2023 AIME I Problems/Problem 7
Contents
[hide]Problem
Call a positive integer extra-distinct if the remainders when
is divided by
and
are distinct. Find the number of extra-distinct positive integers less than
.
Solution 1
can either be
or
(mod
).
Case 1:
Then, , which implies
and
, and therefore
. Using CRT, we obtain
, which gives
values for
.
Case 2:
is then
. If
,
, a contradiction. Thus,
, which implies
.
can either be
, which implies that
by CRT, giving
cases; or
, which implies that
by CRT, giving
cases.
The total number of extra-distinct numbers is thus .
~mathboy100
Solution 2 (Simpler)
Because the LCM of all of the numbers we are dividing by is , we know that all of the remainders are
again at
, meaning that we have a cycle that repeats itself every
numbers.
After listing all of the remainders up to , we find that
,
, and
are extra-distinct. So, we have
numbers every
which are extra-distinct.
and
, so we have
extra-distinct numbers in the first
numbers. Because of our pattern, we know that the numbers from
thru
will have the same remainders as
thru
, so we have
other extra-distinct number (
).
.
~Algebraik
Solution 3
.
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
.
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
.
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
.
The condition implies
,
.
Because is extra-distinct,
for
is a permutation of
.
Thus,
.
However, conflicts
.
Therefore, this case has no solution.
.
The condition implies
and
.
Because is extra-distinct,
for
is a permutation of
.
Because , we must have
. Hence,
.
Hence, .
Hence,
.
We have .
Therefore, the number extra-distinct
in this case is 16.
.
The condition implies
and
.
Because is extra-distinct,
and
are two distinct numbers in
.
Because
and
is odd, we have
.
Hence,
or 4.
,
,
.
We have .
We have .
Therefore, the number extra-distinct
in this subcase is 17.
,
,
.
.
We have .
Therefore, the number extra-distinct
in this subcase is 16.
Putting all cases together, the total number of extra-distinct is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4 (Small addition to solution 2)
We need to find that ,
, and
are all extra-distinct numbers smaller than
Let Denote the remainder in the division of
by
as
is extra-distinct.
is extra-distinct.
We need to check all of the remainders up to
and remainders
is extra-distinct.
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~MathProblemSolvingSkills.com
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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