Approximate while you can

We can solve this question with continued fractions. The following algorithm can be used to find continued fraction approximations.

Take the integer part of the number. This is the first term in the sequence. Then take the reciprocal of the fractional part. Repeat indefinitely, appending terms to the continued fraction as necessary.

You can read more about continued fractions here .

---

$\begin{array}{|c|c|c|c|} \hline n_k & \lfloor n_k\rfloor & n_k-\lfloor n_k \rfloor & n_{k+1}=\frac{1}{n_k-\lfloor n_k\rfloor}\\ \hline \sqrt{14} & 3 & \sqrt{14}-3 & \frac{\sqrt{14}+3}{5} \\ \hline \frac{\sqrt{14}+3}{5} & 1 & \frac{\sqrt{14}-2}{5} & \frac{\sqrt{14}+2}{2} \\ \hline \frac{\sqrt{14}+2}{2} & 2 & \frac{\sqrt{14}-2}{2} & \frac{\sqrt{14}+2}{5} \\ \hline \frac{\sqrt{14}+2}{5} & 1 & \frac{\sqrt{14}-3}{5} & \sqrt{14}+3 \\ \hline \sqrt{14}+3 & 6 & \sqrt{14}-3 & \frac{\sqrt{14}+3}{5} \\ \hline \end{array}$

Note the repetition of $\frac{\sqrt{14}+3}{5}$ here. The terms will repeat, and the integer part of $n_k$ will repeat according to the pattern $\{1,2,1,6\}$ after the initial $3.$ Thus, $\sqrt{14}=[3;\overline{1,2,1,6}].$ We can use this to approximate $\sqrt{14}$ with arbitrary precision. The first few approximations are as follows: $\frac{3}{1},\frac{4}{1},\frac{11}{3},\frac{15}{4},\frac{101}{27},\frac{116}{31},\frac{333}{89}.$ $\frac{116}{31}$ is the closest approximation that still has a numerator and denominator sum to less than $150,$ so the answer is $\boxed{31}.$

To find rational approximation of $\sqrt{14}$ , we first find the rational number closest to fractional part of $\sqrt{14}$ or 0.7416573. . . or some ratio slightly less than $0.742 = \frac{742}{1000} = \frac{371}{500}.$ Obviously accepting the ratio $\frac{371}{500}$ would violate the given restriction on total of (m + n). Let us tackle this problem by finding a solution for $371x - 500y = 1$ by well known techniques which works out to $x = 31, y = 23$ , meaning that $\frac{23}{31}$ is nearest approximation of fractional part. The answer is 31.

$Using~ a~ calculator,~~ \dfrac{x-n} n- \sqrt{14}=N.\\ Since~\dfrac{120}{30}=4, ~we~start~our~trials~with~biggest~possible~x~for~best~result.\\ x=149,~~n=31, ~N=.06479.~We~are~sure~that~this~is~the~best~result~since~there ~is~a~sign~change~of~N~from~n=30~to~n=32.\\ x=148,~~n=31, ~N=.0325....~We~are~sure~that~this~is~the~best~result~since~there ~is~a~sign~change~of~N~from~n=30~to~n=32.\\ x=147,~~n=31, ~N=.0027809.~We~are~sure~that~this~is~the~best~result~since~there ~is~a~sign~change~of~N~from~n=30~to~n=32.\\ x=146,~~n=31, ~N=.03....~We~are~sure~that~this~is~the~best~result~since~there ~is~a~sign~change~of~N~from~n=30~to~n=32.\\ It~ is~ clear~ that~ x<146~~ will~not~give~better~results.\\ So~n=\color{#D61F06}{31}.$