the man from Bhopal.
* The footballer sat at the far right, next to the man from Guwahati.
So, who got the job?
#Puzzle 37.2 ABACAD + FGH = FGABCA + HDA This is an anagram in code, formed by a letter-for-letter substitution. Once the correct letters are in place, they will spell out four different numbers, and the sum depicted will be arithmetically correct. What are the four numbers? |
Mailbox: Last week’s solvers #Puzzle 36.1 In the first column, repeated division by 2 generates a binary representation of multiplier 1, with 0s for the crossed numbers and 1s for the uncrossed numbers. The second column has the products of multiplier 2 and various powers of 2. When we add the uncrossed numbers in the second column, we are adding those products which correspond to 1s in the binary representation of multiplier 1. For the given example, this representation is 100111010. We have 314 = (1 x 2⁸) + (0 x 2⁷) + (0 x 2⁶) + (1 x 2⁵) + (1 x 2⁴) + (1 x 2³) + (0 x 2²) + (1 x 2¹) + (0 x 2⁰). Therefore, 314 x 159 = (159 x 2⁸) + (159 x 2⁵) + (159 x 2⁴) + (159 x 2³) + (159 x 2¹) — Anshul Kumar While Anshul is correct, he does not address the reason we ignore remainders during the division. When you halve a binary number, and ignore the remainder if the dividend is odd, you are effectively removing the last digit of the binary dividend. In this example, halving 100111010 successively gives you 10011101, 1001110, 100111, 10011, 1001, 100, 10 and 1. At each step, the last digit is being removed. In your next step, when you cross out the even numbers on the left and keep the odd ones, you are effectively counting the 1s and discounting the 0s in the binary form. |
#Puzzle 37.1 Hi Kabir, The combination is 5 6 5 3 + 9 7 8 2 5 ____________________________________________________ 1 0 3 4 7 8 — Akshai Bakhai, Mumbai Madhuri Patwardhan offers 5254 + 68935 = 074189, which I am accepting as correct although a total beginning with 0 was not in my thoughts originally. |
| Solved both puzzles: Anshul Kumar, Akshai Bakhai (Mumbai), Madhuri Patwardhan (Thane), Amardeep Singh (Meerut), Sunita & Naresh Dhillon (Gurgaon), Rajesh Bansal (Noida), Sandeep Bhateja (Hoshiarpur), Amar Lal Miglani (Mohali), Shishir Gupta (Indore), Mayobhav Pathak (Gurgaon) |
| Solved #Puzzle 36.2: Kanwarjit Singh (Delhi), Dr Sunita Gupta (Delhi), Mohit Taak (Gurgaon), Shri Ram Aggarwal (Delhi), Anju Rai (Delhi), Shrey Gujral (Delhi), Geetansha Gera (Faridabad), Anshul Siwariya (Mohali), Tosheeba Naidu (Noida), Valen Naidu (Delhi), Chiranjeev Khurana (Pataudi), Avni Nayak, Mihir Joshi |
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com
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