2012 AIME II Problems/Problem 12
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
has one solution if
n
3 (mod 7)$$ (Error compiling LaTeX. Unknown error_msg)3 (mod 11)$$ (Error compiling LaTeX. Unknown error_msg)7 (mod 13)
n$.
This means that by the Chinese Remainder Theorem,$ (Error compiling LaTeX. Unknown error_msg)n2\cdot 6 \cdot 8 = 96
7 \cdot 11 \cdot 13 = 1001
960
n
0 \le n < 10010
10000
10007
10006
\fbox{958}$ 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 |
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