1990 AHSME Problems/Problem 19
Contents
Problem
For how many integers between
and
is the improper fraction
in lowest terms?
Solution 1
What we want to know is for how many is
We start by setting
for some arbitrary
. This shows that
evenly divides
. Next we want to see under which conditions
also divides
. We know from the previous statement that
and thus
Next we simply add
to get
However, we also want
which leads to
from the previous statement. Since from that statement
divides
evenly,
must be of the form
, for some arbitrary integer
. After this, we can set
and
Finally, we must find the largest
such that
This is a simple linear inequality for which the answer is
, or
.
Solution 2
Rearranging the expression we get
Thus,
Thus,
Hence, for some arbitrary number
Solving for in
we find the answer is
or
~coolmath2017
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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