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Difference between revisions of "2019 AIME II Problems/Problem 14"
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==Solution==
 
==Solution==
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By the Chicken McNugget theorem, the least possible value of <math>n</math> such that <math>91</math> cents cannot be formed satisfies <math>5n - 5 - n = 91</math>, so <math>n = 24</math>. For values of <math>n</math> greater than <math>24</math>, notice that if <math>91</math> cents cannot be formed, then any number <math>1 \mod 5</math> less than <math>91</math> also cannot be formed. The proof of this is that if any number <math>1 \mod 5</math> less than <math>91</math> can be formed, then we could keep adding <math>5</math> cent stamps until we reach <math>91</math> cents. However, since <math>91</math> cents is the greatest postage that cannot be formed, <math>96</math> cents is the first  number that is <math>1 \mod 5</math> that can be formed, so it must be formed without any <math>5</math> cent stamps. There are few <math>(n, n + 1)</math> pairs, where <math>n \geq 24</math>, that can make <math>96</math> cents. These are cases where one of <math>n</math> and <math>n + 1</math> is a factor of <math>96</math>, which are <math>(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)</math>, and <math>(96, 97)</math>. The last two obviously do not work since <math>92</math> through <math>94</math> cents also cannot be formed, and by a little testing, only <math>(24, 25)</math> and <math>(47, 48)</math> satisfy the condition that <math>91</math> cents is the greatest postage that cannot be formed, so <math>n = 24, 47</math>. <math>24 + 47 = \boxed{071}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2019|n=II|num-b=13|num-a=15}}
 
{{AIME box|year=2019|n=II|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:28, 22 March 2019

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

By the Chicken McNugget theorem, the least possible value of $n$ such that $91$ cents cannot be formed satisfies $5n - 5 - n = 91$, so $n = 24$. For values of $n$ greater than $24$, notice that if $91$ cents cannot be formed, then any number $1 \mod 5$ less than $91$ also cannot be formed. The proof of this is that if any number $1 \mod 5$ less than $91$ can be formed, then we could keep adding $5$ cent stamps until we reach $91$ cents. However, since $91$ cents is the greatest postage that cannot be formed, $96$ cents is the first number that is $1 \mod 5$ that can be formed, so it must be formed without any $5$ cent stamps. There are few $(n, n + 1)$ pairs, where $n \geq 24$, that can make $96$ cents. These are cases where one of $n$ and $n + 1$ is a factor of $96$, which are $(24, 25), (31, 32), (32, 33), (47, 48), (48, 49), (95, 96)$, and $(96, 97)$. The last two obviously do not work since $92$ through $94$ cents also cannot be formed, and by a little testing, only $(24, 25)$ and $(47, 48)$ satisfy the condition that $91$ cents is the greatest postage that cannot be formed, so $n = 24, 47$. $24 + 47 = \boxed{071}$.

See Also

2019 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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