Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

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.

Let the ten's and unit digit be x and	8	respectively.
	x

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 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.

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.