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 | This gives us | ||
− | <math>\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor | + | <math>\left\lfloor \dfrac{998}{1} \right\rfloor+\left\lfloor \dfrac{999}{1} \right\rfloor+\left\lfloor \dfrac{1000}{1}\right \rfloor = \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3</math> which is divisible by 3. |
− | Therefore, the case < | + | Therefore, the case <math>n = 1</math> does not work. |
Line 35: | Line 35: | ||
− | <b>Case 1:</b> < | + | <b>Case 1:</b> <math>n</math> divides <math>998</math> |
− | The first case does not work, as the three terms in the expression must be < | + | The first case does not work, as the three terms in the expression must be <math>(a, a, a)</math>, as mentioned above, so the sum becomes <math>3a</math>, which is divisible by <math>3</math>. |
− | <b>Case 2:</b> < | + | <b>Case 2:</b> <math>n</math> divides <math>999</math> |
− | Because < | + | Because <math>n</math> divides <math>999</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>999</math>, excluding <math>1</math>. |
− | < | + | <math>999</math> = <math>3^3 \cdot 37^1</math> |
− | So, the total number of factors of < | + | So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</math>. |
− | However, we have to subtract < | + | However, we have to subtract <math>1</math>, because the case <math>n = 1</math> doesn't work, as mentioned previously. |
− | < | + | <math>8 - 1 = 7</math> |
We now do the same for the third and last case. | We now do the same for the third and last case. | ||
− | <b>Case 3:</b> < | + | <b>Case 3:</b> <math>n</math> divides <math>1000</math> |
− | Because < | + | Because <math>n</math> divides <math>1000</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>1000</math>, excluding <math>1</math>. |
− | < | + | <math>1000</math> = <math>5^3 \cdot 2^3</math> |
− | So, the total number of factors of < | + | So, the total number of factors of <math>1000</math> is <math>4 \cdot 4 = 16</math>. |
− | Again, we have to subtract < | + | Again, we have to subtract <math>1</math>, for the reason stated in Case 2. |
− | < | + | <math>16 - 1 = 15</math> |
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Now that we have counted all of the cases, we add them. | Now that we have counted all of the cases, we add them. | ||
− | < | + | <math>7 + 15 = 22</math>, so the answer is <math>\boxed{\textbf{(A)}22}</math>. |
Revision as of 21:25, 1 February 2020
Contents
[hide]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
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 |
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