# Representing Any $n \geq 4$ as a Sum of 2's and 5's

Use induction on $n$ to prove that for all integers $n\geq 4$, postage of $n$ cents can be realized using only $2$ cent and $5$ cent stamps.

I thinks it is little bit different. How can I use induction to this kind of problem?

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It may be useful when attempting to make $n+1$ cents to realize that if you've made $n$ cents out of $2$ and $5$ cent coins, then removing a $5$ cent coin and adding three $2$ two cent coins gives you $n+1$ cents, or removing two $2$ cent coins and adding a $5$ cent coin gives you $n+1$ cents. You just need to be careful that you have at least two $2$ cent coins or at least one $5$ cent coin already. –  yunone Jan 17 '12 at 0:00
@cardinal first. it is not homework. second. I didnot know about it and I did it. –  Q123 Jan 17 '12 at 2:20
BTW: Induction is way overkill in this problem. If $n=2k$ then, $k$ 2-cent stamps will do it, while if $n=2k+1$, then $k \geq 2$ and one $5$-cent stamp and $(k-2)$ two-cents stamps will do... If you check both solutions below, both of them basically lead by induction to exactly this solution ;) –  N. S. Jan 17 '12 at 6:19

It is trivial that $4$ and $5$ are achievable. Now it's all over. From those two, you can get any greater postage you want by adding a suitable number of $2$ cent stamps.

More formally, we want to show that for every $n \ge 2$, both $2n$ and $2n+1$ are achievable. Since every integer is even or odd, that will do it. We prove this by induction on $n$. The result is certainly true for the base case $n=2$.

Now suppose that for a certain $k\ge 2$, both $2k$ and $2k+1$ are achievable. We want to show that $2(k+1)$ and $2(k+1)+1$ are achievable. That is easy: add a $2$ cent stamp.

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The base case is 4 cents, which can be realized using two 2-cent stamps.

Suppose now that we can realize $n$ cents. We want to prove that we can realize $n + 1$ cents.

Consider two cases: either a 5-cent stamp is used in our hypothetical realization of $n$ cents or not.

If there is, remove it and add three 2-cent stamps. We now have $n - 5 + 3 \cdot 2 = n+1$; a realization of $n+1$ cents.

If there is not, then it must be that only 2-cent stamps are used. Moreover, since $n \geq 4$, at least two 2-cent stamps are used. Remove two of these and add a single 5-cent stamp. We now have $n - 2 \cdot 2 + 5 = n+1$; a realization of $n+1$ cents.

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