"1) Solution:Let n be a positive integer,By division algorithm there exists a positive integer qand r such that n = 6 q + r 0<=r<=6Put r = 0,1,2,3,4,5 So that we get n = 6q,6q+1,6q+2,6q+3,6q+4,6q+5,Out of these 6q,6q+2,6q+3, 6q+4 are compositenumbers corresponding to any integral value of qThe prime numbers can be represented only by 6q+1and 6q+5 Let us assume the contrary that there exist only finitelymany primes of the form 6 m + 5Let them be q1q2……………………..qsConsider the positive integer N=6q1q2…………..qs-1 = 6 (q1q2………..qs-1)+5and let N = r1r2………rt be its prime factorization since N is an odd integer, we have rx ?2 for all x so that rk is either of the form 6 m +1, 6m + 5 The product of any number of prime of the form 6m + 1,is again an integer of this typeFor N to take the form of 6 m + 5, N must contain at least one prime factor ri of the form 6m + 5 But ri cannot be found among the listing ofq1q1……………….qsThis will lead to the contradiction that ri /1Hence the only possible conclusion is that there areinfinitely many primes of the form 6 m + 5. Hence the proof 2)Solution : The sum of a series of consecutive integers is the productof the number of terms and the "middle value" (the middleterm, or the average of the two middle terms). In order forthis product to be a power of 2, then both the middle valueand the number of terms must be a power of 2. But if thereare at least two terms and the number of terms is a power of2 then there are an even number of terms, so the middlevalue is not even an integer much less a power of 2.If z is not a power of 2 then it has an odd prime factor, n.Let n be the number of terms, and let z/n be the middlevalue. So any number z, that is not a power of 2 can beexpressed as a sum of consecutive integers.3) Solution : It is given that ad – bc = ± 1Let x’ = ax + dy and y’ = cx+ by -----------(1) Where a, b, c, d are positive integersSolving (1) x = (dx’ – b y’) / (ad – bc)And y = (cx’- ay’) / (ad – bc)It is given that ad – bc = ± 1We know that this is the necessary and sufficientcondition that the transformation (1) should transform? into itself is that ? = ± 1Where ? = ad – bc Here ?/a, ?/b, ?/ c and ?/ d And ?^ 2 /a+b and ?^ 2 /c+d -----(1) It is given that ? = ± 1Hence ? ^2 = 1From (1) 1/(a+ b) and 1 / (c + d)Hence gcd (a+ b, c+ d) = 1Hence the proof. 4) solution : Here it is given that n = 2 ^ ?(3)Here we have to use the following theoremFor every positive integer n, there exists a prime suchthat n < p,=2nProof:every prime which dividesmust be ? 2n/3 let p be such prime suppose since / ?where it follows that thus if e >= 2 then p <= ?2nso there are at most ?2n primes appear in the primepower factorization of which exponents larger than 1 and in each case hence <= (2n) ^ ? 2n ? p butis the largest of 2n + 1 terms in the expansionof (1 + 1) ^ 2ntherefore 4^ n < (2n+ 1) so 4^ n / (2n + 1) (2n) ^ ? 2n ? p4^ n / (2n + 1) (2n) ^ ? 2n 4 ^ (2n/ 3)or 4 ^(n/3)< (2n) ^ ? 2n+ 2 taking log on both sides n log 4 / 3 < ( ? 2n+ 2) log 2 this is false for > 512 and so there is a prime between n and 2n for n > 512each number is smaller than twice the one preceding it,so there is also such a prime for all n < = 512and hencefor all n >= 1from the above theorem 3< 2^ ? <=2^ ?(3)Taking log on both sides we get log {3/2^ ?<1<=3}Log (3) – log (2^ ?)< log 1 = 0.4771- ? (log 2) <0.4771- ?(.3010)?= .4771/.3010=1.5855)Solution :It is given that the pair (a,b) is amicable and that m is evenand n is odd Two numbers are said to be amicable numbers if one isequal to the sum of the proper divisors of the otherThus if s(n) us the sum of the proper divisors of n then nand m are amicable if s (n) = m and s (m) = n Now s (n) = ? (n) – n and s (m) = ? (m) – mm= ? (n) – n and n= ? (m) – m Therefore ? (n) = m + n and ? (m) = m + nLet n = p1^a1 p2^a2…………pk^ak be the primedecomposion of n .If n is a perfect square then each of the powersa1,a2,………ak is evenTherefore each a1+1, a2+1 ……ak+1 is oddTherefore (a1+1)(a2+1)…(ak+1) is odd. Conversly if (a1+1)(a2+1)…(ak+1) is odd.Therefore each a1,a2,………ak are eventTherefore n = p1^a1 p2^a2…………pk^ak is a perfectsquare Hence the proof6) Solution :If ?? (d) ?(d) = n then ? (n) = ? and n = ? prove d/n Solution :If n = p1 ^ k1p2^ k2………pr^kr is the standardfactorization of the integer n then ?(n) = r is the numberof distinct prime divisors in n We know that the mobius function ? (n) is defined as{ 0 if p ^ 2 /n where p is a prime? (n) =? 1 if n = 1{ (-1) r if n = p1p2p3…………….prwhere pi’s are the distinct primesLet p1,p2,.pk be the distinct prime factors ofn n Now the divisors of p1 are p1 and 1? ? (d) ?(d) =?(1) ?(1) + ?(p1)?(p1)d/p1 we know that ?(1) = 1 from the definition of ?.And ?(1)= p1^2 – 1 /(p1 – 1) = (p1+1) (p1-1) / (p1-1) = (p1+ 1) ((p1-1) gets cancelled)and the second term is ?(p1) ?(p1)= (-1) ^ 1(p2+1)Hence If ? ? (d) ?(d) =1 (p1+1) + (- 1) (p 2 + 1) d/n…………………..(- 1) ^ r (pr + 1)Finally we get (- 1) ^ r p1.p2.p3…………..prThen ?(n) = rAnd n = (- 1) ^ r p1.p2.p3…………..prHence the proof7) Solution:We have to calculate M (n) = ??(d) n/d d/n We know that? is an arithmetic function and ? ?(d) = nd/n By Mobius inversion formula if F (n) = ? ?(d)d/nthen?(n) = ? ? (d) F (n / d)d/n Hecne F(n) = n Therefore,?(n) = ?? (d) (n / d)d/n Hence the value of M(n) =?(n) = ?? (d) (n / d)d/n 8) Suppose that A and B are the lattice point (a, b)and (c,d) Here it is given that A and B are visible from each otherif there is no other point of I on which the line segmentjoining A and BIt is given that gcd (a-b, c-d) >1Let gcd (a-b,c-d) >1It is given that d >1, So d has prime factor say PTherefore p / d Hence p / a ,p/b, p/c and p/d Hence p^2 / a-b and p^2 / c-dSince gcd (a-b, c-d) >1-----------(1)Let x’ = ax + dy and y’ = cx+ by -----------(2) Where a, b, c, d are positive integersSolving (2) x = (dx’ – b y’) / (ad – bc)And y = (cx’- ay’) / (ad – bc)From (1) and (2) Hence (x, y) >1?1We know the general result that a point A of I is visibleif there is no point of I on OA between A and BIn order that (x, y) should be visible it is necessary andsufficient that x/y should be in its lowest terms or (x, y) = 1 b) There is at least one such point C, if D is the fourthvertex of the parallelogram and if there exists the fifthpoint E and CE is parallel and equal to OD but with theopposite sense(Since the properties of lattices are symmetrical andindependent of a particular lattice point chosen as theorigin) Hence E is also the point of I and there are at least 2points of I inside I unless E coincides with DHence E coincides with D Hence there cannot be 5 pointsHence the proof.9) It is given that n= p1^ a1 p2^a2……pk^ak and that?(n) /? (n) = n/2 We know that ?(n)= ?d------(1) d/n and? (n) = ?1------------------(2) d/nit is given that ?(n) /? (n) = n/2 From (1) and (2) we get (1) / (2) = n/2?d / ?1 = n / 2 cross multiplying we getn ( ?1) = 2 ( ?d)d/n d/nHence n (number of divisors) = 2 (sum of the divisors)Hence n^2 = 2 (k) Hence the value of k = 2 10)Solution :Let A = (n !) !)/((n!) ^ (n – 1) ! (1.2.3.4…………….n)!/ (n!) ^ ((n – 1) ! 1! 2 ! 3! …n!/(n !) ^ (n – 1) !{1! (1! 2)(2! . 3)………(n – 1) ! n} /{n!^ (n – 1) !Taking log on both sides we getLog{ (n – 1) ! n / n! ^ (n – 1) ! }Log (n-1)! + log n / (n-1) ! log n!By one theorem we know that if a1+ a2+………ar.= n Then n! / a1! a2!…………ar!.ia an integerHence A is an integerSince A = (n !) !)/((n!) ^ (n – 1) ! (n !) ^ (n – 1) ! / (n !) !Hence the proof.11)Solution :It is given that 3 ^ m – 2 ^ n = 1----------(1) An integer x is said to be a solution of the congruence ifaxi ? b (mod m) that is m/axi-b The given problem can be solved by 3^m ?2^n (mod 1)If m= 1 and n = 1 then (1) satisfiesIt is given that m is even and n is odd Let m = 2 and since n is odd n = 3Then 3^2 – 2^3=9 – 8 = 1If m= 2 and n = 5 this does not holds If m = 4 and n = 3Then 3 ^ 4 –2^3 = 81-8?1Since n is an odd none of the odd power of 2 gives 80 (Since 81-1 = 80)Similarly proceeding in the same way the only solutionto 3 ^ m – 2 ^ n = 1 is (1, 1) and (2, 3) Hence the result12) Solution: To prove this problem we are using the following theorem(If m1, m2 ,………..m k are co prime then the system x ?c1(mod m1), x ? c2 (mod m2) ……………………. and x ?ck (mod mk) has a unique solGiven by x = n1M1 c1 + n2M2c2+………………..nkMkck)If the day in the question is the x-th (counting from andincluding the first Monday) Then x = 1 + 2 k 1 = 2 + 3 k 2=3 + 4 k 3 = 4+ k4 = 5 + 6 k5 = 6 + 5 k 6 = 7 k 7Where the k are the integersThat is 1) x ? 1 (mod 2) 2) x ? 2 (mod 3)3) x ? 3 (mod 4)4) x ? 4 (mod 1)5) x ? 5 (mod 6)6) x ? 6 (mod 5)7) x ? 0 (mod 7)Among the above congruence (4) is of no restriction and(1) and (2) are included in (3) and (5) Of the latter two (3)shows that x is congruent to 3, 7 or 11 mod (12)and (5) thatx is congruent to 5 or 11 So that (3) and (5) together are equivalent to x ? 11 (mod 12) Hence the problems is that of solving x ? 11 (mod 12), x ? 6 (mod 5) and x ? 0 (mod 7)or x ? -1 (mod 12), x ? 1 (mod 5) and x ? 0 (mod 7)Here m1 = 12, m2 = 5 and m3 = 7 M = 420M1 = 35, M2 = 84, M3 = 60 Then n are given by 35n1 ? 1 (mod 12),84n2 ? 1 (mod 5) and 60n3 ? 1 (mod 7) Or -n1 ? 1 (mod 12), -n2 ? 1(mod 5) and 4n3 ? 1 (mod 7)We can take n1 = -1, n2 = - 1, n 3 = 2Hence x=(–1) (-1) (35) + (-1) (1) (84) +2 (0) (60) = - 49 ? 371 (mod 420)The first x satisfying the condition is 371.13) solution :We know that ? (n)= ? (p1^ a1p2^a2………..pk^ak)= ? (p1^a1) ? (p2^a2) ………….. ? (pk^ak)= p1^a1((1-(1/p1) p2^a2((1-(1/p2)………………………..pk^ak((1-(1/pk)p1^a1p2^a2………pk^ak ((1-(1/p1) ((1-(1/p2)…………………..((1-(1/pk) = n((1-(1/p1)((1-(1/p2)………..((1-(1/pk)--------(1) Here each term ((1-(1/pi) <1Hence ?(n) < n……………….(2)We have to find out ? (n ^ 2) + ?((n+1) ^ 2) From (1) ? (n ^ 2)= n^2 ((1-(1/p1) ((1-(1/p2)………… ((1-(1/pk)………………..(3) ?((n+1) ^ 2)= (n+1)^2 ((1-(1/p1) ((1-(1/p2)………… ((1-(1/pk)………………..(4) Adding (3) + (4) gives? (n ^ 2) + ?((n+1) ^ 2) < n^ 2 + (n + 1) ^ 2 { from (2)< n ^ 2 +n^ 2 + 2n + 1< 2 n ^ 2 + 2n + 1 it is given that n > 2 hence n ^ 2 > 4therefore n^ 2 > 1 or 1 < n ^ 2 ?1<(n ^ 2)/2? (n ^ 2) + ?((n+1) ^ 2) < = 2n^ 2 + 2 n + n ^ 2= (3 n ^ 2 + 2n) /2Hence the proof.14) Prove that ??(d) ? (n / d) = n? (n)d/nWe know that ?? (n / d) = n? ?(d) = ? ?(pi^ki+1)-1/(pi-1)= =?( ?i + 1)=? (n) d/n d/n 1<=i<=rand? (n) =?( ?i + 1) 1<=i<=rHence ? ?(d) ? (n / d) = n? (n)b) We know that ? (n) = n ?1/d--------------(1) and ? (n) = n ? ?(d) / d------------------------(2) (1) + (2) gives ?(n) + ?(n) =n ?(1/d)+ n ? ?(d) / d= n?(n) 15) solution :We know that k ? - (p – k) mod p Substituting k as 2 we get2 ? - (p – 2) mod pSubstituting k as 4 we get4 ? - (p – 4) mod pSubstituting k as 6 we get6 ? - (p – 6) mod p …………………….Substituting k as (p – 1) we getp-1 ? - (p – (p – 1)) mod pWe get p-1 ? 1 mod pMultiplying all of the above we get 2.4.6.8…………..(p – 1) ?{(-1)^ (p –1) / 2} 1.3.5……………………………………………….(p – 2) (mod p)Multiplying by 1 . 3 . 5.(p – 2) on both sides we get1.2.3.4……………(p – 1) ?{(-1)^ (p –1) /2}(1^2)(3^2)………………….. (p – 2) ^ 2 (mod p)…………………………………………..(1)we know that (p – 1) ! = 1.2.3.4..(p – 1)Substituting the value of (p – 1) ! in (1) we get(p – 1) ! ? (- 1) ^ (p – 1) / 2 (1^2)(3^2)…………………..(p – 2) ^ 2 (mod p) ………………(3) By wilson’s theorem we know that (p – 1) ! ? - 1 (mpd p)………. …………..(2)usingf (2) in (3) we get 1^2.3^2.5^2……((p–2) ^ 2 ?(- 1) ^ (p + 1) / 2 (mod p)Hence the proof "