practice problems pending
Q1. How many two digit numbers have exactly 4 positive factors? (Here 1 and the number (n) are also considered as factors of (n).) S1. Only 2 possibilities: 1. p^3 where p is prime 2. p*q where p,q are prime. p^3 only one 3^3: 1 p*q we can construct 2*(5 to 47 primes): 13 3*(5 to 31): 9 5*(7 to 19): 5 7*(11 to 13): 2 Total: 30 = answer. Q2. Find the number of pairs ((a,b)) of natural numbers such that (b) is a 3-digit number, (a+1) divides (b-1), and (b) divides (a^2+a+2). S2. b-1 = k(a+1) b = k(a+1) + 1 k(a+1) + 1 divides a^2 + a +2 => ak + k + 1 divides k(a^2 + a +2) = 2k + ak + a^2.k = 2k + a(k + ak + 1 - 1) = 2k - a + a(k + ak + 1) => ak + k + 1 divides 2k - a Now 3 cases from here: 2k - a = 0 => a = 2k 2k - a > 0 => ak + k + 1 < 2k - a => a(k+1) -k + 1 < 0 => a(k+1) <= k-1 => a <= (k-1)/(k+1) which is not possible since a > 1 and RHS is < 1. 2k - a < 0 => ak + k + 1 < a - 2k => a(k-1) + 3k + 1 < 0 which is not possible ...