Question

Find a positive number such that it's fractional part, Integer part, and itself are in Geometric Progression

Answer 1

Here is your answer



For a positive number x, {x} is the fractional part of x. {x} = x-[x] , where [x] is the greatest integer less than or equal to x. The fractional part of 12.8, for example is 0.8. What is the value of {z}, if z is positive and has exactly 4 digits after the decimal?

(1) {z/3} = 0.4175. This implies that the fractional part of z/3 is 0.4175, thus:

z3=integer+0.4175z3=integer+0.4175;

z=3∗integer+3∗0.4175z=3∗integer+3∗0.4175;

z=3∗integer+1.2525z=3∗integer+1.2525;

z=(3∗integer+1)+0.2525z=(3∗integer+1)+0.2525;

z=integer+0.2525z=integer+0.2525.

Therefore, {z}, the fractional part of z is 0.2525. Sufficient.

(2) {3z} = 0.7575. This implies that the fractional part of 3z is 0.7575, thus:

3z=integer+0.75753z=integer+0.7575;

z=integer3+0.2525z=integer3+0.2525;

z=(multiple of 3)3+0.2525z=(multiple of 3)3+0.2525;

z=integer+0.2525z=integer+0.2525.

Therefore, {z}, the fractional part of z is 0.2525. Sufficient.

Answer 2

Let the positive number be x.

And its integral part = m
its fractional part = n

So x = m+n

Now, it's given that its fractional part, Integer part, and itself are in Geometric Progression.

Or n, m and x are in GP. So we can write

$\frac{m}{n} = \frac{x}{m} \\ \\ \Rightarrow \frac{m}{n} = \frac{n + m}{m} \\ \\ \Rightarrow m^2 = n^2 + nm\\ \\ \Rightarrow n^2 + nm - m^2 = 0$

Now solve for n(fractional part)

$n = \frac{ - m \pm \sqrt{ {m}^{2} - 4 \times 1 \times {( - m)}^{2} } }{2 \times 1}$

$n = \frac{ - m \pm \sqrt{5 {m}^{2} } }{2} \\ \\ n = \frac{ - m \pm m\sqrt{5 } }{2}$

$n = ( \frac{ - 1 \pm \sqrt{5} }{2} )m$

Since n is less than 1 and m is a positive integer, we have to rule out n = (-1 -√5)/2 m

$thus \: \: n = ( \frac{ - 1 + \sqrt{5} }{2})m$

$for \: \: m = 1 \\ \\ n = \frac{ - 1 + \sqrt{5} }{2} \\ \\ x = n + m =\frac{ - 1 + \sqrt{5} }{2} + 1 = \frac{ 1 + \sqrt{5} }{2}\\ \\ The\ number\ is\ \frac{ 1 + \sqrt{5} }{2}$