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Difference between revisions of "1995 AIME Problems/Problem 10"

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== Problem ==
 
== Problem ==
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What is the largest positive integer that is not the sum of a positive integral multiple of 42 and a positive composite integer?
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What is the largest positive integer that is not the sum of a positive integral multiple of <math>42</math> and a positive composite integer?
  
 
== Solution ==
 
== Solution ==
−
The requested number mod 42 must be a prime number. Also, every number that is a multiple of 42 greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to 42 and the numbers that are multiples of 42 greater than them, until they reach a composite number.
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The requested number <math>\mod {42}</math> must be a prime number. Also, every number that is a multiple of <math>42</math> greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to <math>42</math> and the numbers that are multiples of <math>42</math> greater than them, until they reach a composite number.
  
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2:44
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<cmath>
 +
\begin{tabular}{|r||r|r|r|r|r|}
 +
\hline
 +
1 & 43 & 85&&& \\
 +
2&44&&&& \\
 +
3&45&&&& \\
 +
5&47&89&131&173&215 \\
 +
7&49&&&& \\
 +
11&53&95&&& \\
 +
13&55&&&& \\
 +
17&59&101&143&& \\
 +
19&61&103&145&& \\
 +
23&65&&&& \\
 +
29&71&113&155&& \\
 +
31&73&115&&& \\
 +
37&79&121&&& \\
 +
41&83&125&&& \\
 +
\hline
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\end{tabular}</cmath>
  
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3:45
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Since <math>\boxed{215}</math> is the greatest number in the list, it is the answer. Note that considering <math>\mod {5}</math> would have shortened the search, since <math>\text{gcd}(5,42)=1</math>, and so within <math>5</math> numbers at least one must be divisible by <math>5</math>.
  
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5:47,89,131,173,215
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~minor edit [[User: Yiyj1|Yiyj1]]
 +
====Afterword====
  
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7:49
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Basically, we are looking for a number where when a multiple of 42 is subtracted from it, the result is a prime number. Any number that ends in a 5 is not prime, except for 5 itself. Since 42 keeps the parity the same and advances odd numbers in the unit digit, then we can conclude that the sought number is <math>5 \mod 42</math>. Specifically, <math>5 * 42 + 5</math>.
  
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11:53,95
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-jackshi2006
  
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13:55
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== Solution 2 ==
  
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17:59,101,143
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Let our answer be <math>n</math>. Write <math> n = 42a + b </math>, where <math>a, b</math> are positive integers and <math> 0 \leq b < 42 </math>. Then note that <math> b, b + 42, ... , b + 42(a-1) </math> are all primes.
  
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19:61,103,145
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If <math>b</math> is <math>0\mod{5}</math>, then <math>b = 5</math> because <math>5</math> is the only prime divisible by <math>5</math>. We get <math> n = 215</math> as our largest possibility in this case.
  
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23:65
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If <math>b</math> is <math>1\mod{5}</math>, then <math>b + 2 \times 42</math> is divisible by <math>5</math> and thus <math>a \leq 2</math>. Thus, <math>n \leq 3 \times 42 = 126 < 215</math>.
  
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29:71,113,155
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If <math>b</math> is <math>2\mod{5}</math>, then <math>b + 4 \times 42</math> is divisible by <math>5</math> and thus <math>a \leq 4</math>. Thus, <math>n \leq 5 \times 42 = 210 < 215</math>.
  
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31:73,115
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If <math>b</math> is <math>3\mod{5}</math>, then <math>b + 1 \times 42</math> is divisible by <math>5</math> and thus <math>a = 1</math>. Thus, <math>n \leq 2 \times 42 = 84 < 215</math>.
  
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37:79,121
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If <math>b</math> is <math>4\mod{5}</math>, then <math>b + 3 \times 42</math> is divisible by <math>5</math> and thus <math>a \leq 3</math>. Thus, <math>n \leq 4 \times 42 = 168 < 215</math>.
  
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41:83,125
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Our answer is <math>\boxed{215}</math>.
  
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215 is the greatest number in the list, so it is the answer.
+
== See also ==
 +
{{AIME box|year=1995|num-b=9|num-a=11}}
  
−
== See also ==
+
[[Category:Intermediate Number Theory Problems]]
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* [[1995 AIME Problems/Problem 9 | Previous problem]]
+
{{MAA Notice}}
−
* [[1995 AIME Problems/Problem 11 | Next problem]]
 
−
* [[1995 AIME Problems]]
 

Latest revision as of 11:34, 14 June 2024

Problem

What is the largest positive integer that is not the sum of a positive integral multiple of $42$ and a positive composite integer?

Solution

The requested number $\mod {42}$ must be a prime number. Also, every number that is a multiple of $42$ greater than that prime number must also be prime, except for the requested number itself. So we make a table, listing all the primes up to $42$ and the numbers that are multiples of $42$ greater than them, until they reach a composite number.

\[\begin{tabular}{|r||r|r|r|r|r|} \hline 1 & 43 & 85&&& \\ 2&44&&&& \\ 3&45&&&& \\ 5&47&89&131&173&215 \\ 7&49&&&& \\ 11&53&95&&& \\ 13&55&&&& \\ 17&59&101&143&& \\ 19&61&103&145&& \\ 23&65&&&& \\ 29&71&113&155&& \\ 31&73&115&&& \\ 37&79&121&&& \\ 41&83&125&&& \\ \hline \end{tabular}\]

Since $\boxed{215}$ is the greatest number in the list, it is the answer. Note that considering $\mod {5}$ would have shortened the search, since $\text{gcd}(5,42)=1$, and so within $5$ numbers at least one must be divisible by $5$.

~minor edit Yiyj1

Afterword

Basically, we are looking for a number where when a multiple of 42 is subtracted from it, the result is a prime number. Any number that ends in a 5 is not prime, except for 5 itself. Since 42 keeps the parity the same and advances odd numbers in the unit digit, then we can conclude that the sought number is $5 \mod 42$. Specifically, $5 * 42 + 5$.

-jackshi2006

Solution 2

Let our answer be $n$. Write $n = 42a + b$, where $a, b$ are positive integers and $0 \leq b < 42$. Then note that $b, b + 42, ... , b + 42(a-1)$ are all primes.

If $b$ is $0\mod{5}$, then $b = 5$ because $5$ is the only prime divisible by $5$. We get $n = 215$ as our largest possibility in this case.

If $b$ is $1\mod{5}$, then $b + 2 \times 42$ is divisible by $5$ and thus $a \leq 2$. Thus, $n \leq 3 \times 42 = 126 < 215$.

If $b$ is $2\mod{5}$, then $b + 4 \times 42$ is divisible by $5$ and thus $a \leq 4$. Thus, $n \leq 5 \times 42 = 210 < 215$.

If $b$ is $3\mod{5}$, then $b + 1 \times 42$ is divisible by $5$ and thus $a = 1$. Thus, $n \leq 2 \times 42 = 84 < 215$.

If $b$ is $4\mod{5}$, then $b + 3 \times 42$ is divisible by $5$ and thus $a \leq 3$. Thus, $n \leq 4 \times 42 = 168 < 215$.

Our answer is $\boxed{215}$.

See also

1995 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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