The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?

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 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 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 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 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.