Difference between revisions of "2005 PMWC Problems/Problem I3"

Revision as of 11:01, 18 September 2008

Problem

Let $x$ be a fraction between $\frac{35}{36}$ and $\frac{91}{183}$ . If the denominator of $x$ is $455$ and the numerator and denominator have no common factor except $1$ , how many possible values are there for $x$ ?

Solution

We let $x=\frac{y}{455}$ .

We find the ranges for $y$ , disregarding the restriction that $y$ must be relatively prime with 455.

$\frac{91}{183}\leq \frac{y}{455}$

$183y\geq 91*455=41405$ , $y\geq 227$ .

$36y\leq 35*455=15925$ , $y\leq 441$

Since $y$ must be relatively prime to 455, it cannot have any prime factors equal to any in $455=5*7*13$ . Thus we use the Principle of Inclusion-Exclusion and complementary counting:

There are 43 multiples of 5 in that range.

There are 31 multiples of 7 in that range.

There are 16 multiples of 13 in that range.

There are 6 multiples of 7*5=35 in that range.

There are 3 multiples of 5*13=65 in that range.

There are 2 multiples of 7*13=91 in that range.

There aren't any multiples of 455 in that range.

$43+31+16-6-3-2+0=79$

Now there are $441-227+1=215$ numbers total, so subtracting the unsuccessful numbers(79) from that we get $\boxed{136}$ possible values of $x$ .

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Difference between revisions of "2005 PMWC Problems/Problem I3"

Revision as of 11:01, 18 September 2008

Problem

Solution

See also

@@ Line 16: / Line 16: @@
 There are 43 multiples of 5 in that range.
 There are 31 multiples of 7 in that range.
 There are 16 multiples of 13 in that range.
 There are 6 multiples of 7*5=35 in that range.
 There are 3 multiples of 5*13=65 in that range.
 There are 2 multiples of 7*13=91 in that range.
 There aren't any multiples of 455 in that range.