Question 1198341
Given that  {{{ a +  1/( b+2/c ) = 17/5  }}}, where a is the integer part of the mixed number, and c is the numerator of the fraction part of the mixed number when the mixed number is written in the least terms, find the product abc.

A) 9
B) 18
C) 6
D) 30
E) 3
<pre>             {{{highlight(a) + 1/(b + 2/c)}}}
{{{matrix(1,5, 17/5, "=", 3&2/5, or, highlight(3) + 2/5)}}}
            
We see clearly that a = 3 

In addition, since c is the numerator of the reduced fraction in the mixed number, then c = 2

Now, we need to change the 2 in the numerator in the mixed number on the right ({{{highlight(2)/5}}}) to <font color = red><font size = 4><b>MATCH</font></font></b>
the 1 in the numerator in the mixed number on the left {{{highlight(1)/(b + 2/c)}}}.
How do we achieve that? 
We simply divide the 2 in the numerator in the mixed number on the right by 2 to get 1. We then do the same to the 
denominator (divide it by 2), as follows: {{{matrix(1,7, 2/5, "=", (2/2)/(5/2), "=", 1/highlight((5/2)), "=", 1/highlight((b + 2/c)))}}}
                                       We now see that: {{{matrix(1,3, 5/2, "=", b + 2/c)}}} 
                                                        {{{matrix(1,3, 5/2, "=", b + 2/2)}}} ---- Substituting 2 for c
                                                        {{{matrix(1,3, 5/2, "=", (2b + 2)/2)}}}
                                                         5 = 2b + 2 --- Denominators are equal and so are the numerators
                                                        {{{matrix(2,3, 3, "=", 2b, 3/2, "=", b)}}}

                                      With {{{matrix(3,3, a, being, 3, b, being, 3/2, c, being, 2)}}}, we finally get: {{{highlight_green(matrix(1,7, highlight(abc), "as:", 3(3/2)(2), "=", highlight(9), "(CHOICE", "A)"))}}}

=======
<font color = red><font size = 4><b>CHECK</font></font></b>
=======
{{{matrix(6,3, a + 1/(b + 2/c), "=", 17/5, 3 + 1/(3/2 + 2/2), "=", 17/5, 3 + 1/(5/2), "=", 17/5, 3 + 1 * (2/5), "=", 17/5, 3 + 2/5, "=", 17/5, 17/5, "=", 17/5)}}}</pre>