Hi,
Well you need an output that is either 7 bits or 8 bits, probably 7 bits. This means you need 7 flip flops between pin 4 and 3. That's because that will divide the output signal by 128 and provide the feedback with a signal that is the output divided by 128 which is your input signal.
I guess we didnt explain that part yet. What happens is you provide divide counters between 4 and 3 that divide the output down to the same frequency that you intend to use as input. So if you intend to input 1k and get out 128k then you use a divide counter set that has N flip flops between pins 4 and 3, which N=7 because 2^N=2^7=128 in your case.
The chip compares the divided down signal to your input signal and tries to make them the same phase which means they end up being the same frequency if everything goes right. So when it makes the divided down frequency (via phase) the same as your input frequency (1kHz) that means the output to the counters (input into the counters first stage) must be 128 times the input frequency.
See how easy that is to figure out? So if you need 64 times your input you use N=6 stages because 2^N=2^6=64.
If you only needed twice the frequency you would only need one flip flop stage because 2^N=2^1=2 where the 2 here is the multiplication factor and N=1 so you only need one stage.
If you need four times the frequency then you use two stages (two flip flops).
If you need eight times the frequency then you use three stages (three flip flops).
etc,. etc.
Be aware that some of those big divide chips do not expose all of the bits to the output pins. Sometimes you'll only get certain ones. In your case if you need 6, 7, or 8 flip flops then you are better off using an 8 bit counter chip or use two 4 bit counter chips strung end to end to make up an 8 bit counter and use possibly only 6 or 7 flip flops instead of all 8 internal flip flops.
Make more sense now?
ADDED:
It appears that the 4040 chip you talked about exposes all of the bit outputs to physical pins so it looks like the 4040 would do it yes. Try using the Q7 output. Might also have to change a cap or resistor also from that circuit i posted.