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 should not be able to form the value $91,$ but we should be able to form the values from $92$ to $96$ . We can form any value after $96$ by adding $5$ cent stamps.

We need to form the values $96$ while also not forming the value $91.$ If we used a $5$ cent stamp here, then we could just remove 1 to get down to $91.$ So we need to be able to form $96$ only using denominations of $n$ and $n + 1.$

Now we can figure out the working $(n, n+1)$ pairs that can used to obtain $96$ . Note that we can use at most $\frac{96}{24}=4$ stamps without going over. Let $na + (n + 1)b = 96,$ where $a$ is the number of $n$ stamps used and $b$ is the number of $n + 1$ stamps used. The possible $(a,b)$ pairs are: $(a,b) \rightarrow (4,0), (3,1), (2,2), (1,3), (0,4), (3,0), (2,1), (1,2), (0,3), (2,0), (1,1), (0,2), (1,0), (0,1).$ Solving for $n$ in each $(a,b)$ pair we find that the potential solutions are: $(n, n + 1) \rightarrow (\xcancel{23}, \xcancel{24}), (24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96), (96, 97).$

The last two do not work because they are too large to form the values $92$ through $95$ . By inspection, we find that $(24, 25)$ and $(47, 48)$ are the only pairs that can form $96$ without forming $91,$ so $n \in \{24, 47\}$ . The requested sum is $24 + 47 = \boxed{071}$ .

~Revision by emerald_block ~Minor Revision by Mathkiddie, ~Revisions by User:grogg007

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, using the number $5$ and the two numbers in the pair, it is impossible to form a sum of $91$ , and it is possible to form sums of $92$ , $93$ , 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.)

~emerald_block

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 McNugget 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

Solution 4 (standard)

Consider a postage that gives $96$ . We cannot use a $5$ -stamp as otherwise simply removing it yields a postage that gives $91$ . Additionally, there cannot be at least $5$ of $n$ -stamps or $n + 1$ -stamps, as else we can convert $5$ of the same valued stamp into a positive number of $5$ -stamps, then remove one to get a postage of $91$ .

From here consider integers $0 \le a, b, \le 4$ where $an + b(n + 1) = 96 \implies n = \dfrac{96 - b}{a + b}$ . The only pairs $(a, b)$ that yield an integer value are $(a, b) = (0, 2), (0, 3), (0, 4), (1, 0), (2, 0), (3, 0), (4, 0), (4, 1)$ which generate the values $n = 47, 31, 23, 96, 48, 32, 24, 19$ respectively. It is easy to find counterexamples of postages that evaluate to $91$ besides $n = 24, 47$ .

Now for $n = 24$ clearly $91$ is unobtainable since we need a $4 \pmod {5}$ amount of $24$ -stamps which exceeds a value of $96$ . A similar result holds for $n = 47$ as any evaluation $\le 2 \cdot 47$ can only be $0, 2, 3 \pmod{5}$ . In both cases it is easy to construct a postage for $92, 93, 94, 95, 96$ , to which repeatedly adding $5$ -stamps makes all postages worth $>91$ possible. The requesteed sum is $24 + 47 = \boxed{71}$ .

- blueprimes

Video Solution

Video solution by Dr. Osman Nal: https://www.youtube.com/watch?v=fTZP2e-_rjA

2019 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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