Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

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.

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 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.

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 5 x 4 = 20), so there will be 6 significant digits to the right of the decimal point.