# Thread: [SOLVED] find the value of k

1. ## [SOLVED] find the value of k

Hey again,
This is on page 851 #15 on the SAT OG Book.

If $\frac{n}{n-1}$ multiply $\frac{1}{n}$ multiply $\frac{n}{n+1}$ = $\frac{5}{k}$
for positive integers n and k, what is the value of k?

a) 1
b)5
c) 24
d) 25
e)26

2. Originally Posted by fabxx
Hey again,
This is on page 851 #15 on the SAT OG Book.

If $\frac{n}{n-1}$ multiply $\frac{1}{n}$ multiply $\frac{n}{n+1}$ = $\frac{5}{k}$
for positive integers n and k, what is the value of k?

a) 1
b)5
c) 24
d) 25
e)26

when you multiply the left side out, you get $\frac{n}{n^2-1}$

So you end up with $\frac{n}{n^2-1}=\frac{5}{k}$

For these fractions to be equal, $n=5$, and $k=n^2-1$

So $k=\dots$

I'm sure you take it from here.

--Chris

3. Hello, fabxx!

If $\frac{n}{n-1} \cdot \frac{1}{n} \cdot \frac{n}{n+1} \:= \:\frac{5}{k}$ for positive integers $n$ and $k$, what is the value of $k$ ?

. . $(a)\;1 \qquad(b)\:5 \qquad (c)\;24 \qquad (d)\;25 \qquad (e)\;26$

We have: . $\frac{n}{n^2-1} \:=\:\frac{5}{k}$

We see that $n$ is either 5 or a multiple of 5 (and the fraction is reduced).

The fraction is: . $\frac{n}{(n-1)(n+1)}$
It is comprised of three consecutive numbers, like $\frac{7}{6\cdot8}\,\text{ or }\,\frac{10}{9\cdot11}$
. . Hence, the fraction cannot be reduced.

Therefore: . $n = 5\quad\Rightarrow\quad \frac{5}{5^2-1} \:=\:\frac{5}{k} \quad\Rightarrow\quad k = 24$ ... answer (c)