The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

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.

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

Then, 3x = 2(x + 4) + 3   ⟺   x = 11.

∴ Third integer = x + 4 = 15.

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 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 number be x.

Then, x + 17 =	60
	x

⟹ x² + 17x - 60 = 0

⟹ (x + 20)(x - 3) = 0

⟹ x = 3.

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.

4.036	=	403.6	= 100.9
0.04		4

Given expression	= (0.96)3 - (0.1)3 (0.96)2 + (0.96 x 0.1) + (0.1)2
	= ❨ a3 - b3 ❩ a2 + ab + b2
	= (a - b) = (0.96 - 0.1) = 0.86

3.14 x 10⁶ = 3.14 x 1000000 = 3140000.