Problematics | How to bring equality to two unequal piles
Another party trick with playing cards, and a set of movie anagrams. Which one do you find easier?
This will be familiar territory for regular readers: someone is playing a party trick, and you figure out why the trick works. So, once again, you are at a party where a smart alec is playing card tricks. Yes, he is the same smart alec who got you working on probability theory five weeks ago (#Puzzle 41.1). This time, however, it’s a different trick he’s playing.
The trickster now asks you to give him any 20 of the 52 cards. Holding the pack beneath the table so that he cannot see, you select 20 cards out of 52, again at random, and hand these over.
Obviously, these 20 cards may include some face-up cards. Although the figure shows six face-up cards out of 20, that’s just an illustrative example; the number of face-up cards in this subset can be anything between 0 and 20.
His hands still under the table, the trickster says: “I don’t know how many cards are face-up among the 20 you gave me, nor how many are face-up among the 32 still with you. But I will flip some of the cards in my hand so that there is an equal number of face-up cards in both piles.”
He does something to the cards under the table, then places all 20 on the table. You place your 32 cards too. You count. Indeed, each pile has the same number of face-up cards.
How did the trickster do it?
Mailbox: Last week’s solvers
If one cow eats 1 unit of grass in one day, 70 cows in 24 days will eat 70 x 24 = 1,680 units, and 30 cows in 60 days will eat 30 x 60 = 1,800 units. So, 1,800 – 1,680 = 120 additional units grow in 60 – 24 = 36 days. We need to find the number of cows that will eat the grass in 96 days, or another 36 days after 60 days. During this period, an additional 120 units will grow. Thus, the total grass will be 1,800 + 120 = 1,920 units, to be eaten in 96 days. Hence 1,920 ÷ 96 = 20 cows will be able to finish this grass.
— Sunita & Naresh Dhillon, Gurgaon
Amardeep Singh identifies my source for this puzzle: Yakov Pereleman’s Mathematics Can Be Fun. Ajay Ashok adds that the puzzle also appears in Perelman’s Algebra Can Be Fun.
Initially, if there were only 4 tyres, every tyre would have to travel 240 km, which would be a total of 960 km.
Now, rather than 4 tyres, we divide this among 5 tyres. Hence, each tyre must cover 192 km.
— Shruti M Sethi, Ludhiana
Solved both puzzles: Sunita & Naresh Dhillon, Shruti M Sethi, Amardeep Singh, Ajay Ashok, Gaurav Kabir Gupta, Dr Sunita Gupta, Abhishek Garg, Yadvendra Somra, Kruti Marthak, Sabornee Jana, Anil Khanna, Hurditya Dand, Geeta Arora, Rahul RB, Prakash Bhate, Sachin Gupta, Ekansh Singh, Mahipal Singh, Amar Lal Miglani, Kanwarjit Singh, Jawahar Lal Aggarwal, Subrata Chakraborty, Rajender Parsad Agarwal, Shri Ram Aggarwal, Navneet Singh Jasrotia, Vinod Mahajan, Madhuri Patwardhan, Vaibhav Aggarwal, Vivek Aggarwal, Shishir Gupta, Sandeep Bhateja, Akshai Bakhai, Saurish Seksaria, Prof Anshul Kumar, Akshay Khanna, Yash Chillar, Shankar Subramanian, Ravish Iyer, Sri Ramachandra Sai Bapatla, Bipin Das, Aziza, Chitra Nitsure, Narendra Prasad, Nipun Kumar & Pooja Thakkar, Sonia Jain, YK Munjal
Solved #Puzzle 45.1: Jobin Babu, Vikas Nanda, Sanghamitran, Aryan Kumar, Dipanjan Kajuri, Mahi Dutt
Solved #Puzzle 45.2: Fateh Pal Singh Malhi, GV Ravishankar, HP Choudhury, Umesh CS Bisht, Geetansha Gera, Shriya Seshia, Navneet Rathi, Yana Kapoor, Sanjay Bhatia, Neeraj Kumar, Saarth Karkera, Joydeep Mahato
Problematics will be back next week. Please send in your replies by Friday noon to firstname.lastname@example.org