Math IOQM

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 ...

 Q1. What is parametrization of line passing through (2,0) and (0,1)

Q1. 100 in base 10 needs to be converted to base 2,3,4,5,6,7,8,9. S2. Base 9: 100 = 81 + 18 + 1 = 121 Base 8: 100 = 64 + 4.8 + 4 = 144 Base 7: 100 = 98 + 2 = 202 Base 6: 100 = 72 + 24 + 4 = 244 Base 5: 100 = 25*4 = 400 Base 4: 100 = 64.1 + 16.2 + 4 = 1210 Base 3: 100 = 3^4.1 + 3^2.2 + 1 = 10201 Base 2: 100 = 2^6.1 + 2^5 + 2^2.1 = 1100100  Q2. S2. Let's try. Last digit is 1. Then 5+4 = 9 but it says 2. So it is not base 10. Looks like base 7. And since there is enough gap in the beginning of the first number, 2 numbers will fit there. Answer: 23451 + 15642. Q3. Let n be a positive integer and d be a digit such that the value of the numeral 32d ​ in base n equals 263, and the value of the numeral 324 ​ in base n equals the value of the numeral 11d1 in base six. What is n+d? S3. 3n^2 + 2n + d = 263 3n^2 + 2n + 4 = 216 + 36 + 6d + 1 Solving for d we get d = 2 And solving for n we get n = 9 and another non integer. Answer n + d = 11 Q4. Using the digits 1,2,3,4,5,6,7, and 9 , form ...

Q1. If x^n + y^n is a prime number for some x,y > 1 then n is necessarily: Options: Power of 2, Multiple of 4, Multiple of 3, Odd S1. Answer: Power of 2. Proof by contradiction: Assume that n has an odd factor k > 1. n = m.k x^n + y^n = (x^m)^k + (y^m)^k A^k + B^k = (A+B)(A^k-1 -A^k-2.B ....+ B^k-1) for odd 'k'. Here A = x^m and B = y^m x,y > 1 m>=1 => A+B >= 4 And the other factor is also more than 1 since A^k + B^k > A + B since k > 1. So A^k + B^k is not a prime. So n can't have any odd factor. So n has to be power of 2. H.P. For e.g. n = 2 x = 2, y = 3 4 + 9 = 13 Q2. A natural number has last digit 8 in base 16 representation. What will be the remainder, when it is divided by 8 , in decimal system S2. 0 Q3. If a prime can be written as sum of two perfect square, then the remainder when it is divided by 8 cannot be S3. Any square upon division by 8 will leave 0,1,4 as remainders. So adding 2 of them can only give 0,1,2,4,5 as remainders. And if th...

 Q1. Prove that for every positive integer (n), the number [ 3^{3^n}+1 ] is the product of at least (2n+1) (not necessarily distinct) primes. S1. Prove by induction. Q2. Prove that for every positive integer (n), there exists an (n)-digit number divisible by (5^n) whose all digits are odd. S2. Proof by induction: Let's assume it holds true for k  Now we construct 5 (n+1) digit numbers using A: All 5 of them leave different remainders modulo 5 so at least one of them will leave 0 and that would give us divisibility by 5^(n+1). Why will all be different modulo 5? Let's try to prove by contradiction: 2^n + a = 3.2^n + a mod 5 => 2.2^n = 0 mod 5 false. And each time you do this you will get some even number = 0 mod 5. Which is false. H.P. Q3. Let n be a positive integer. Let 0 < a1 < a2 ... an be real numbers. Prove that at least {n+1}C{2}) of the sums +-a1 +-a2 ... +-an are distinct. S3. First we will show that this problem maps to another simpler problem. And then solve...

Q1. Show that (1 - 1/2^2)(1 - 1/3^2)....(1 - 1/(n+1)^2) = n+2/2n+2 S1. Easy to show via induction. Other way via telescopic cancellations: Kth term is (k^2 - 1)/k^2 = (k-1)(k+1)/k^2 = (k-1)/k.(k+1)/k Now separate out k-1/k and k+1/k terms and multiply: 2-1/2 * 3-1/3 * 4-1/4 .... n/n+1 = 1/n+1 2+1/2 * 3+1/3 ... n+2/n+1 = n+2/2 Hence proved. Q2. Show that: S2. Short method: r*nCr = n*(n-1)C(r-1) and take sigma simply. Longer method:  2^(n-1) = (n-1)C0 + (n-1)C1... (n-1)C(n-1) (n-1)Ck *n = nCk*(n-k) How? n * (n-1)!/k!(n-1-k)! = n! /k!(n-k-1)! But (n-k-1)! = (n-k)!/(n-k) So n * (n-1)!/k!(n-1-k)! = (n-k)n! /k!(n-k)! = nCk*(n-k) So n*2^(n-1) = n *2^(n-1) = n*[(n-1)C0 + (n-1)C1... (n-1)C(n-2)+ (n-1)C(n-1)] = nC0 * (n-0) + nC1*(n-1) ... nC(n-2)*(2)+ nC(n-1)*(1) It can be rewritten as: = nCn*n + nC1 * (n-1) ... nC2 * 2 + nC1 * 1 = (r = 1 to n)Sigma(r*nCr) = (r = 0 to n)Sigma(r*nCr) Hence proved. Q3. S3. Quite simple. Let the given sum be S. 3S = 1.3^2 + 2.3^3... n.3^(n+1) S-3S = 3 + 3^2 ......