Question

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?

Answer 1

8/5, 1.6

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.

Answer 2

Possible answer:
`\displaystyle c = (16)/(10) = (8)/(5) = 1.6`.

Explanation:

Rewrite the bounds of
`c` as fractions:

The simplest fraction for
`1.5` is
`\displaystyle (3)/(2)`. Write the upper bound
`2` as a fraction with the same denominator:

`\displaystyle 2 = 2 * 1 = 2 * (2)/(2) = (4)/(2)`.

Hence the range for
`c` would be:

`\displaystyle (3)/(2) < c < (4)/(2)`.

If the denominator of
`c` is also
`2`, then the range for its numerator (call it
`p`) would be
`3 < p < 4`. 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
`2`.

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.)

`\displaystyle (3)/(2) = (2 * 3)/(2 * 2) = (6)/(4)`.

`\displaystyle (4)/(2) = (2* 4)/(2 * 2) = (8)/(4)`.

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

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

`\displaystyle (3)/(2) = (3 * 3)/(3 * 2) = (9)/(6)`.

`\displaystyle (4)/(2) = (3* 4)/(3 * 2) = (12)/(6)`.

`\displaystyle (3)/(2) = (4 * 3)/(4 * 2) = (12)/(8)`.

`\displaystyle (4)/(2) = (4* 4)/(4 * 2) = (16)/(8)`.

`\displaystyle (3)/(2) = (5 * 3)/(5 * 2) = (15)/(10)`.

`\displaystyle (4)/(2) = (5* 4)/(5 * 2) = (20)/(10)`.

It is important to note that some expressions for
`c` can be simplified. For example,
`\displaystyle (16)/(10) = (2 * 8)/(2 * 5) = (8)/(5)` because of the common factor
`2`.

Apparently
`\displaystyle c = (16)/(10) = (8)/(5)` works.
`c = 1.6` while
`\displaystyle (1)/(c) = (5)/(8) = 0.625`.