Divided we rule ...blame it on them Brits!

Remember when you were in school and a smarter than normal aleck asked you, “How far can a dog run into the woods?” And while you were wondering which shadow to slink away in he answered, “Halfway! Be

Remember when you were in school and a smarter than normal aleck asked you, “How far can a dog run into the woods?” And while you were wondering which shadow to slink away in he answered, “Halfway! Because the rest of the time he’s running out.” Can you find anything wrong with this tortuous logic?
Not hot? Okay then challenge a friend or foe to put the digits 1-9 in order so that the first two digits formed a number that was a multiple of 2, the first three digits were a multiple of 3, and so on. Remember though that leaving the digits in the conventional order 1234356789 doesn’t work: 12 is divisible by 2 and 123 by 3, but 1234 isn’t evenly divisible by 4.

THROUGHPUT
(The problem was: “There is a particular integer of unknown number of digits consisting only of 7s which is divisible by 199. Find the last four digits of the quotient without working out the complete quotient.”)
The last 4 digits of the quotient is 6823. Solution is got by writing multiplicand of 199 to give 7 as last digit. Last digit had to be 3 to give 7. 3*199 = 597. With carry of 59 the next digit must be 2 giving 199*2 + 59 = 457. With carry of 45 the next digit must be 8 giving 199*8 + 45 = 1637. With carry of 163 the next digit must be 6 giving 199*6 + 163 = 1357. Oh! I can stop here I have the last 4 digits. -- Raghavendra Rao Hebbani, rao.raghavendrah@gmail.com
A counter-question: how many digits will the smallest number (consisting of only 7’s and divisible by 199) have? -- Rajagopalan K T, ktremail@gmail.com

(Among the first five who also got it right are: Prof S Manikutty, manikuti@iima.ac.in; Manuel Jeyamoney, orcontactme@gmail.com; Abhay Prakash, abhayprakash@hotmail.com; Ramesh Kumar, rameshkumarthayyil@gmail.com; Balagopalan Nair, balagopalannair@gmail.com.)
(The second one was: “How many pairs of prime numbers are there whose sum is 999?”)
Only one pair of prime numbers whose sum is 999 is possible, namely, 2 & 997 because 999 is a odd number and sum of even and odd numbers make an odd number, but any even number except 2 will not be a prime number. -- Arvind Shanker Karan, askaran2@rediffmail.com
The answer is 1 (one), because only one pair of prime numbers: 2 and 997 will give a sum of 999. This is because the sum of any two odd numbers will give an even number (odd + odd = even).The sum of an odd number and an even number is odd (odd + even = odd). And the only even prime number is 2. So, the answer is one pair. -- Rekha G, g.rekhapai@gmail.com

(Among the first five who also got it right are: Ganesh Ram Palanisamy, 1969ganram@gmail.com; Dr P Gnanaseharan, gnanam.chithrabanu@gmail.com; Gopunatarajan Natarajan, natarajangopunatarajan3@gmail.com; Shashi Shekher Thakur, shashishekher@yahoo.com; Srinivasa Yoganandarao, netrakanti71@gmail.com.)
(The third problem was: “In a certain country it’s legal for second cousins to marry. So can third cousins, fourth, fifth and higher cousins. Anita and Bob are third cousins but cannot marry. Why?’)
Because they’re brother and sister. Their parents are second cousins who married. That makes Anita and Bob both siblings (because they have the same parents) and third cousins (because they have the same great-great-grandparents). -- Dhruv Narayan, dhruv510@gmail.com
Anita and Bob cannot marry because they are siblings. Their parents are second cousins who married. In that way they are siblings, as well as third cousins (because they have the same great-great-grandparents). -- Saifuddin S F Khomosi, Dubai.

BUT GOOGLE THIS NOW
1. Find the next term in the series that completes the set: 15, 91, 82, 46, 61, 84, 59, 78, ?​ (Submitted by Ajit Athle, ajitathle@gmail.com)
2. A convex regular polyhedron can stand stably on any face, because its centre of gravity is in the middle. On the other hand, it’s easy to construct an irregular polyhedron that’s unstable on certain faces, so that it topples over. Okay, pay attention to the question now: Why is it impossible to make a model of an irregular polyhedron that’s unstable on every face? (No math please; just a straight, clear and simple reason understandable to all.)

Sharma is a scriptwriter and former editor of Science Today magazine.(mukul.mindsport@gmail.com)

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