Let the number be x.

Then,	1	of	1	of x = 15      x = 15 x 3 x 4 = 180.
	3		4

So, required number =		3	x 180		= 54.
		10

Let the three integers be x, x + 2 and x + 4.

Then, 3x = 2(x + 4) + 3      x = 11.

 Third integer = x + 4 = 15.

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

Then, (10x + y) - (10y + x) = 36

 9(x - y) = 36

 x - y = 4.

Since the number is greater than the number obtained on reversing the digits, so the ten's digit is greater than the unit's digit.

Let ten's and unit's digits be 2x and x respectively.

Then, (10 x 2x + x) - (10x + 2x) = 36

 9x = 36

 x = 4.

 Required difference = (2x + x) - (2x - x) = 2x = 8.

Then,		10x +	8		+ 18 = 10 x	8	+ x
			x			x

 10x² + 8 + 18x = 80 + x²

 9x² + 18x - 72 = 0

 x² + 2x - 8 = 0

 (x + 4)(x - 2) = 0

 x = 2.

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

Then, x + y = 15 and x - y = 3   or   y - x = 3.

Solving x + y = 15   and   x - y = 3, we get: x = 9, y = 6.

Solving x + y = 15   and   y - x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

Let the numbers be a, b and c.

Then, a² + b² + c² = 138 and (ab + bc + ca) = 131.

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca) = 138 + 2 x 131 = 400.

 (a + b + c) = √400 = 20.

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.

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 number be x.

 x² + 17x - 60 = 0

 (x + 20)(x - 3) = 0

 x = 3.

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 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 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, 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 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.