Hi RAA,
The old 74xx series hasn't been produced by many manufacturers for years so I had a hard time finding their datasheets.
You don't say anything about a 5V voltage regulator for those old IC's. They will burnout with a 9V supply.
Each LED has a slightly different voltage, so when they are in parallel, their brightness will all be different. It might be so different that one LED will hog all the current and burnout. Then the rest will burnout one-at-a-time because the current for only a few will be too high. Are you feeling lucky that they all might be the same? :lol:
The current in so many LEDs will kill a little 9V battery in a short time. Plus the high operating current of the old 74xx IC's. You are going to need an AC to 9VDC adapter for power.
But anyway, you could use 2N3906 PNP transistors to drive each group of paralleled LEDs in series with a current-limiting resistor. Connect the transistor's emitter to the positive battery terminal, its collector to the current-limiting resistor, the other end of the resistor to the anodes of the LEDs and the cathodes of the LEDs to ground. The base of each transistor should have a 4.7k resistor in series that is driven by the output of the 7442.
The 1st time you try it use a current-limiting resistor for a single group of LEDs of 270 ohms 1/4W. If all the LEDs in the group are about the same brightness (very dim) without a bright one, add another 270 ohm resistor in parallel with the 1st one and try again. If all LEDs are the same, you can use five 270 ohm 1/4W resistors in parallel for 12 LEDs, or four 270 ohm 1/4W resistors in parallel for 9 LEDs.
Then try the single 270 ohm resistor and parallel more again with every other group of paralleled LEDs.
Next time, think about the theory stuff before the wiring stuff. :lol: