Difference between revisions of "2021 Fall AMC 12B Problems/Problem 25"
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
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+ | ==Solution 3== | ||
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+ | To get from <math>n</math> to <math>n+1</math>, <math>R(n)</math> would add by <math>9</math> for each remainder <math>2, 3, 4, 5, 6, 7, 8, 9, 10</math>. However, given that some of these remainders can "round down" to <math>0</math> given the nature of mods, we must calculate the possible values of <math>n</math> such that the remainders in <math>R(n+1)</math> "rounds down" by a total of <math>9</math>, effectively canceling out the adding by <math>9</math> initially. | ||
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+ | To do so, we will analyze the "rounding down" for each of <math>2, 3, 4, 5, 6, 7, 8, 9, 10</math>: | ||
+ | |||
+ | |||
+ | <math>n+1 \equiv 0 \pmod 2</math>: subtract by <math>2</math> | ||
+ | <math>n+1 \equiv 0 \pmod 3</math>: subtract by <math>3</math> | ||
+ | <math>n+1 \equiv 0 \pmod 4</math>: subtract by <math>4</math>, but this also implies mod <math>2</math>, so subtract by <math>6</math>. | ||
+ | <math>n+1 \equiv 0 \pmod 5</math>: subtract by <math>5</math> | ||
+ | <math>n+1 \equiv 0 \pmod 6</math>: subtract by <math>6</math>, but this also implies mod <math>2</math> and <math>3</math>, so subtract by <math>11</math>: too much | ||
+ | <math>n+1 \equiv 0 \pmod 7</math>: subtract by <math>7</math> | ||
+ | <math>n+1 \equiv 0 \pmod 8</math>: subtract by <math>8</math>, but this also implies mod <math>2</math> and <math>4</math>, so subtract by <math>14</math>: too much | ||
+ | <math>n+1 \equiv 0 \pmod 9</math>: subtract by <math>9</math>, but this also implies mod <math>3</math>, so subtract by <math>12</math>: too much | ||
+ | <math>n+1 \equiv 0 \pmod 10</math>: subtract by <math>10</math>: too much | ||
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+ | Notice that <math>9 = 7+2 = 6+3 = 5+4 = 4+3+2</math>. By testing these sums, we can easily show that the only time when the total subtraction is <math>9</math> is when <math>n+1 \equiv 0 \pmod 2</math> AND <math>n+1 \equiv 0 \pmod 7</math>. By CRT, <math>n+1 \equiv 0 \pmod 14</math>: | ||
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+ | As in solution 1, then, only <math>n+1=14 \cdot 1</math> and <math>n+1=14 \cdot 7</math> give possible solutions, so our answer is <math>\boxed{\textbf{(C) }2}</math>. | ||
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+ | ~xHypotenuse | ||
Revision as of 01:13, 21 August 2024
Contents
Problem
For a positive integer, let
be the sum of the remainders when
is divided by
,
,
,
,
,
,
,
, and
. For example,
. How many two-digit positive integers
satisfy
Solution 1
Note that we can add to
to get
, but must subtract
for all
. Hence, we see that there are four ways to do that because
. Note that only
is a plausible option, since
indicates
is divisible by
,
indicates that
is divisible by
,
indicates
is divisible by
, and
itself indicates divisibility by
, too. So,
and
is not divisible by any positive integers from
to
, inclusive, except
and
. We check and get that only
and
give possible solutions so our answer is
.
- kevinmathz
Solution 2
Denote by the remainder of
divided by
.
Define
.
Hence,
Hence, this problem asks us to find all , such that
.
:
.
We have .
Therefore, there is no in this case.
:
and
.
The condition implies
.
This further implies
.
Hence,
.
To get , we have
.
However, we have .
Therefore, there is no in this case.
:
for
and
.
The condition implies
with
.
Hence,
and
.
To get , we have
.
However, we have .
Therefore, there is no in this case.
:
for
and
.
To get , we have
.
Hence, we must have and
for
.
Therefore, .
:
for
and
.
The condition implies
with
.
Hence,
and
.
To get , we have
.
However, we have .
Therefore, there is no in this case.
:
for
and
.
To get , we have
.
This can be achieved if ,
,
.
However, implies
. This implies
. Hence,
.
We get a contradiction.
Therefore, there is no in this case.
:
for
and
.
The condition implies
with
.
Hence,
.
To get , we have
. This implies
.
Because and
, we have
.
Hence,
.
However, in this case, we assume
.
We get a contradiction.
Therefore, there is no in this case.
:
for
and
.
To get , we have
. This is infeasible.
Therefore, there is no in this case.
:
for
.
To get , we have
. This is infeasible.
Therefore, there is no in this case.
Putting all cases together, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3
To get from to
,
would add by
for each remainder
. However, given that some of these remainders can "round down" to
given the nature of mods, we must calculate the possible values of
such that the remainders in
"rounds down" by a total of
, effectively canceling out the adding by
initially.
To do so, we will analyze the "rounding down" for each of :
: subtract by
: subtract by
: subtract by
, but this also implies mod
, so subtract by
.
: subtract by
: subtract by
, but this also implies mod
and
, so subtract by
: too much
: subtract by
: subtract by
, but this also implies mod
and
, so subtract by
: too much
: subtract by
, but this also implies mod
, so subtract by
: too much
: subtract by
: too much
Notice that . By testing these sums, we can easily show that the only time when the total subtraction is
is when
AND
. By CRT,
:
As in solution 1, then, only and
give possible solutions, so our answer is
.
~xHypotenuse
Video Solution
~MathProblemSolvingSkills.com
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=Fy8wU4VAzkQ
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
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