Mathematical Discussions

QUESTION: Find all pairs of natural numbers (a,b) such that: 1/a + 1/b =3/2018 SOLUTION: We start solving the equation: 1/a + 1/b =3/2018 To make it a linear equation, we multiply both the sides of the equation by 2018ab to get      2018a+2018b=3ab or, 3ab–2018a–2018b=0 To factorise the LHS, we multiply both sides of the equation by 3 to get      3.3ab–3.2018a–3.2018b=0 Adding both sides by 2018² and rearranging, we have      (3a)(3b)–(3a)2018–2018(3b)+2018²=2018² or, 3a(3b–2018)–2018(3b–2018)=(2×1009)² or, (3a–2018)(3b–2018)=2²×1009², where 2 and 1009 are prime numbers. Case1: 3a–2018=1 and 3b–2018=2018² Solving them,  we get (a,b)=(673,673×2018) Note that second equation is solved by taking 2018 in RHS Case2: 3a–2018=2 and 3b–2018=2×1009² Solving this, we get a=2020/3, which is not a natural number.  Case3: 3a–2018=2² and 3b–2018=1009² Solving them, we get (a,b)=(674,337×1009) Note that second equation is solved by writing 2018 as...

Question: Prove that there are no integers a, b and c for which a²+b²–8c=6 Detailed solution: a²+b²–8c=6 gives a²+b²=6+8c. Since, the RHS is even, so LHS is even, which means a²+b² is even. So we get two cases in which the first one is that a² and b² both are even and in the second one, a² and b² both are odd. Case 1: a² and b² are both even. This gives that a and b both are even. So, a=2p and b=2q for some integers p and q. Substituting the values of a and b in the equation a²+b²–8c=6, we get (2p)²+(2q)²–8c=6 or, 4p²+4q²–8c=6 or, 2p²+2q²–4c=3 (Dividing both sides by 2) Clearly, the LHS is even and RHS is odd, which is not not possible. So we get a contradiction in this case. Case 2: a² and b² both are odd. This gives that a and b are both odd. So, a=2m+1 and b=2n+1 for some integers m and n. Substituting the values of a and b in the equation a²+b²–8c=6, we get (2m+1)²+(2n+1)²–8c=6 or, 4m²+4m+1+4n²+4n+1–8c=6 or, 4m(m+1)+4n(n+1)=8c+4 or, m(m+1)+n(n+1)=2c+1   ....(1) (Dividing b...

Question: If a,b,c and d are positive integers such that abcd=1, then show that (1+a)(1+b)(1+c)(1+d)≥16 Detailed solution: To solve the question, you just need basic knowledge of AM-GM inequality. Using the AM-GM inequality for the terms of LHS inside brackets, we have (1+a)/2≥√(1×a), which gives (1+a)≥2√a    …….(1) (1+b)/2≥√(1×b), which gives (1+b)≥2√b    …….(2) (1+c)/2≥√(1×c), which gives (1+c)≥2√c   …….(3) (1+d)/2≥√(1×d), which gives (1+d)≥2√d    …….(4) Multiplying equations (1),(2),(3) and(4), we have (1+a)(1+b)(1+c)(1+d)≥16√(abcd) or,(1+a)(1+b)(1+c)(1+d)≥16 [Since abcd=1] Hence proved

Question: Find all two-digit positive integers such that the sum of the number and the number obtained by reversing the order of digits is a perfect square. Detailed solution: Let xy be any such two-digit number (Note: Here xy is just the form of a two-digit number like 25,43,etc., where x and y are digits) Therefore, xy can be written as 10x+y (in expanded form). Number obtained by reversing the order of digits of xy is yx=10y+x According to question, The sum of xy and the number obtained by reversing the order of digits of yx is a perfect square, which means (10x+y)+(10y+x)=Perfect square or, 11x+11y=Perfect square or, 11(x+y)=Perfect square .....(1) But, maximum value of x and y are 9 each, so maximum value of x+y is 9+9=18. Observe that, to satisfy equation (1), x+y=11k², for some positive integer k. For k=1,x+y=11 and for k>1 i.e., for k≥2, x+y≥44, which is not possible. Therefore, we get that x+y=11 is the only choice to satisfy equation (1). Hence, (x,y)=(2,9),(3,8),(4,7),(5,...