Difference between revisions of "2023 AIME I Problems/Problem 7"
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Because the LCM of all of the numbers we are dividing by is 60, we know that all of the remainders are 0 again at 60, meaning that we have a cycle that repeats itself every 60 numbers. | Because the LCM of all of the numbers we are dividing by is 60, we know that all of the remainders are 0 again at 60, meaning that we have a cycle that repeats itself every 60 numbers. | ||
− | After listing all of the remainders up to 60, we find that 35, 58, and 59 are extra-distinct. So, we have 3 numbers every 60 which are extra-distinct. <math>60\cdot16</math> = 960 and <math>3\cdot16</math> = 48, so we have 48 extra-distinct numbers in the first 960 numbers. Because of our pattern, we know that the numbers from 961-1000 will act the same as 1-40, where we have 1 extra-distinct number. 48 + 1 = \boxed{\textbf{(49) }}$ | + | After listing all of the remainders up to 60, we find that 35, 58, and 59 are extra-distinct. So, we have 3 numbers every 60 which are extra-distinct. <math>60\cdot16</math> = 960 and <math>3\cdot16</math> = 48, so we have 48 extra-distinct numbers in the first 960 numbers. Because of our pattern, we know that the numbers from 961-1000 will act the same as 1-40, where we have 1 extra-distinct number. 48 + 1 = \boxed{\textbf{(49) }}$. |
==See also== | ==See also== |
Revision as of 22:28, 8 February 2023
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
. By CRT,
, and therefore
. Using CRT again, we obtain
, which gives
values for
.
Case 2:
is then
. If
, then by CRT,
, a contradiction. Thus,
, which by CRT implies
.
can either be
, which implies that
,
cases; or
, which implies that
,
cases.
.
~mathboy100
Solution 2
.
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 3
Because the LCM of all of the numbers we are dividing by is 60, we know that all of the remainders are 0 again at 60, meaning that we have a cycle that repeats itself every 60 numbers.
After listing all of the remainders up to 60, we find that 35, 58, and 59 are extra-distinct. So, we have 3 numbers every 60 which are extra-distinct. = 960 and
= 48, so we have 48 extra-distinct numbers in the first 960 numbers. Because of our pattern, we know that the numbers from 961-1000 will act the same as 1-40, where we have 1 extra-distinct number. 48 + 1 = \boxed{\textbf{(49) }}$.
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.