Difference between revisions of "1995 AIME Problems/Problem 10"
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== Second Solution == | == Second Solution == | ||
− | Let our answer be <math>n</math>. Write <math> n = 42a + b </math>, where <math>a, b</math> are positive integers and <math> 0 | + | 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. |
If <math>b</math> is <math>0\mod{5}</math>, then <math>b = 5</math> because 5 is the only prime divisible by 5. We get <math> n = 215</math> as our largest possibility in this case. | If <math>b</math> is <math>0\mod{5}</math>, then <math>b = 5</math> because 5 is the only prime divisible by 5. We get <math> n = 215</math> as our largest possibility in this case. |
Revision as of 16:30, 27 February 2018
Contents
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.
is the greatest number in the list, so it is the answer. Note that considering
would have shortened the search, since
, and so within
numbers at least one must be divisible by
.
Second Solution
Let our answer be . Write
, where
are positive integers and
. Then note that
are all primes.
If is
, then
because 5 is the only prime divisible by 5. We get
as our largest possibility in this case.
If is
, then
is divisible by 5 and thus
. Thus,
.
If is
, then
is divisible by 5 and thus
. Thus,
.
If is
, then
is divisible by 5 and thus
. Thus,
.
If is
, then
is divisible by 5 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.