Posts

practice problems pending 1,2,4,7

Image
  Homework(proofs): Q1. S1. n = 1 => (2k+1)^(2^1) = 4k^2 + 4k + 1 = 4k(k+1) + 1 = 8p + 1 = 1 mod 2^3(since k(k+1) is even) So works for n = 1 Assume (2k+1)^(2^n) = 1 mod 2^(2^(n+2)) _____[1] and using this prove for n+1: (2k+1)^(2^(n+1)) = 1 mod 2^(2^(n+3)) LHS = (2k+1)^(2^n.2^1) = [(2k+1)^(2^n)]^2 = [2^(n+2).m + 1]^2 using [1] = 1 + 2.m.2^(n+2) + m^2.[2^(n+2)]^2 = 1 + m.2^(n+3) + m^2.2^(2n+4) If you take mod 2^(n+3), it's clearly 1. H.P. Q2. Call a natural number n convenient, if n^2 + 1 is divisible by 1000001 . Prove that among the numbers   1,2,…,1000000, there are evenly many "convenient" numbers. S2. So essentially we need to find some sort of complement for each 'n' satisfying this. If we can show that for each 'n' satisfying this, there is another number satisfying this, we are done. Simplest 'complement' here is -n but that's negative. So let's think circular and try N-n where N = 1000,001 (N-n)^2 +1 = N^2 + n^2 - 2N.n + 1 = n^2...
Post a Comment
Read more

practice problems

 Homework multiple choice: (done) Q1. True/False For a polynomial f(x) with integer coefficients and integers a,b,l  we have a = b mod l => f(a) = f(b) mod l? S1. True Q2. The 2-digit integers from 19 to 92 are written consecutively to form the integer  N = 192021⋯9192. Suppose that 3^k is the highest power of 3 that is a factor of N. What is k? S2. Method 1: Since 10^m = 1 mod 9 for any m => 192021⋯9192 = 19 + 20 .. 92 mod 9 = 74/2[19 + 92] = 37*111 mod 3 = 0  mod 9 = 1 * 3 = 3 So k = 1. Method 2: Sum of digits =  (2+3+4+5+6+7+8)*10 + (1+2... 9)*7 + sum_digits(19 + 90 + 91 + 92) = 35*10 + 45*7 + 40 Doing mod 3 gives 2 + 0 + 1 = 0 mod 3 Doing mod 9 gives 8 + 0 + 4 = 3 mod 9 So 3 divides it but 9 doesn't. k = 1 = answer. Q3. Remainder of 3^89*7^86 mod 17? S3. Method 1: = 21^86*27 21 mod 17 = 4 21^2 mod 17 = 16 = -1 (-1)^43 = -1 => -1*27 = -27 mod 17 = -10 mod 17 = 7 mod 17 Method 2: 3^4 = 81 = -4 m 17 3^8 = 16 m 17 = -1 3^89 = 3^88.3 = (-1)^11.3 mod 17 =...
Post a Comment
Read more

calculus quick notes pending

double derivative for concavity check. single derivative for slope(rate of change check). Rolle's theorem: f(a) = f(b) => for some c in (a,b) f'(c) = 0 Mean value theorem, for some c, f'(c) = (f(b) - f(a)/(b-a)
Post a Comment
Read more

calculus pending

Q1. The position of an object is given by s(t) = sin⁡(3t) − 2t + 4. Determine where in the interval [0,3] the object is moving to the right and moving to the left. Given arccos(2/3) = 0.841 radians. and 2.pie = 6.283 radians. S1. s'(t) = 3cos(3t) - 2 First find critical points: 3cos(3t) - 2 = 0 => cos(3t) = 2/3 => Since cos(3t) is positive, there are 2 solutions (1st and 4th quadrant). 3t = acos(2/3) + 2.pie.k = 0.841 + 2.pie.k and 3t = [2.pie - acos(2/3)] + 2.pie.k = 5.442 + 2.pie.k How many valid solutions are there for 3t in [0,9] since t in [0,3]. Apart from the 2 above, there is only 1 more when 3t = .841 + 6.283 = 7.124 So these 3 are our critical points. Let's evaluate at each of them. Case 1: 3t between [0,.841) s'(0) = 1 => +ive => moving right Case 2: 3t between (.841, 5.442), try 3t = 3(t = 1) => 3cos(3t) - 2 = 3cos3 - 2 cos3 is close to cos(3.14), i.e. -1 so value is negative, moving left. Case 3: 3t between (5.442,7.124), try 3t = 6.283 = 2.pi =...
Post a Comment
Read more

week 2: number theory practice questions pending

remainders/residues/residue classes Q1. If a = b mod m and c = d mod m, then: Prove that: (1) a +- c = b +- d mod m (2) ac = bd mod m (3) ax + cy = bx + dy mod m Q2. if a = b mod m, pr. th. a^n = b^n mod m. Q3. If a = b mod m then a = b mod d if d | m. Q4.  Remainder of 13^73 + 14^3 mod 11. Q5. Pr. th. ax = ay mod m iff x = y mod (m/gcd(a,m)) S5. m = k.q1, a = k.q2 where k = gcd(a,m) m/k = q1 x = y mod q1 => ax = ay mod q1 => ax = ay mod m since q1| m. ----------------------------------------------------------------
Post a Comment
Read more

week 1 practice problems pending from Day 4 Q3

Image
Day 1: Q1. The repeating decimals x = 0.ababab... and y = 0.abcabc.... satisfy x + y = 33/37 Find the 3 digit number abc. S1. (10a+b)(1/100 + 1/10000...) = (10a+b)/100[1/1-1/100] = (10a + b)/99 (100a + 10b +c)(1/1000 + 1/1000,000....) = (100a+10b+c)/999 (10a + b)/99 + (100a+10b+c)/999 = 33/37 => 999(10a+b) + (100a+10b+c).99 = 33.99.999/37 = 27.33.99 => 111(10a+b) + (100a+10b+c).11 = 99^2 = 9801 => 2210a + 221b + 11c = 9801______[1] => a <= 4 2210*4 = 8840 => 221b + 11c = 961 b = 4 => 11c = 961 - 884 = 77 => c = 7 abc = 447 Also in [1] if we take mod 11 then: 2210a + 221b = 0 mod 11 221(-a +b) = 0 mod 11 => a = b mod 11 Since a,b are single digit => a = b 221a.11 + 11c = 9801 => 221a + c = 891 a = 4 => c = 7 Q2. Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. S2. Let's do step by step. Subset size: 3 => There are 10 ways. Total: 10 Size: 4 => 10 ways to pick 4...
Post a Comment
Read more

practice problems

Q1. if p | ab then either p divides a or p divides b. 'p' is a prime. S1. Case 1: p | a: done Case 2: p doesn't divide a: gcd(a,p) = 1 => au + pv = 1 Multiply by 'b': bau + bpv = b p divides both terms on LHS => p | RHS => p | b H.P. Corollary: If p | a1a2....an then p divides some ai. Q2. Theorem: A composite number has at least one prime divisor. Prove this. S2. A set of all its +ve divisors except 1 and itself. Let (S) be the set. As the no. must have some other divisors than 1 & itself, then (S) is non-empty and S is subset of N(natural numbers). So by well ordering property (S) has a least element (l). If (l) is a prime then we are done. But if (l) is not a prime then there exists d | l (d divides l). C is the composite number we are talking about. d | l and l | C => d | C => d is in S and d < l which is a contradiction. Q3. find all possible values of 'p' s.t. (p) and (p^2 + 8) both are prime. S3. Case 1: p mod 3 = 0 => p can ...
Post a Comment
Read more