Problematics | A gift for packing things correctly

This week’s puzzles are somewhat less challenging, but interesting nevertheless. It is, incidentally, a landmark week of sorts: the 25th in an unbroken series of puzzling weeks.

Last week’s puzzles were deceptively tough, with a number of readers sending incomplete or flawed solutions. Once you know the answers, however, they are ridiculously easy, as we shall soon see.

Welcome to Problematics! (Shutterstock)

This week’s puzzles are somewhat less challenging, but interesting nevertheless. It is, incidentally, a landmark week of sorts: the 25th in an unbroken series of puzzling weeks.

#Puzzle 25.1.

In a certain school, before Class 12 passes out, the principal decides to reward the toppers of all six sections: striped ties for the boys, chequered ones for the girls.

The topper of Section-A and her brother in Section-B both like red, so the siblings’ chequered and striped ties are both red. The sister in Section-C and the brother in Section-D both love blue and get blue ties, one chequered and the other striped. The topper of Section-E likes red while her brother in Section-F likes blue, which means the chequered tie here is red and the striped one is blue.

Each gift goes into a small box, which is unmarked, so you don’t know its contents until you unpack it. These six boxes are paired into three larger boxes, which the teachers do correctly.

The larger boxes are to be labelled with the respective siblings’ names. For convenience, let these be A & B, C & D, and E & F. This is where the teachers get mixed up. When they realise it, they know that the label on each large box is wrong, but they don’t know which large box contains which pair of siblings’ gifts.

The teachers obviously want to avoid the bother of unpacking all nine boxes. What’s the minimum number of boxes, big and small, that they must open in order to find out which label should go where? And which ones should they open first?

#Puzzle 25.2:

I AM A HUMBLE BAR STEWARD IN CHINA The above is a composite anagram of four Indian cities, their letter counts being (7), (6), (3, 5), (6). Which cities?

Mailbox: Last week’s solvers:

Solved #Puzzle 24.1.

Solved #Puzzle 24.2.

In the illustration, the words in any row, any column and any diagonal have exactly one letter in common. The card values, as Rohit points out, add up to 15 in each row, column and diagonal.

Solved both puzzles: Ravi Sondhi & Rudra Sondhi (Gurgaon), Rohit Khanna (Noida), Jasvinder Singh (Nabha)

For a flawless solution, it’s important to: (a) identify the parallel with tic-tac-toe; (b) arrange the cards and the words in separate layouts; and (c) note the strategy that that the first player must begin with a corner and the second must respond with the centre. Vasu Handa, Sunita & Naresh Dhillon, Amardeep Singh, Sishir Gupta, Jaikumar Bhatia and Lucky Singh Randhawa have missed out on one or more of these aspects.

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com