Problems on Numbers - General Questions (2)

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 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.

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.

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 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 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