The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:

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

Then, (10x + y) - (10y + x) = 36

⟹ 9(x - y) = 36

⟹ x - y = 4.

Let the numbers be x and y.

Then, xy = 9375 and	x	= 15.
	y

xy	=	9375
(x/y)		15

⟹ y² = 625.

⟹ y = 25.

⟹ x = 15y = (15 x 25) = 375.

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

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 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 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