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House guests can be like fish, says Sophy McHannot, the originator of the following problem.

By Mukul Sharma

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Published: Fri 26 Sep 2014, 2:05 PM

Last updated: Fri 3 Apr 2015, 6:27 PM


That’s because after more than a few days, they start to go bad. Her immediate family apparently found that out when seven of their friends came to stay one summer, and for 11 nights in a row their guest room, and at times even the sofas, were occupied. Now McHannot wants to know if you can discover 
the full names of all of their visitors (one surname was Newton) and their days of arrival and departure from the following 13 clues.

(1) Each of the two married couples arrived and left together as a pair; each husband and wife share the same last name. (2) The first guest or guests arrived on Monday, August 20, and the last left on Friday the 31st. (3) Nobody left on a Wednesday. (4) Nobody stayed exactly three nights, or more than four nights. (5) A single man arrived on Saturday. (6) Bea and her husband arrived before Lisa 
(possibly on the same day), and left after her. (7) Ms McCarthy (who isn’t Kathy) arrived the day that Jon was leaving. (8) Veronica didn’t arrive on a Thursday or Friday. (9) Andrew stayed over Sunday night. (10) Ms Duval arrived on a Tuesday. (11) Kathy arrived the day that Mr Vogel was leaving. (12) There was at least one female guest in the house for each of the 11 nights. (13) Ralph stayed longer than Mr Soames.

(And no guessing please!)


(The problem was: “You have two plastic cubes. Every day you arrange both so that the front faces show the current date. What numbers are on the faces of the cubes to allow this?” — MS)


You can arrange the cubes in two ways, depending on whether you want to represent the dates with a leading zero or without. If without, then create one cube with digits 1, 2, 3, 4, 5, 6 and the other with 0, 1, 2, 7, 8, 9. If you do want the leading zero, then you will have to have zero on both cubes to represent each date from 01 to 09. However, this presents a problem: you need 0, 1 and 2 on each cube, but that only leaves room for six more digits. The trick is to note that 6 and 9 look pretty much the same when turned upside down, so you can omit one of them! So you have cubes with 0, 1, 2, 3, 4, 5 and 0, 1, 2, 6, 7, 8.

  • Vanessa Fernandes,

Among others who also got it right are: Azza Goltay,; Syed Omar Ghayas Rizvi,; Ramesh S Mahalingam,; Jayesh M P,; Kiran Anand R,; Harmeet Kohli,

(The other problem was: “If you write from 1 to 1,000,000 in order, how many digits would you use and what would be the millionth digit?” — MS)


There are nine one-digit numbers, 90 two-digit numbers, and so on up to... 900,000 six-digit numbers. Therefore, the total number of digits from 1 to 1,000,000 would be 9 x 1 + 90 x 2 + 900 x 3 + 9,000 x 4 + 90,000 x 5 + 900,000 x 6 + 7 = 5,888,896. Also, we can deduce that there are 488,889 digits from 1 to 99,999 using the same method. Thus, starting from 100,000, we should move 1,000,000 – 488,889 = 511,111 digits further; getting us in the range of six-digit numbers. Dividing 511,111 by 6, we get 85,185; which we need to add to 100,000 to get 185,185. The total number of digits so far is 488,889 + 6 x 85,185 = 999,999. The next digit, which will be the millionth digit, is 1 (which is the first digit of 185,185).

  • Saifuddin S F Khomosi,


1. What is the most “striking” anagram of: MAN LAUNCHES HARD STRIKE?

2. A number of cats killed 1,111,111 mice, every cat being responsible for an equal number of mice. As there were more mice killed by each cat than there were cats, how many cats were there in all?

(To get in touch, email Mukul at

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