Just get a boxful of 2.4 K 5% resistors and check them with a DMM
Was hoping to find a way without ordering, and waiting for a delivery....
when in series, resistor's resistances add up (so if there are two resistors in series, R1 + R2 gives you total resistance)
Thank you 
You could just use a 5K pot and set it with your DVM - no?
good plan, will see if I have on
I agree with Sparks about using a pot if you need that kind of accuaracy. How about posting a schematic, or describing it?
This is one way to sift the leakage from true gain. You hook up a couple of resistors and a DMM to the device, and the resistors set up conditions you can control to see what is what. If you really want to do this, get a 2.2M resistor and a 2.4K ; better, get one each 2.2M and 2.49K metal film 1% resistors. This will set you back about US$0.30 if you get them from Mouser, and slightly more or less than that from other sources. If you're going to do much of this, get a transistor socket to, so you can easily test a large number of devices.
If you are satisfied with an indication of gain but are willing to settle for lower accuracy, you can carbon film at 5%, but recognize that the accuracy will be less. If you can, get several 2.4K resistors and measure them. You may find one that's closer to 2.472 ohms, which would be ideal. I'm being picky about the ohms because if you get exactly 2.2M and 2472 ohms, and use a 9.0V battery, you'll find that the voltage across the resistor will be numerically equal to the indicated gain! That's why the somewhat odd resistor values, and the discussion on the values. It makes the final numbers on your DMM come out about right - multiply the voltage by 100, and that's the gain.
To do the test, stick the transistor in the socket, and read the DC voltage across the 2.4K resistor. The resistor will convert any leakage current from the transistor into a voltage that you can then read on your meter. A 2472 ohm resistor is 2.472 volts per milliamp, so a milliamp of leakage will cause 2.472 volts to display. That is incredibly too much leakage, so any transistor that does that is not going to be useful for a FF. In fact, although it will differ a bit, any transistor that shows more than a few micro amps of leakage is suspect. Because of the resistor scaling, the indicated value on your meter is "false leakage gain" and will have to be subtracted from the total reading that you do next.
To test the total gain, press the switch that connects the 2.2M resistor to the base. This causes a touch more than 4 microamps of base current to flow in the base. The transistor multiplies this by its internal gain, and the sum of the leakage (which doesn't change with base current) and the amplified base current. If the transistor has a gain of 100 and no leakage, the voltage across the 2.4K resistor is then (4uA)(100)(2472) = 0.9888V - which is almost exactly 1/ 100 of the actual gain. Pretty neat, huh?
But we know that germanium really does have leakage - that's why were doing this little dance in the first place. So, let's say that the device leaks 100uA to start with. We stick the device into the socket, and read the voltage before we press the switch. It reads (100E-6)*(2472) = 247mV. So the leakage is making the meter believe that there's a "gain" of almost 25 with no current into the base at all.
How much leakage is too much? 100uA is common, 200 happens pretty often. More than 300uA means the device is suspicious, and more than 500uA I would say is bad.
Let's say the device really leaks 93uA, and has a gain of 110 - a prime specimen. What happens when we test? We chuck the thing in the socket, and read (93uA)*(2472) = .229V. Then we press the switch, and read 1.330V. To get the real gain, we subtract 0.229V from 1.330V and get 1.101V. The true gain is just 100 times the reading.