Difference between revisions of "2022 AIME II Problems/Problem 8"
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==Solution 1== | ==Solution 1== | ||
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+ | We need to find all numbers between <math>1</math> and <math>600</math> inclusive that are multiples of <math>4</math>, <math>5</math>, and/or <math>6</math> which are also multiples of <math>4</math>, <math>5</math>, and/or <math>6</math> when <math>1</math> is added to them. | ||
+ | |||
+ | We begin by noting that the LCM of <math>4</math>, <math>5</math>, and <math>6</math> is <math>60</math>. We can therefore simplify the problem by finding all such numbers described above between <math>1</math> and <math>60</math> and multiplying the quantity of such numbers by <math>10</math> (<math>600</math>/<math>60</math> = <math>10</math>). | ||
+ | |||
+ | After making a simple list of the numbers between <math>1</math> and <math>60</math> and going through it, we see that the numbers meeting this condition are <math>4</math>, <math>5</math>, <math>15</math>, <math>24</math>, <math>35</math>, <math>44</math>, <math>54</math>, and <math>55</math>. This gives us <math>8</math> numbers. <math>8</math> * <math>10</math> = <math>\boxed{080}</math>. ~burkinafaso | ||
+ | |||
+ | ==Solution 1.5== | ||
+ | |||
+ | This is Solution 1 with a slick element included. Solution 1 uses the concept that <math>60k+l</math> is a solution for <math>n</math> iff <math>60k+l</math> is a multiple of <math>3</math>, <math>4</math>, and/or <math>5</math> and <math>60k+l+1</math> is a multiple of <math>3</math>, <math>4</math>, and/or <math>5</math> for positive integer values of <math>l</math> and essentially any integer value of <math>k</math>. But keeping the same conditions in mind for <math>k</math> and <math>l</math>, we can also say that if <math>60k+l</math> is a solution, then <math>60k-l-1</math> is a solution! Therefore, one doesn't have to go as far as determining the number of values between <math>1</math> and <math>60</math> and then multiplying by <math>10</math>. One only has to determine the number of values between <math>1</math> and <math>30</math> and then multiply by <math>20</math>. The values of <math>n</math> that work between <math>1</math> and <math>30</math> are <math>4</math>, <math>5</math>, <math>15</math>, and <math>24</math>. This gives us <math>4</math> numbers. <math>4</math> * <math>20</math> = <math>\boxed{080}</math>. ~burkinafaso | ||
+ | |||
+ | ==Solution 2== | ||
1. For <math>n</math> to be uniquely determined, <math>n</math> AND <math>n + 1</math> both need to be a multiple of <math>4, 5,</math> or <math>6.</math> Since either <math>n</math> or <math>n + 1</math> is odd, we know that either <math>n</math> or <math>n + 1</math> has to be a multiple of <math>5.</math> We can state the following cases: | 1. For <math>n</math> to be uniquely determined, <math>n</math> AND <math>n + 1</math> both need to be a multiple of <math>4, 5,</math> or <math>6.</math> Since either <math>n</math> or <math>n + 1</math> is odd, we know that either <math>n</math> or <math>n + 1</math> has to be a multiple of <math>5.</math> We can state the following cases: | ||
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~Lucasfunnyface | ~Lucasfunnyface | ||
− | ===Solution | + | ===Solution 2 Supplement=== |
Here is a detailed solution for Solution 1. | Here is a detailed solution for Solution 1. | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
− | ==Solution | + | ==Solution 3== |
The problem is the same as asking how many unique sets of values of <math>\lfloor\frac{n}{4}\rfloor</math>, <math>\lfloor\frac{n}{5}\rfloor</math>, and <math>\lfloor\frac{n}{6}\rfloor</math> can be produced by one and only one value of <math>n</math> for positive integers <math>n</math> less than or equal to 600. | The problem is the same as asking how many unique sets of values of <math>\lfloor\frac{n}{4}\rfloor</math>, <math>\lfloor\frac{n}{5}\rfloor</math>, and <math>\lfloor\frac{n}{6}\rfloor</math> can be produced by one and only one value of <math>n</math> for positive integers <math>n</math> less than or equal to 600. | ||
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Plugging in the values of <math>j</math> and <math>k</math>, finding the solutions to the 6 systems of linear congruences, and correcting for the repeated values, we find that there are <math>\boxed{080}</math> solutions of <math>n</math>. | Plugging in the values of <math>j</math> and <math>k</math>, finding the solutions to the 6 systems of linear congruences, and correcting for the repeated values, we find that there are <math>\boxed{080}</math> solutions of <math>n</math>. | ||
− | ===Solution | + | ===Solution 3.5 Supplement=== |
By [https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem Chinese Remainder Theorem], the general solution of systems of <math>2</math> linear congruences is: | By [https://artofproblemsolving.com/wiki/index.php/Chinese_Remainder_Theorem Chinese Remainder Theorem], the general solution of systems of <math>2</math> linear congruences is: | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
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==Sidenote== | ==Sidenote== |
Revision as of 13:06, 20 February 2022
Contents
Problem
Find the number of positive integers whose value can be uniquely determined when the values of
,
, and
are given, where
denotes the greatest integer less than or equal to the real number
.
Solution 1
We need to find all numbers between and
inclusive that are multiples of
,
, and/or
which are also multiples of
,
, and/or
when
is added to them.
We begin by noting that the LCM of ,
, and
is
. We can therefore simplify the problem by finding all such numbers described above between
and
and multiplying the quantity of such numbers by
(
/
=
).
After making a simple list of the numbers between and
and going through it, we see that the numbers meeting this condition are
,
,
,
,
,
,
, and
. This gives us
numbers.
*
=
. ~burkinafaso
Solution 1.5
This is Solution 1 with a slick element included. Solution 1 uses the concept that is a solution for
iff
is a multiple of
,
, and/or
and
is a multiple of
,
, and/or
for positive integer values of
and essentially any integer value of
. But keeping the same conditions in mind for
and
, we can also say that if
is a solution, then
is a solution! Therefore, one doesn't have to go as far as determining the number of values between
and
and then multiplying by
. One only has to determine the number of values between
and
and then multiply by
. The values of
that work between
and
are
,
,
, and
. This gives us
numbers.
*
=
. ~burkinafaso
Solution 2
1. For to be uniquely determined,
AND
both need to be a multiple of
or
Since either
or
is odd, we know that either
or
has to be a multiple of
We can state the following cases:
1. is a multiple of
and
is a multiple of
2. is a multiple of
and
is a multiple of
3. is a multiple of
and
is a multiple of
4. is a multiple of
and
is a multiple of
Solving for each case, we see that there are possibilities for cases 1 and 3 each, and
possibilities for cases 2 and 4 each. However, we over-counted the cases where
1. is a multiple of
and
is a multiple of
2. is a multiple of
and
is a multiple of
Each case has possibilities.
Adding all the cases and correcting for over-counting, we get
~Lucasfunnyface
Solution 2 Supplement
Here is a detailed solution for Solution 1.
![]()
,
![]()
,
,
,
,
,
,
,
,
, 30 integers.
![]()
,
![]()
,
,
,
,
,
,
,
,
, 20 integers.
![]()
,
![]()
,
,
,
,
,
,
,
,
, 30 integers.
![]()
,
![]()
,
,
,
,
,
,
,
,
, 20 integers.
Over-counted cases:
![]()
,
![]()
,
,
,
,
,
,
,
,
,
,
, 10 integers.
![]()
,
![]()
,
,
,
,
,
,
,
,
,
,
, 10 integers.
Solution 3
The problem is the same as asking how many unique sets of values of ,
, and
can be produced by one and only one value of
for positive integers
less than or equal to 600.
Seeing that we are dealing with the unique values of the floor function, we ought to examine when it is about to change values, for instance, when is close to a multiple of 4 in
.
For a particular value of , let
,
, and
be the original values of
,
, and
, respectively.
Notice when
and
, the value of
will be 1 less than the original
. The value of
will be 1 greater than the original value of
.
More importantly, this means that no other value less than or greater than will be able to produce the set of original values of
,
, and
, since they make either
or
differ by at least 1.
Generalizing, we find that must satisfy:
Where and
are pairs of distinct values of 4, 5, and 6.
Plugging in the values of and
, finding the solutions to the 6 systems of linear congruences, and correcting for the repeated values, we find that there are
solutions of
.
Solution 3.5 Supplement
By Chinese Remainder Theorem, the general solution of systems of linear congruences is:
,
,
Find
and
such that
,
Then
![]()
, we solve the number of values for
, then multiply by
to get the number of values for
. We are going to solve the following
systems of linear congruences:
![]()
,
![]()
No solution
![]()
,
![]()
![]()
,
![]()
No solution
![]()
,
![]()
, there are
values for
. For
, the answer is
.
Sidenote
For solving a system of linear congruences, see https://youtu.be/-a88u99nmkw
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.