Saturday, August 30, 2003
M I N D  G A M E S


Crazy Daisy and the Mask
Aditya Rishi

Unsung heroes

Inderpreet Kaur, Gurbaksh Singh, Shalin Ved, Manjit Singh, Parveen Bajaj, Seema Budhwar from Pathankot, Shimul Sachdeva and Mansha Singh from Ghanauli in Ropar were among the first ones to save Harker from Count Dracula. In office, there’s always a high chance of mail getting lost, but chances that it will be recovered are higher still.

TRAFFIC in the easy problem zone is always heavy and drivers get time to look back in anger at the long bumpy ride before that. Dirt bikers have a lot to teach us. The higher the bump, the further up they go. When down, they are up and running in no time. They know their course and like their roughs. There’s only one that I don’t like about them: they throw a lot of dirt. I had many of them complaining this time that the road had been too smooth for comfort for the past few weeks.

"It was a real easy question you gave this time and the week before that. Your standard has gone down (It’s slightly abridged here. No insult could be greater). I’d like to see some of the good stuff again. Well, for this time, the answer is 301.

 


First, I thought that the number might be a multiple of 5 and a remainder of 1 could be either 16 or 21. If it was even like 16, it would give all with 2, so it’s more like 21. As 7*3=21 and 7*13=91 both ending with 1, getting on with taking multiples of 7 with 3 at the end and comparing divisibility with 2, 3, 4 and 6 only, as 5, 2 and 7 were known gets us to an easy mental task to mull with 3, 13, 23, 33, 43. Within eight minutes, I got that 7*43=301, which fulfilled all conditions. This is how I got the measly answer. I promise to set a hat-trick record like my brother is claiming this time," says Nipun Khanna of Chandigarh, a typical dirt biker. However, in my long innings, I have seen a lot many hat-tricks, so there goes your record, biker!

Dirt bikers are too cold; perhaps, they come from North Pole. As we move towards the Equator, we discover life in the easy street. This solution comes all the was from there: The chinaman problem is to solve p1: x = 2 (mod 3); p2: x = 3 (mod 5); p3: x = 2 (mod 7); From p1, x = 3t + 2, for some integer t. Substituting this into p2 gives 3t = 1 (mod 5). Looking up 1/3 in the division table modulo 5 reduces it to a simpler equation: p4: t = 2 (mod 5). This, in turn, is equivalent to t = 5s + 2 for an integer s. Substitution into x = 3t + 2 yields x = 15s + 8. This now goes into p3: 15s + 8 = 2 (mod 7). Casting out 7 gives s = 1 (mod 7). From here, s = 7u + 1 and, finally, x = 105 u + 23. As 105 = lcm(3,5,7), we have solutions 23, 128, 233, ...

The Mask problem is to solve q1: x = 1 (mod 2); q2: x = 1 (mod 3); q3: x = 1 (mod 4); q4: x = 1 (mod 5); q5: x = 1 (mod 6); q6: x = 0 (mod 7). With the experience acquired so far, the combination of q1-q5 is deduced to be equivalent to q7: x = 1 (mod 60); x = 60t + 1. Plugging this into q6 gives 60t = -1 (mod 7). Casting out 7 simplifies this to 4t = 6 (mod 7) and then 2t = 3 (mod 7). From division tables modulo 7, 3/2 = 5 (mod 7). Therefore, t = 7u + 5. Finally, x = 420u + 301. The most likely number of eggs Ms Daisy might expect to be compensated for is 301.

"The key (formula) to your puzzle is: [20(3x-1)]7 (where x is any number)," says Neeraj Sharma of Monal Hostel at Palampur. Among the first ones to send in solutions this time were Puneet Goyal (Nabha), Ajit Partap Singh (Ludhiana), Amarjeet Kaur (Ludhiana), Arshdeep Singh (Jalandhar); Sameer Madan (Panchkula), Saurabh Sood (Ludhiana), Nitin Khanna, Ravneet Kotwal (Jalandhar), Abhishek, Shalin and Shalini Ved (Yamunanagar, with half points); Sanjay Chawla, Harish Gautam. Harmik Singh (Pathankot), Dr Tarsem Lal (Khanna), Kamal Sethi (Hisar), Suhail Singh Shergill, Vikas Vashisht, Jaskirat Singh Jassal (Amritsar), Aman Aggarwal (Ambala Cantt) and Vrinda Prasad Tiwari from Panipat. Gurminder Singh, Rahul Khanna, Navneet Kad (Ludhiana) and Tarandip Singh had bike trouble close to the finish line.

Problems of this kind are examples of what universally became known as the Chinese Remainder Theorem. The problems can be stated as finding n, given its remainders of division by several numbers m1, m2,...,mk: n = n1 (mod m1); n = n2 (mod m2)... n = nk (mod mk).