Difference between revisions of "2012 AIME II Problems/Problem 12"
Tempaccount (talk | contribs) (Remove extra problem section) |
Farnoodensan (talk | contribs) (→Solution) |
||
Line 15: | Line 15: | ||
so since <math>gcd(7,11,13)</math>=1, there is 1 solution for n for this case of residues of <math>n</math>. | so since <math>gcd(7,11,13)</math>=1, there is 1 solution for n for this case of residues of <math>n</math>. | ||
− | This means that by the Chinese Remainder Theorem, <math>n</math> can have <math>2\cdot 6 \cdot 8 = 96</math> different residues mod <math>7 \cdot 11 \cdot 13 = 1001</math>. Thus, there are <math>960</math> values of <math>n</math> satisfying the conditions in the range <math>0 \le n < 10010</math>. However, we must now remove any values greater than <math>10000</math> that satisfy the conditions. By checking residues, we easily see that the only such values are <math> | + | This means that by the Chinese Remainder Theorem, <math>n</math> can have <math>2\cdot 6 \cdot 8 = 96</math> different residues mod <math>7 \cdot 11 \cdot 13 = 1001</math>. Thus, there are <math>960</math> values of <math>n</math> satisfying the conditions in the range <math>0 \le n < 10010</math>. However, we must now remove any values greater than <math>10000</math> that satisfy the conditions. By checking residues, we easily see that the only such values are <math>10004</math> and <math>10005</math>, so there remain <math>\fbox{958}</math> values satisfying the conditions of the problem. |
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=11|num-a=13}} | {{AIME box|year=2012|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:50, 15 March 2019
Problem 12
For a positive integer , define the positive integer
to be
-safe if
differs in absolute value by more than
from all multiples of
. For example, the set of
-safe numbers is
. Find the number of positive integers less than or equal to
which are simultaneously
-safe,
-safe, and
-safe.
Solution
We see that a number is
-safe if and only if the residue of
is greater than
and less than
; thus, there are
residues
that a
-safe number can have. Therefore, a number
satisfying the conditions of the problem can have
different residues
,
different residues
, and
different residues
. The Chinese Remainder Theorem states that for a number
that is
(mod b)
(mod d)
(mod f)
has one solution if
. For example, in our case, the number
can be:
3 (mod 7)
3 (mod 11)
7 (mod 13)
so since
=1, there is 1 solution for n for this case of residues of
.
This means that by the Chinese Remainder Theorem, can have
different residues mod
. Thus, there are
values of
satisfying the conditions in the range
. However, we must now remove any values greater than
that satisfy the conditions. By checking residues, we easily see that the only such values are
and
, so there remain
values satisfying the conditions of the problem.
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.