Difference between revisions of "2019 AIME II Problems/Problem 14"

Revision as of 23:25, 21 July 2025

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 $96$ . After $96,$ we can form any value by adding additional $5$ cent stamps ( $92 + 5x$ , $93 + 5x$ , $94 + 5x$ , $95 + 5x$ , and $96 + 6x$ , where $x$ is a positive integer, together make all positive integers after 96). We also need to be able to form $96$ using only denominations of $n$ and $n + 1,$ because if we also used denominations of $5,$ then we could just remove a $5$ cent stamp to get back to $91.$

Now we can figure out the working $(n, n+1)$ pairs that can be used to obtain $96$ but not $91$ . (Keep in mind that this is a good way to narrow down possible values of $n,$ but it still doesn't guarantee that all numbers from $92 - 95$ can also be formed, we will have to double check this later.) We can use no more than a total of $\frac{96}{24}=4$ stamps. Let $an + b(n + 1) = 96,$ 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: $(n, n + 1) \rightarrow (23,24), (24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96), (96, 97).$

The first pair doesn't work because $n$ has to be $\geq 24,$ and the last two don't work either because they are too large to form the values $92$ through $95$ . Finally, $(31,32), (32, 33),$ and $(48, 49)$ don't work because they can also form $91$ with denominations of $5$ (for example, $32 \cdot 0 + 33\cdot 2 + 5 \cdot 5 = 91$ ). We are left with $(24, 25)$ and $(47, 48).$ Now we need to make sure that $92 - 95$ can be formed. We know from the Chicken McNugget Theorem that $91$ is the largest unformable number with denominations $24, 25$ and $5,$ so this means that all numbers after $91$ should be formable, and we can bash to confirm that all numbers from $92-95$ can be formed with denominations of $47, 48,$ and $5.$ So $n \in \{24, 47\}$ . The requested sum is $24 + 47 = \boxed{071}$ .

~Revisions by grogg007, emerald_block, Mathkiddie

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

@@ Line 8: / Line 8: @@
 For a value of <math>n</math> to work, we should not be able to form the value <math>91,</math> but we should be able to form the values from <math>92</math> to <math>96</math>. After <math>96,</math> we can form any value by adding additional <math>5</math> cent stamps (<math>92 + 5x</math>, <math>93 + 5x</math>, <math>94 + 5x</math>, <math>95 + 5x</math>, and <math>96 + 6x</math>, where <math>x</math> is a positive integer, together make all positive integers after 96). We also need to be able to form <math>96</math> using only denominations of <math>n</math> and <math>n + 1,</math> because if we also used denominations of <math>5,</math> then we could just remove a <math>5</math> cent stamp to get back to <math>91.</math>
-Now we can figure out the working <math>(n, n+1)</math> pairs that can be used to obtain <math>96</math> but not <math>91</math>. Note that we can use no more than a total of <math>\frac{96}{24}=4</math> stamps. Let <math>an + b(n + 1) = 96,</math> where <math>a</math> is the number of <math>n</math> stamps used and <math>b</math> is the number of <math>n + 1</math> stamps used. The possible <math>(a,b)</math> pairs are: <cmath>(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).</cmath>Solving for <math>n</math> in each <math>(a,b)</math> pair we find that the potential solutions are:<cmath>(n, n + 1) \rightarrow (23,24), (24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96), (96, 97).</cmath>
+Now we can figure out the working <math>(n, n+1)</math> pairs that can be used to obtain <math>96</math> but not <math>91</math>. (Keep in mind that this is a good way to narrow down possible values of <math>n,</math> but it still doesn't guarantee that all numbers from <math>92 - 95</math> can also be formed, we will have to double check this later.) We can use no more than a total of <math>\frac{96}{24}=4</math> stamps. Let <math>an + b(n + 1) = 96,</math> where <math>a</math> is the number of <math>n</math> stamps used and <math>b</math> is the number of <math>n + 1</math> stamps used. The possible <math>(a,b)</math> pairs are: <cmath>(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).</cmath>Solving for <math>n</math> in each <math>(a,b)</math> pair we find that the potential solutions are:<cmath>(n, n + 1) \rightarrow (23,24), (24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96), (96, 97).</cmath>
-The first pair doesn't work because <math>n</math> has to be <math>\geq 24,</math> and the last two don't work either because they are too large to form the values <math>92</math> through <math>95</math>. Finally, <math>(31,32), (32, 33),</math> and <math>(48, 49)</math> don't work because they can also form <math>91</math> with denominations of <math>5</math> (for example, <math>32 \cdot 0 + 33\cdot 2 + 5 \cdot 5 = 91</math>). We are left with <math>(24, 25)</math> and <math>(47, 48).</math> We can just use the Chicken McNugget Theorem again to confirm that <math>91</math> is the smallest unformable number with denominations <math>24</math> and <math>5</math> and we can bash to confirm that all numbers from <math>92-95</math> can be formed with denominations of <math>47, 48,</math> and <math>5.</math> So <math>n \in \{24, 47\}</math>. The requested sum is <math>24 + 47 = \boxed{071}</math>.
+The first pair doesn't work because <math>n</math> has to be <math>\geq 24,</math> and the last two don't work either because they are too large to form the values <math>92</math> through <math>95</math>. Finally, <math>(31,32), (32, 33),</math> and <math>(48, 49)</math> don't work because they can also form <math>91</math> with denominations of <math>5</math> (for example, <math>32 \cdot 0 + 33\cdot 2 + 5 \cdot 5 = 91</math>). We are left with <math>(24, 25)</math> and <math>(47, 48).</math> Now we need to make sure that <math>92 - 95</math> can be formed. We know from the Chicken McNugget Theorem that <math>91</math> is the largest unformable number with denominations <math>24, 25</math> and <math>5,</math> so this means that all numbers after <math>91</math> should be formable, and we can bash to confirm that all numbers from <math>92-95</math> can be formed with denominations of <math>47, 48,</math> and <math>5.</math> So <math>n \in \{24, 47\}</math>. The requested sum is <math>24 + 47 = \boxed{071}</math>.
 ~Revisions by [[User:grogg007|grogg007]], [[User:emerald_block|emerald_block]], [[Mathkiddie|Mathkiddie]]

2019 AIME II (Problems • Answer Key • Resources)
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