Question 398063
Given that {{{ sqrt( 6.4*10^n ) }}} is a rational number, what can you say about the value of n?
<pre><font face = "batangche" color = "indigo" size = 4><b>

{{{sqrt(6.4*10^n)}}}
{{{(6.4*10^n)^(1/2)}}}

{{{(expr(64/10)*10^n)^(1/2)}}}

{{{(64*10^(n-1))^(1/2)}}}

{{{(8^2*10^(n-1))^(1/2)}}}

{{{drawing(100,80,0,1,-1,1,locate(0,0,8*10^((n-1)/2))))}}}

Only integer powers of 10 are rational. So the exponent
must be an integer, say k

{{{(n-1)/2=k}}}

n-1 = 2k

n = 2k+1

Therefore n must be odd in order that {{{sqrt(6.4*10^n)}}}
be rational.
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What if both {{{ sqrt( 6.4*10^n ) }}} and {{{ root(3, 6.4*10^n ) }}} are rational? What can you say about the value of n?
<pre><font face = "batangche" color = "indigo" size = 4><b>
For

{{{root(3,6.4*10^n)}}}
{{{(6.4*10^n)^(1/3)}}}

{{{(expr(64/10)*10^n)^(1/3)}}}

{{{(64*10^(n-1))^(1/3)}}}

{{{(4^3*10^(n-1))^(1/3)}}}

{{{drawing(100,80,0,1,-1,1,locate(0,0,4*10^((n-1)/3))))}}}

Only integer powers of 10 are rational. So the exponent
must be an integer, say p

{{{(n-1)/3=p}}}

n-1 = 3p

n = 3p+1

Therefore n must be 1 more than a multiple of 3

Now we set the exponents equal:

n = 3p+1 = 2k+1

3p = 2k

2p+p = 2k

p + {{{p/2}}} = 2k

{{{p/2}}} = 2k-p

The right side is an integer, so the left side must be also

Let that integer be A

Then 

{{{p/2}}} = A 

p = 2A

Substitute in

3p = 2k

3(2A) = 2k

6A = 2k

3A = k

Now supstitute in

n = 3p+1 = 2k+1

n = 3(2A)+1 = 2(3A)+1

n = 6A+1 = 6A+1 

So n must be 1 more than a multiple of 6.

Edwin</pre>