Difference between revisions of "2022 AIME II Problems/Problem 8"
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For a particular value of <math>n</math>, let <math>a</math>, <math>b</math>, and <math>c</math> be the original values of <math>\lfloor\frac{n}{4}\rfloor</math>, <math>\lfloor\frac{n}{5}\rfloor</math>, and <math>\lfloor\frac{n}{6}\rfloor</math>, respectively. | For a particular value of <math>n</math>, let <math>a</math>, <math>b</math>, and <math>c</math> be the original values of <math>\lfloor\frac{n}{4}\rfloor</math>, <math>\lfloor\frac{n}{5}\rfloor</math>, and <math>\lfloor\frac{n}{6}\rfloor</math>, respectively. | ||
− | Notice when <math>n</math> <math>\equiv0\mod4</math> | + | Notice when <math>n</math> <math>\equiv0\mod4</math> and <math>n</math> <math>\equiv4\mod5</math>, the value of <math>\lfloor\frac{n-1}{4}\rfloor</math> will be 1 less than the original <math>a</math>. The value of <math>\lfloor\frac{n+1}{5}\rfloor</math> will be 1 greater than the original value of <math>b</math>. |
More importantly, this means that no other value less than or greater than <math>n</math> will be able to produce the set of original values of <math>a</math>, <math>b</math>, and <math>c</math>, since they make either <math>a</math> or <math>b</math> differ by at least 1. | More importantly, this means that no other value less than or greater than <math>n</math> will be able to produce the set of original values of <math>a</math>, <math>b</math>, and <math>c</math>, since they make either <math>a</math> or <math>b</math> differ by at least 1. |
Revision as of 21:38, 18 February 2022
Contents
[hide]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. 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 overcounted 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 overcounting, we get
~Lucasfunnyface
Side note: solution does not explain how we found the 20 possibilities, 30, possibilities, etc. It would be great if somebody added that in.
Solution 2
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
.
Sidenote
The mod computation can be more easily done by first finding the solutions in the range 1-60, correcting for overcounting, and multiplying by 10.
Alternatively, before taking the time to consider a systematic solution, you can notice that the pattern “repeats” every 60 positive integers. From there, bash to see how many of the first 60 numbers work and multiply by 10.
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.