If $\frac ab$ rounded to the nearest trillionth is $0.008012018027$, where $a$ and $b$ are positive integers, what is the smallest possible value of $a+b$?
I don't see any strategies here for solving this problem, any help? Thanks in advance!
$\begingroup$
$\endgroup$
0
Add a comment
|
$\begingroup$
$\endgroup$
2
The continued fraction representations of the limits of the interval are $$ 0.0080120180265 = [0; 124, 1, 4, 2, 1, 463872, 1, 1, 12, 1, 1, 41] \\ 0.0080120180275 = [0; 124, 1, 4, 3, 545777, 2, 13, 1, 1, 1, 1, 2] $$
The simplest continued fraction (and therefore also the simplest ordinary fraction!) in that interval is $$ [0;124,1,4,3] = \frac{16}{1997} = 0.00801201802704056084\ldots $$ and the sum of its numerator and denominator is $2013$.
(I used Wolfram Alpha to expand the continued fractions fully. For a pencil-and-paper solution one only needs to carry out the expansion until they start differing, which requires only a handful of long divisions with remainder.)
answered Aug 22 '16 at 20:09
hmakholm left over Monicahmakholm left over Monica
268k21 gold badges373 silver badges626 bronze badges
-
1$\begingroup$ I wonder if this is a contest-math problem from a few years ago ... $\endgroup$ – hmakholm left over Monica Aug 22 '16 at 20:13
-
$\begingroup$ I don't think so...searched the web and there was no results. Anyway, thanks for the answer, Henning Makholm! $\endgroup$ – user359548 Aug 22 '16 at 20:16
$\begingroup$
$\endgroup$
2
$\frac 1{0.008012018027} = 124.8125000006\\ 0.8125 = \frac {13}{16}\\ 124.8125 = \frac {1997}{16}\\ a+b = 1997+16=2013$
-
1
-
$\begingroup$ If I could figure out how to do it with just a picture, no numbers and no words, then I would. $\endgroup$ – Doug M Aug 23 '16 at 15:13
$\begingroup$
$\endgroup$
Of course, the question is not exciting; that would be interesting is the following:
I give a decimal expansion with $3k$ or $4k$ digits of a rational $\dfrac{a}{b}\in (0,1)$ where $b<10^k$. Find this rational. Solution: convert the decimal expansion in a continued fraction and write its convergents; stop writing the previous sequence when you notice a large increase of the denominator.
Example. You want to calculate $z=\dfrac{5783}{6612}+\dfrac{7037}{8004}$. With $14$ significant digits, $z\approx 1.7538073068729$. Its convergents are: $[1, 2, 7/4, 114/65, 691/394, 21535/12279, 22226/12673,6701582829/3821162566]$.
Thus $z=22226/12673$.
EDIT. I forgot the required answer. The convergents of $u=0.008012018027$ are
$[0, 1/124, 1/125, 5/624, 11/1373, 16/1997, 98913787/12345677040]$.
Thus the solution is $16/1997$.
