Showing that 49¢ is not makeable using the given conditions While going through 6.042J from MITOCW, in the text Mathematics for Computer Science, I came across the following problem at which I'm stuck. 
Now, I proceeded doing the proof in the following manner.
Suppose $C$ be the set of counterexamples to $P(n)$ where
$$P(n)::=\text{It is possible to make a postage stamp of n}¢$$
So, $$C = \bigg\{n \in \mathbb{N}, n \geq50 \mid \text{P(n) is false} \bigg\}$$ where $\mathbb{N}$ denotes the set of non-negative integers (i.e., integers $\geq 0$)
We assume that $C$ has a least element $m \in C \text{ such that P(m) doesn't hold}$.
Assuming that $\text{51¢, 52¢, 53¢, ..., 56¢}$ are all makeable (given in the question), we can say that $$m > 56$$
$$\implies m-6 > 50$$
Now, as $\text{m is the least element in C, P(m$-6$) has to be true}$
So, $(m-6)¢$ is makeable. Thus, adding $6¢$ to it, $(m-6)¢+6¢$ is also makeable, and thus $m¢$ is also makeable.
Thus we arrive at a contradiction, i.e., $\text{P(m) is actually true}$. So, $C$ must be empty. Thus it is possible to make any amount of postage over $50¢$.
Is this proof complete and correct? Why did they say to assume that all denominations from $51¢ \text{ to } 100¢ $ were makeable, when in fact, assuming upto $56¢$ would suffice?
Also, for part $\text{(b)}$, isn't $49¢$ actually makeable? Because $49¢ = 14¢+14¢+21¢$
So why does it say to prove that it is not? Is this part incorrect?
 A: One way of explaining the problem composers intent, on the first part of the problem is that
Since $50$ is makeable, and since the range $51$ through $100$ is makeable, you have an immediate proof by induction.
That is, assuming that the range $(50k + 1)$ through $(50k + 50)$ is makeable, and that $(50)$ is also makeable, you immediately conclude that
the range $(50[k+1] + 1)$ through $(50[k+1] + 50)$ is also makeable.
So, the problem composer is trying to stretch the intuition of the Math student into visualizing the induction principle, applied in as simple a way as possible.
As you indicated, you can conclude that since $51$ through $56$ is  makeable, and $6$ is makeable,  then an alternative approach is feasible.  The problem composer is attempting to coerce you into following the specific induction path, that he has laid out.

As you indicated, $(49)$ is makeable, so this is a clear mistake on the part of the problem composer.  As it turns out, each of $44$ through $48$ is also makeable.  So, the problem composer, having overreached, should have settled for asking you to show that $43$ is not makeable.
Let $a,b,c$ denote the non-negative integer scalars to be applied to $6, 14,$ and $21$, respectively.
If $c = 2$, clearly $(43- [2 \times 21])$ will not be makeable.
Suppose that $c = 1$.
Then, you have to find scalars $a,b$ so that $6a + 14b = (43 - 21) = 22.$
Neither $b = 0$, nor $b = 1$ will work in this situation.
Suppose that $c = 0$.
Then, you have to find scalars $a,b$ so that $6a + 14b = (43).$
This is impossible because $(43)$ is odd, while $6$ and $14$ are even.
Therefore, there is no satisfying combination that has $c \in \{0,1,2\}$
Therefore, $(43)$ is not makeable.
