5 times
Step 1: 2 times => reduce 24 to 6 balls by 4 groups A,B,C, and D
(A=B & B=C) => D? or (A=B & B!=C) => C? or (A!=B & B=C) => A? or (A!=B & B=C) => B?
Step2: 1 or 2 times => reduce 6 to 2 balls by 3 groups E, F, and G
(E=F) => G? or (E!=F & F=G) => E? or (E!=F & E=G) => F?
Step3: two times
case 1: two ball H and I from G, H is same as normal one, and one more time to know I is too light or too heavy
case 2:two ball H and I from E or F, we already knew the abnormal group is too light or too heavy from step 2. only one time is needed to know which one from the abnormal group is too light or too heavy
