Saturday, August 17, 2002 

RIDDLER: "Now, riddle me that, a weight of 11,111 g is placed on the right hand plate of a balance. Someone places weights on the two plates of the balance; the first weight is 1 g; and each successive weight is twice as heavy as the previous one. At some moment, the two plates are in equilibrium. Where was the 16 g weight placed (on the left or on the right)?" BATMAN: "Since the
two plates are in equilibrium, the sum of weights on the lefthand plate
is equal to the sum of weights on the righthand plate. These sums are
equal modulo any number 'n'. By assumption, all weights (except 11,111)
are powers of 2: 1, 2, 4,... 2^n. Modulo n=4 all weights are zero,
except 1, 2 and 11,111 which is congruent to 3 mod 4, which makes it
clear that 1 must be on the opposite side of 11,111. This reduces the
problem to a situation where we have weight 11,110 on the righthand
plate and the other weights are positive powers of 2: 2, 4,...2^n." 
"Modulo n=32 all weights are zero, except 8, 16 and 11,112=8 mod 32. Thus, 8 must be on the same side as 11,112. This means that we have weight 11,120 on the right hand plate and the other are powers of 2, beginning with 16: 16, 32,... 2^n." "Modulo n=64 all weights are zero, except 16, 32 and 11,120=48 mod 64, which shows that 16 must be on the opposite side of 11,120 (on the lefthand plate)." Batman turns the knob labelled "left", quickly presses the "delay" switch on the bomb and cuts Robin loose, all in one action. BATMAN: "I see without seeing. To me, darkness is as clear as daylight. What am I?" RIDDLER: "Oh please. You're blind as a bat." BATMAN: "Exactly!" Robin and Batman duck, as the bomb goes off under the Riddler's chair. He had failed to see in the dark that Batman had thrown the device under his seat. The blastedoff chair takes Riddler to the sea below, from where, the police boat picks him up. Edward's enigma is solved. Several persons who helped Batman complete his mission were Rupinder Sayal of Patiala ("The exact amount of the balanced weights is: ON THE LEFT PAN  1, 2, 16, 32, 128, 256, 1024, 4096, 8192; ON THE RIGHT PAN: 11111, 512, 2048, 64, 4, 8. When the pans are balanced, each has a weight of 13747 g." Minor mistake, major cost), Dr Lokesh Handa ("The weights will be added thus: R 11111, 4, 8, 64, 512, 2048; L 1, 2, 16, 32, 128, 256, 1024, 4096, 8192; with 13747 weight on both sides." A hitch at the assumption stage), Ravinder Mittal ("The highest weight will be 8192 g and the gross weight will be 16383 g, one less than twice the weight of the greatest weight. This gives excess of 5272 g in the lefthand side plate. To balance, 2636 g need to be shifted to the right." Assumption again), Tarun Bedi ("1+2+4+8+...=(2 to the power an integer)  1 = z (say); Even for high values of z like 32767, 65535, except 8192, the other weights will stay as these are.") and Rohit Pardasani, our Robin, night's second guardian. Two years of Mind Games: your opinion at The Tribune or adityarishi99@yahoo.co.in. — Aditya Rishi
