Write 100 as the sum of two positive integers 
Write $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.

Since $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then finding the complement that is also a multiple of the other. So then 
$100 = 44 + 56 = 4 \times 11 + 8 \times 7$. 
But is it the smart way of doing it? Is it the way I was supposed to solve it? I'm thinking here about a situation with a really large number that turns my plug-in method sort of unwise.
 A: From Bezout's Lemma, note that since $\gcd(7,11) = 1$, which divides $100$, there exists $x,y \in \mathbb{Z}$ such that $7x+11y=100$.
A candidate solution is $(x,y) = (8,4)$.
The rest of the solution is given by $(x,y) = (8+11m,4-7m)$, where $m \in \mathbb{Z}$. Since we are looking for positive integers as solutions, we need $8+11m > 0$ and $4-7m>0$, which gives us $-\frac8{11}<m<\frac47$. This means the only value of $m$, when we restrict $x,y$ to positive integers is $m=0$, which gives us $(x,y) = (8,4)$ as the only solution in positive integers.

If you do not like to guess your candidate solution, a more algorithmic procedure is using Euclid' algorithm to obtain solution to $7a+11b=1$, which is as follows.
We have
\begin{align}
11 & = 7 \cdot (1) + 4 \implies 4 = 11 - 7 \cdot (1)\\
7 & = 4 \cdot (1) + 3 \implies 3 = 7 - 4 \cdot (1) \implies 3 = 7 - (11-7\cdot (1))\cdot (1) = 2\cdot 7 - 11\\
4 & = 3 \cdot (1) + 1 \implies 1 = 4 - 3 \cdot (1) \implies 1 = (11-7 \cdot(1)) - (2\cdot 7 - 11) \cdot 1 = 11 \cdot 2-7 \cdot 3
\end{align}
This means the solution to $7a+11b=1$ using Euclid' algorithm is $(-3,2)$. Hence, the candidate solution $7x+11y=100$ is $(-300,200)$. Now all possible solutions are given by $(x,y) = (-300+11n,200-7n)$. Since we need $x$ and $y$ to be positive, we need $-300+11n > 0$ and $200-7n > 0$, which gives us
$$\dfrac{300}{11} < n < \dfrac{200}7 \implies 27 \dfrac3{11} < n < 28 \dfrac47$$
The only integer in this range is $n=28$, which again gives $(x,y) = (8,4)$.
A: 
But is it the smart way of doing it?

You are asked to find a and b so that $7a+11b=100\iff7(a+b)+4b=100\iff$
$4\Big[(a+b)+b\Big]+3(a+b)=100\iff3\Big\{\Big[(a+b)+b\Big]+(a+b)\Big\}+\Big[(a+b)+b\Big]=$ $100$. 
But $100=99+1=3\cdot33+1$, so $a+2b=1\iff2a+4b=2$, and $2a+3b=33$. It follows  that $b_0=-31$ and $a_0=63$ is one solution. But, then again, so are all numbers of the form $a_k=63-11k$ and $b_k=-31+7k$, with $k\in$ Z. All we have to do now is pick one pair whose components are both positive. The first equation implies $k<6$, and the latter $k>4$.
A: Since $\operatorname{gcd}(7,11)=1$ , you can find $a,b \in \mathbb Z$ with $7a+11b=1$. Now multiply both sides of the equation by $100$ to get one (and so all possible) results:
$$700a+1100b=100$$
Once you have a solution for $700a+1100b=100$, you have all solutions:
$$700(a+11k)+1100(b-7k)=100$$
You can then find $k$ so that both coefficients are positive.
A: An effort to avoid any enumeration or hiding an inductive leap to the answer.
$7a + 11b = 100: a,b \in N$
$ 11b \leq 100 - 7 = 93$
$\implies 1 \leq b \leq 8$
$ 7(a+b) = 100 - 4b$
$\implies 100 - 4b \equiv 0 \mod 7$
$\implies 25 - b \equiv 0 \mod 7 $
$\implies b \equiv 25 \mod 7 $
$\implies b \equiv 4 \mod 7$
$ \implies b = 4 + 7n$
We know $ 1 \leq b \leq 8  $. 
So we have $b \equiv 25 \mod 7$, so $ b = 4$  and hence $a = 8$.
A: While certainly not the ideal solution, this problem is certainly in the realm of Integer Programming. As plenty of others have pointed out, there are more direct approaches. However, I suspect ILP solvers would operate quite efficiently in your case, and requires less 'thought capital'.
A: $7a+11b=100$ so modding both sides by $11$ you get $11 \mid (7a-1)$. Let $11a'=7a-1$. Then the problem is transformed to finding $a',b$ such that $a'+b=9$ and $7 \mid (11a'+1)$, which (in my opinion) is somewhat easier to test for.
A: $7x+11y=100$
$7x=100-11y$
$x=\frac{100-11y}7=14-2y+\frac{2+3y}7$
$a=\frac{2+3y}7$
$7a=2+3y$
$3y=-2+7a$
$y=\frac{-2+7a}3=-1+2a+\frac{1+a}3$
$b=\frac{1+a}3$
$3b=1+a$
$a=3b-1$
$y=\frac{-2+7(3b-1)}3=\frac{-9+21b}3=-3+7b$
$x=\frac{100-11(-3+7b)}7=\frac{133-77b}7=19-11b$
$\begin{matrix}
x\gt 0&\to&19-11b\gt 0&\to&11b\lt 19&\to&b\lt\frac{19}{11}&\to&b\le 1&\\
&&&&&&&&&b=1\\
y\gt 0&\to&-3+7b\gt 0&\to&7b\gt 3&\to&b\gt\frac37&\to&b\ge 1&\\
\end{matrix}$
$1$ is the only integral value of $b$ for which both $x$ and $y$ yield positive values, as required by the problem. So, $x=8$ and $y=4$.
A: Use the Extended Euclidean Algorithm to solve
$$
7x+11y=100
$$
Using the implementation detailed in this answer
$$
\begin{array}{r}
&&1&1&1&3\\\hline
1&0&1&-1&\color{#C00000}{2}&\color{#0000F0}{-7}\\
0&1&-1&2&\color{#00A000}{-3}&\color{#E0A000}{11}\\
11&7&4&3&1&0\\
\end{array}
$$
we get that
$$
(\color{#00A000}{-3})\,7+(\color{#C00000}{2})\,11=1
$$
multiply by $100$ and use the last column from the algorithm to get the general answer, we get
$$
(-300+\color{#E0A000}{11}k)\,7+(200\color{#0000F0}{-7}k)\,11=100
$$
set $k=28$ (the only $k$ that works) to make the coefficients positive, we get that
$$
(8)\,7+(4)\,11=100
$$

Larger than $\boldsymbol{100}$
Suppose we want to write $1000000$ as a sum of a multiple of $7$ and a multiple of $11$. We can use the result of the algorithm above. That is
$$
(-3000000+11k)7+(2000000-7k)11=1000000
$$
We can use any $272728\le k\le 285714$ to make both coefficients positive. 
$k=272728$ gives
$$
(8)\,7+(90904)\,11=1000000
$$
$k=285714$ gives
$$
(142854)\,7+(2)\,11=1000000
$$
