Difference between revisions of "1995 AIME Problems/Problem 10"
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Ahuang080411 (talk | contribs) (→Solution) |
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== Solution == | == Solution == | ||
− | The requested number <math>\mod {42}</math> must be a | + | 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. |
<cmath> | <cmath> | ||
\begin{tabular}{|r||r|r|r|r|r|} | \begin{tabular}{|r||r|r|r|r|r|} | ||
\hline | \hline | ||
+ | 1 & 43 & 85&&& \\ | ||
2&44&&&& \\ | 2&44&&&& \\ | ||
3&45&&&& \\ | 3&45&&&& \\ | ||
Line 24: | Line 25: | ||
\end{tabular}</cmath> | \end{tabular}</cmath> | ||
− | <math>\boxed{215}</math> is the greatest number in the list, | + | 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>. |
+ | ~minor edit [[User: Yiyj1|Yiyj1]] | ||
====Afterword==== | ====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 <math>5 \mod 42</math>. Specifically, <math>5 * 42 + 5</math>. | |
-jackshi2006 | -jackshi2006 | ||
− | == | + | == Solution 2 == |
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. | 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. |
Latest revision as of 12:34, 14 June 2024
Problem
What is the largest positive integer that is not the sum of a positive integral multiple of and a positive composite integer?
Solution
The requested number must be a prime number. Also, every number that is a multiple of
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
and the numbers that are multiples of
greater than them, until they reach a composite number.
Since is the greatest number in the list, it is the answer. Note that considering
would have shortened the search, since
, and so within
numbers at least one must be divisible by
.
~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 . Specifically,
.
-jackshi2006
Solution 2
Let our answer be . Write
, where
are positive integers and
. Then note that
are all primes.
If is
, then
because
is the only prime divisible by
. We get
as our largest possibility in this case.
If is
, then
is divisible by
and thus
. Thus,
.
If is
, then
is divisible by
and thus
. Thus,
.
If is
, then
is divisible by
and thus
. Thus,
.
If is
, then
is divisible by
and thus
. Thus,
.
Our answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.