2020 AMC 10A Problems/Problem 22
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
= \frac{998}1 + \frac{999}1 + \frac{1000}1 = 2997 = 999 \cdot 3
n = 1$does not work.
Now, we test the three cases mentioned above.
<b>Case 1:</b>$ (Error compiling LaTeX. Unknown error_msg)n998
(a, a, a)
3a
3$.
<b>Case 2:</b>$ (Error compiling LaTeX. Unknown error_msg)n999
n
999
n
999
1
999
3^3 \cdot 37^1
999
4 \cdot 2 = 8$.
However, we have to subtract$ (Error compiling LaTeX. Unknown error_msg)1n = 1
8 - 1 = 7$We now do the same for the third and last case.
<b>Case 3:</b>$ (Error compiling LaTeX. Unknown error_msg)n1000
n
1000
n
1000
1
1000
5^3 \cdot 2^3
1000
4 \cdot 4 = 16$.
Again, we have to subtract$ (Error compiling LaTeX. Unknown error_msg)116 - 1 = 15
7 + 15 = 22
\boxed{\textbf{(A)}22}$.
~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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