CSE Blog - quant, math, computer science puzzles

Source: Discussion with Ashwin Rao ( http://zlemma.com , Ex-MD Morgan Stanley) and Gaurav Kumar (Credit Suisse IB, Ex-MS, GS, IITB CSE 2006, Booth 2012) Problem: Suppose teams A and B play in the world series of up to 7 games in which the first team to win 4 games wins the series and then no more games are played. Suppose that you want to bet on each individual game in such a way that when the series ends you will be ahead by exactly $100 if your team wins the series, or behind by exactly $100 if your team loses the series, no matter how many games it takes. How much would you bet on the first game? Update (22 June 2014): Solution posted by Alex, JDGM, Arnab in comments!

Source:  http://www.math.utah.edu/~cherk/puzzles.html Problem: Prove that for any natural N, 1000^N - 1 cannot be a divisor of 1978^N - 1 Short and Sweet :) Solution: Highlight the part between the * symbols for the answer. * Look at the power of 3 dividing each number. 1000^N - 1 always contains MORE factors of 3 than 1978^N - 1, so it can never divide it. By the Lifting the Exponent lemma (LTE) for the odd prime 3: if 3 | a - 1 then v_3(a^N - 1) = v_3(a - 1) + v_3(N). For 1000: 1000 - 1 = 999 = 27 x 37, so v_3(1000^N - 1) = 3 + v_3(N). For 1978: 1978 - 1 = 1977 = 3 x 659 with 3 not dividing 659, so v_3(1978^N - 1) = 1 + v_3(N). Hence v_3(1000^N - 1) = 3 + v_3(N) > 1 + v_3(N) = v_3(1978^N - 1) for every N. A divisor cannot contain a higher power of a prime than the number it divides, so 1000^N - 1 never divides 1978^N - 1. (Elementarily: 1000^N - 1 is a string of 3N nines, divisible by 27 as soon as... the digit-sum factorization 9 x 111 x 1001001 x ... pulls out one ...