Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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− | Note that <math>n = 1</math> doesn't work; to prove this, we just have to substitute <math>1</math> for <math>n</math> in the expression | + | Note that <math>n = 1</math> doesn't work; to prove this, we just have to substitute <math>1</math> for <math>n</math> in the expression. |
− | + | This gives us | |
− | < | + | <math></math>\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor<math> = \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3</math> which is divisible by 3. |
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Therefore, the case <math>n = 1</math> does not work. | Therefore, the case <math>n = 1</math> does not work. | ||
Revision as of 22:20, 1 February 2020
Problem
For how many positive integers is
not divisible by
? (Recall that
is the greatest integer less than or equal to
.)
Solution (Casework)
Expression:
Solution:
Let
Notice that for every integer ,
if
is an integer, then the three terms in the expression above must be
,
if
is an integer, then the three terms in the expression above must be
, and
if
is an integer, then the three terms in the expression above must be
.
This is due to the fact that ,
, and
share no common factors collectively (other than 1).
Note that doesn't work; to prove this, we just have to substitute
for
in the expression.
This gives us
$$ (Error compiling LaTeX. Unknown error_msg)\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor
which is divisible by 3.
Therefore, the case
does not work.
Now, we test the three cases mentioned above.
Case 1: divides
The first case does not work, as the three terms in the expression must be , as mentioned above, so the sum becomes
, which is divisible by
.
Case 2: divides
Because divides
, the number of possibilities for
is the same as the number of factors of
, excluding
.
=
So, the total number of factors of
is
.
However, we have to subtract , because the case
doesn't work, as mentioned previously.
We now do the same for the third and last case.
Case 3: divides
Because divides
, the number of possibilities for
is the same as the number of factors of
, excluding
.
=
So, the total number of factors of
is
.
Again, we have to subtract , for the reason stated in Case 2.
Now that we have counted all of the cases, we add them.
, so the answer is
.
~dragonchomper
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.