%solvers

This is not a general solution. I pretty much lucked out since you used $19$ .

Anyways, $\left\lfloor \frac{100N}{T} \right\rfloor = 19$

$\implies 19 \leq \frac{100N}{T} < 20$

$\implies \frac{19}{100} \leq \frac{N}{T} < \frac{20}{100}$

$\implies \frac{19}{100N} \leq \frac{1}{T} < \frac{20}{100N}$

$\implies \frac{100N}{19} \geq T > \frac{100N}{20}$

$\implies 5.26N \geq T > 5N$

Now suppose that $5.26N - 5N < 1$ . That would imply $5.26N < 5N + 1$ . Also, since $T > 5N,$ and since $5N$ and $T$ are integers, we have $T \geq 5N + 1$ . Therefore, $T \geq 5N + 1 > 5.26N \implies T > 5.26N$ . Which is a contradiction since we have $5.26N \geq T$ .

Therefore $5.26N - 5N > 1$

So, $5.26N - 5N \geq 1 \implies .26N \geq 1 \implies N \geq 3.84$ Taking $N = 4$ , we have

$21.04 \geq T > 20$ , giving us a solution of $T = \boxed{21}$

Aim is to solve 100x - 19y = 1 in x = 4 and y = 21, as x/y is nearly 19/100. The method of solving this is by using continued fractions, viz. 100/19 = 5 + 1/{3 + 1/[1 + 1/4}}. Then calculate back after ignoring 1/4 to get 21/4, giving the above mentioned natural numbers y and x. Answer = 21. See my problem "Never seen before" and its solution.

(I forgot to hashtag it so I'm putting it here. Lol) #easymoney

Anyway, here's the solution, it's a little bashy

This is all a game is strategical estimation.

Begin with $\frac{19}{100}$

To reduce this, we add $\pm 1$ to the numerator and get $\frac{20}{100}$ and $\frac{18}{100}$ this increases the value.

Now we have to check cases.

Simplifying and rounding, we will use two techniques. Adding $\pm 1$ to the numerator/denominator and simplifying by dividing through by a common factor. We have to check

$\frac{10}{50}, \frac{5}{25}, \frac{4}{20}~~ \text{and}~~ \frac{9}{50}$

From the first three, they are too large, so we have to either add one to the denominator or subtract one from the numerator. Visa versa for the second.

$\left(\frac{10}{51}, \frac{5}{26}, \frac{4}{21}, \frac{9}{49}\right), ~~ \left(\frac{9}{50},\frac{4}{25},\frac{3}{20},\frac{11}{50}\right)$

Using a calculator, we can start working with the smallest denominator (this way we don't have to check them all. After checking the two smallest, $\frac{4}{21}\approx 19.04\%$ .

Checking all cases less than this for extra measure.

$\frac{3}{16}=18.75\%$

$\frac{3}{15}=20\%$

$\frac{2}{11}=18.18\%$

$\frac{2}{10}=20\%$

$\frac{1}{6}=16.67\%$

$\frac{1}{5}=20\%$

Since $\frac{4}{21}$ is the smallest that works, our answer is $\boxed{21}$

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from math import *
found=False
for i in range(1,101):
    for j in range(1,i+1):
        if floor((j/float(i))*100)==19 and not found:
            print i
            found=True

:)