Problems on Numbers - General Questions

Let the numbers be x and x + 2.

Then, (x + 2)² - x² = 84

 4x + 4 = 84

 4x = 80

 x = 20.

 The required sum = x + (x + 2) = 2x + 2 = 42.

Let the numbers be x and y.

Then, x + y = 25 and x - y = 13.

4xy = (x + y)² - (x- y)²

   = (25)² - (13)²

   = (625 - 169)

   = 456

 xy = 114.

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352.

Since the number increases on reversing the digits, so the hundred's digits is smaller than the unit's digit.

Hence, required number = 253.

Let the numbers be x and y.

Then, xy = 120 and x² + y² = 289.

 (x + y)² = x² + y² + 2xy = 289 + (2 x 120) = 529

 x + y = √529 = 23.

Let the numbers be x and y.

 y² = 625.

 y = 25.

 x = 15y = (15 x 25) = 375.

 Sum of the numbers = x + y = 375 + 25 = 400.

Let the number be x.

 x² + 17x - 60 = 0

 (x + 20)(x - 3) = 0

 x = 3.

Let the ten's digit be x.

Then, unit's digit = x + 2.

Number = 10x + (x + 2) = 11x + 2.

Sum of digits = x + (x + 2) = 2x + 2.

 (11x + 2)(2x + 2) = 144

 22x² + 26x - 140 = 0

 11x² + 13x - 70 = 0

 (x - 2)(11x + 35) = 0

 x = 2.

Hence, required number = 11x + 2 = 24.

Let the ten's digit be x and unit's digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

 (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.