# 2019 AIME II Problems/Problem 14

## Problem

Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed.

## Solution 1

By the Chicken McNugget theorem, the least possible value of $n$ such that $91$ cents cannot be formed satisfies $5n - (5 + n) = 91 \implies n = 24$, so $n$ must be at least $24$.

For a value of $n$ to work, we must not only be unable to form the value $91$, but we must also be able to form the values $92$ through $96$, as with these five values, we can form any value greater than $96$ by using additional $5$ cent stamps.

Notice that we must form the value $96$ without forming the value $91$. If we use any $5$ cent stamps when forming $96$, we could simply remove one to get $91$. This means that we must obtain the value $96$ using only stamps of denominations $n$ and $n+1$.

Recalling that $n \geq 24$, we can easily figure out the working $(n,n+1)$ pairs that can used to obtain $96$, as we can use at most $\frac{96}{24}=4$ stamps without going over. The potential sets are $(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)$, and $(96, 97)$.

The last two obviously do not work, since they are too large to form the values $92$ through $94$, and by a little testing, only $(24, 25)$ and $(47, 48)$ can form the necessary values, so $n \in \{24, 47\}$. $24 + 47 = \boxed{071}$.

~Revision by emerald_block

### Note on finding and testing potential pairs

In order to find potential $(n,n+1)$ pairs, we simply test all combinations of $n$ and $n+1$ that sum to less than $4n$ (so that $n\ge24$) to see if they produce an integer value of $n$ when their sum is set to $96$. Note that, since $96$ is divisible by $1$, $2$, $3$, and $4$, we must either use only $n$ or only $n+1$, as otherwise, the sum is guaranteed to not be divisible by one of the numbers $2$, $3$, and $4$.

$\begin{array}{c|c|c} \text{Combination} & \text{Sum} & n\text{-value} \\ \hline n,n,n,n & 4n & 24 \\ n+1,n+1,n+1 & 3n+3 & 31 \\ n,n,n & 3n & 32 \\ n+1,n+1 & 2n+2 & 47 \\ n,n & 2n & 48 \\ n+1 & n+1 & 95 \\ n & n & 96 \\ \end{array}$

To test whether a pair works, we simply check that it cannot form $91$, and that it can form $92$, $93$, and $94$. ($95$ can always be formed using only $5$s, and the pair is already able to form $96$ because that was how it was found.) We simply need to reach the residues $2$, $3$, and $4$$\pmod{5}$ using only $n$ and $n+1$ without going over the number we are trying to form, while being unable to do so with the residue $1$. As stated in the above solution, the last two pairs are clearly too large to work.

$\begin{array}{c|c|c|c|c} \text{Pair} & \text{Not }91 & 92 & 93 & 94 \\ \hline 24,25 & \checkmark & \checkmark & \checkmark & \checkmark \\ 31,32 & \times & \checkmark & \checkmark & \checkmark \\ 32,33 & \times & \checkmark & \checkmark & \checkmark \\ 47,48 & \checkmark & \checkmark & \checkmark & \checkmark \\ 48,49 & \checkmark & \times & \checkmark & \checkmark \\ \end{array}$ (Note that if a pair is unable to fulfill a single requirement, there is no need to check the rest.)

## Solution 2

Notice that once we hit all residues $\bmod 5$, we'd be able to get any number greater (since we can continually add $5$ to each residue). Furthermore, $n\not\equiv 0,1\pmod{5}$ since otherwise $91$ is obtainable (by repeatedly adding $5$ to either $n$ or $n+1$) Since the given numbers are $5$, $n$, and $n+1$, we consider two cases: when $n\equiv 4\pmod{5}$ and when $n$ is not that.

When $n\equiv 4 \pmod{5}$, we can only hit all residues $\bmod 5$ once we get to $4n$ (since $n$ and $n+1$ only contribute $1$ more residue $\bmod 5$). Looking at multiples of $4$ greater than $91$ with $n\equiv 4\pmod{5}$, we get $n=24$. It's easy to check that this works. Furthermore, any $n$ greater than this does not work since $91$ isn't the largest unobtainable value (can be verified using Chicken McNugget Theorem).

Now, if $n\equiv 2,3\pmod{5}$, then we'd need to go up to $2(n+1)=2n+2$ until we can hit all residues $\bmod 5$ since $n$ and $n+1$ create $2$ distinct residues $\bmod{5}$. Checking for such $n$ gives $n=47$ and $n=48$. It's easy to check that $n=47$ works, but $n=48$ does not (since $92$ is unobtainable). Furthermore, any $n$ greater than this does not work since $91$ isn't the largest unobtainable value in those cases (can be verified using Chicken McNugget Theorem). (Also note that in the $3 \pmod{5}$ case, the residue $2 \pmod{5}$ has will not be produced until $3(n+1)$ while the $1\pmod5$ case has already been produced, so the highest possible value that cannot be produced would not be a number equivalent to $1 \pmod5$)

Since we've checked all residues $\bmod 5$, we can be sure that these are all the possible values of $n$. Hence, the answer is $24+47=\boxed{071}$. - ktong

## Solution 3

Obviously $n\le 90$. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If $n\equiv 0\pmod{5}$, then 91 can be formed by using $n+1$ and some 5's, so there are no solutions for this case. If $n\equiv 1\pmod{5}$, then 91 can be formed by using $n$ and some 5's, so there are no solutions for this case either.

For $n\equiv 2\pmod{5}$, $2n+2$ is the smallest value that can be formed which is 1 mod 5, so $2n+2=96$ and $n=47$. We see that $92=45+47$, $93=48+45$, and $94=47+47$, so $n=47$ does work. If $n\equiv 3\pmod{5}$, then the smallest value that can be formed which is 1 mod 5 is $2n$, so $2n=96$ and $n=48$. We see that $94=49+45$ and $93=48+45$, but 92 cannot be formed, so there are no solutions for this case. If $n\equiv 4\pmod{5}$, then we can just ignore $n+1$ since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that $5n-n-5=91$ meaning $4n=96$ and $n=24$. Hence, the only two $n$ that work are $n=24$ and $n=47$, so our answer is $24+47=\boxed{071}$. -Stormersyle