Find the 2 digits in front of the number 14, so that the 4-digit number we get is divisible by 26. I know that 26=13.2. The last digit of the number is 4, it is divisible by 2. I have to remove the last digit and check out if the remaining digits are divisible by 13 or? 
 A: You are being asked to find an integer $x$ such that $10 \leq x \leq 99$ and
$$ 100 x + 14 \equiv 0 \pmod{26}. $$
Solve this equation, and $x$ will be the two digits you want.
There are multiple possible answers.

A good first step is to rewrite the equation in a more convenient form.
Since it is modulo $26,$ you can replace any of the numbers by something
equivalent modulo $26.$
For example, $100 \equiv 22 \pmod{26},$ so you can write $22$ instead of $100.$
I think it's more convenient to use the fact that $100 \equiv -4 \pmod{26},$
however; it lets you work with smaller numbers.
A: Starting with $100 x + 14 \equiv 0 \pmod{26}$ you can take $100 \pmod{26}$ to get $74, 48, 22, -4$, etc. $-4$ is the smallest, so let's try that. 
Now the equivalence to solve is $-4x + 14 \equiv 0 \pmod{26}$. It'll be easier to see the solution if the LHS is positive, so add $26$ to get $-4x + 40 \equiv 0 \pmod{26}$. Is it possible to subtract 4 from 40 a few times to get $26$? If you subtract $4*3=12$ you still have $2$ to go to get to $26$. 
So add another $26$ to the equation to get $$-4x + 40 \equiv 0 \pmod{26}$$. Note that $66-26=40=4*10$. So $x$ is $10$.
You would then verify $x=10$ in the original problem. Is $1014 \equiv 0 \pmod{26}$?
A: Let $1\le a\le 9$ and $0\le b\le9$ be those two front numbers ($ab14$). So, our desired number is $10^3a+10^2b+14$ where $$10^3a+10^2b+14\equiv0\pmod{26}\implies 12a-4b\equiv12\pmod{26}\implies 3a-3=3(a-1)\equiv b\pmod{13}$$
All choices that satisfy this (just plug in $a=1,2,\ldots9$ and see what happens):
$1014$
$2314$
$3614$
$4914$
$6214$
$7514$
$8814$
Notice how in the congruence, I used $$a\equiv b\pmod{m}\implies \frac{a}{c}\equiv \frac{b}{c}\pmod{\frac{m}{\gcd(m,c)}}$$
as long as $c\vert a$ and $c\vert b$. Also notice that $a\equiv9$ is not possible because that gives an invalid number for $b$
A: One way to find a solution is to start from the useful and memorable fact that $7 \times 11 \times 13 = 1001$.  Hence $26$ divides any even multiple of $1001$, in particular $4004$.
If we can find $x10$ divisible by $26$, then $4004 + x10$ will yield a solution. $x10$ is divisible (for any $x$) by $2$, so it suffices to find $x1$ divisible by $13$. Since $91 = 7 \times 13$, a solution is $4004 + 910 = 4914$.
