Difference between revisions of "2023 AIME I Problems/Problem 7"
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Case 2: <math>n = 1 \pmod{2}</math> | Case 2: <math>n = 1 \pmod{2}</math> | ||
− | <math>n</math> is then <math>3 \pmod{4}</math>. If <math>n</math> is <math>0 \pmod{3}</math>, then by CRT, <math>n = 3 \pmod{6}</math>, a contradiction. Thus, <math>n = 2 \pmod{3}</math>, which by CRT implies <math>n = 5 \pmod{6}</math>. <math>n</math> can either be <math>0 \pmod 5</math>, which implies that <math>n = 35 \pmod 60</math>, <math>17</math> cases; or <math>4 \pmod 5</math>, which implies that <math>n = 59 \pmod 60</math>, <math>16</math> cases. | + | <math>n</math> is then <math>3 \pmod{4}</math>. If <math>n</math> is <math>0 \pmod{3}</math>, then by CRT, <math>n = 3 \pmod{6}</math>, a contradiction. Thus, <math>n = 2 \pmod{3}</math>, which by CRT implies <math>n = 5 \pmod{6}</math>. <math>n</math> can either be <math>0 \pmod{5}</math>, which implies that <math>n = 35 \pmod{60}</math>, <math>17</math> cases; or <math>4 \pmod{5}</math>, which implies that <math>n = 59 \pmod{60}</math>, <math>16</math> cases. |
<math>16 + 16 + 17 = \boxed{49}</math>. | <math>16 + 16 + 17 = \boxed{49}</math>. | ||
~mathboy100 | ~mathboy100 |
Revision as of 17:09, 8 February 2023
Unofficial problem statement: Find the number of positive integers from 1 to 1000 that have different mods in mod 2,3,4,5, and 6.
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.
.
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 2
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
is
, then by CRT,
, a contradiction. Thus,
, which by CRT implies
.
can either be
, which implies that
,
cases; or
, which implies that
,
cases.
.
~mathboy100