I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction with one digit for the numerator and one digit for the denominator. $\bullet$ Both $c$ and $1/c$ can be written as finite (non-repeating) decimals. What is this mystery number?

The simplest fraction for [tex]1.5[/tex] is [tex]\displaystyle \frac{3}{2}[/tex]. Write the upper bound [tex]2[/tex] as a fraction with the same denominator:

[tex]\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}[/tex].

Hence the range for [tex]c[/tex] would be:

[tex]\displaystyle \frac{3}{2} < c < \frac{4}{2}[/tex].

If the denominator of [tex]c[/tex] is also [tex]2[/tex], then the range for its numerator (call it [tex]p[/tex]) would be [tex]3 < p < 4[/tex]. Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than [tex]2[/tex].

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

[tex]\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}[/tex].

[tex]\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}[/tex].

At this point, the difference between the numerators is now [tex]2[/tex]. That allows a number ([tex]7[/tex] in this case) to fit between the bounds. However, [tex]\displaystyle \frac{1}{c} = \frac{4}{7}[/tex] can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

[tex]\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}[/tex].

[tex]\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}[/tex].

[tex]\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}[/tex].

[tex]\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}[/tex].

[tex]\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}[/tex].

[tex]\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}[/tex].

It is important to note that some expressions for [tex]c[/tex] can be simplified. For example, [tex]\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}[/tex] because of the common factor [tex]2[/tex].

Apparently [tex]\displaystyle c = \frac{16}{10} = \frac{8}{5}[/tex] works. [tex]c = 1.6[/tex] while [tex]\displaystyle \frac{1}{c} = \frac{5}{8} = 0.625[/tex].

veronicachaille41 veronicachaille41 · Answer 2 · 2022-07-12T22:45:30+02:00

Answer:

8/5, 1.6

Step-by-step explanation:

Because "c" can be written as a finite decimal, we know it can be written as a fraction whose numerator is an integer and whose denominator is a power of 10. Thus, after simplification, the denominator must still be a divisor of some power of 10. That is, it must be factorable into 2 and 5.

Since this denominator is a single digit, our choices are 1, 2, 4, 5, or 8. We have the same options for the numerator, since we know 1/c also has a finite decimal.

From here we could just test all the possibilities to see if they're between 1.5 and 2 but with a little cleverness we can eliminate some of the remaining possibilities. If we don't use 5 as the numerator or denominator, then "c" is forced to be a power of 2 , so it can't be between 1.5 and 2.

So, we must use 5, and our only plausible choices are 5/2 (which is 2.5), 5/4 (which is 1.25), and 8/5 (which is 1.6 ). Of these, only "c" = 8/5 works.