# Thread: Find a fraction between 97/36 and 96/35 with smallest denominator

1. ## Find a fraction between 97/36 and 96/35 with smallest denominator

Find a fraction between 97/36 and 96/35 which has the smallest denominator. Since 97/36 > 96/35, I formed the following inequality

$\displaystyle \frac {97}{36} < \frac pq < \frac{96}{35} \, .$

Now, appropriately so, we could write the given fractions with common denominators and thusly yield

$\displaystyle \frac {3395}{1260} < \frac pq < \frac{3456}{1260} \, .$

Now this is where I am stuck. The correct answer is p = 19 and q = 7, which corresponds to the fraction 3420/1260.
Anyone got any idea?

2. ## Re: Find a fraction between 97/36 and 96/35 with smallest denominator

Originally Posted by MathCrusader
Find a fraction between 97/36 and 96/35 which has the smallest denominator. Since 97/36 > 96/35, I formed the following inequality

$\displaystyle \frac {97}{36} < \frac pq < \frac{96}{35} \, .$

Now, appropriately so, we could write the given fractions with common denominators and thusly yield

$\displaystyle \frac {3395}{1260} < \frac pq < \frac{3456}{1260} \, .$

Now this is where I am stuck. The correct answer is p = 19 and q = 7, which corresponds to the fraction 3420/1260.
Anyone got any idea?
Hi MathCrusader!

To simplify a fraction, you need to know how the numerators and denominators factorize.
Can you create prime number factorizations of all the numbers you have?

3. ## Re: Find a fraction between 97/36 and 96/35 with smallest denominator

97 is prime.
$\displaystyle 96 = 2^5 \cdot 3$
$\displaystyle 35 = 7 \cdot 5$
$\displaystyle 36 = 3^2 \cdot 2^2$

4. ## Re: Find a fraction between 97/36 and 96/35 with smallest denominator

Originally Posted by MathCrusader
Find a fraction between 97/36 and 96/35 which has the smallest denominator. Since 97/36 > 96/35, I formed the following inequality

$\displaystyle \frac {97}{36} < \frac pq < \frac{96}{35} \, .$

Now, appropriately so, we could write the given fractions with common denominators and thusly yield

$\displaystyle \frac {3395}{1260} < \frac pq < \frac{3456}{1260} \, .$

Now this is where I am stuck. The correct answer is p = 19 and q = 7, which corresponds to the fraction 3420/1260.
Anyone got any idea?
Originally Posted by MathCrusader
97 is prime.
$\displaystyle 96 = 2^5 \cdot 3$
$\displaystyle 35 = 7 \cdot 5$
$\displaystyle 36 = 3^2 \cdot 2^2$
Good!

So you have:

$\displaystyle \frac {97\cdot 35}{35 \cdot 36} < \frac pq < \frac{96 \cdot 36}{35 \cdot 36} \, .$

$\displaystyle \frac {3395}{1260} < \frac pq < \frac{3456}{1260} \, .$

$\displaystyle \frac {3395}{2^2 \cdot 3^2 \cdot 5 \cdot 7 } < \frac pq < \frac{3456}{2^2 \cdot 3^2 \cdot 5 \cdot 7} \, .$

Now we are looking for a number between 3395 and 3456 that has as much factors in common as possible with 1260.

I think this is easiest with just trial and error, combining one or more prime factors into a numbers starting with the lowest possible.
Start with q=2, and see if you can find an acceptable value for p.
Then q=3, q=4, q=5, q=6, q=7... and there you go!