Statement A: Analysis of (12P)3=(123)p
Kisi bhi number system mein, digits hamesha base se chhoti honi chahiye.
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Left side (12P)3 mein, digits {1,2,P} base 3 se chhoti honi chahiye. Iska matlab:
P < 3 \implies P \in \{0, 1, 2\}
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Right side (123)p mein, base p digits {1,2,3} se bada hona chahiye. Iska matlab:
Ab hum equation ko decimal mein convert karte hain:
(1×32)+(2×31)+(P×30)=(1×p2)+(2×p1)+(3×p0)
Agar hum p > 3 ki minimum value p=4 rakhein:
Kyunki P ki value 12 aayi hai jo ki P < 3 ki condition ko todti hai, isliye p ki koi bhi value feasible nahi hai.
Statement A is TRUE.
Statement B: Boolean Simplification
Expression: (P+Q+R)(P+Q+R)(P+Q+R)
Hum Distributive Law ka use karke pehle do terms ko simplify karte hain (P+Q common hai):
(P+Q)+(R⋅R)=(P+Q)+0=(P+Q)
Ab bacha hua expression:
Isse multiply karne par:
P+Q(Q+R)=P+QQ+QR=P+0+QR=P+QR
Statement mein answer (P+Q) diya gaya hai, jo ki galat hai.
Statement B is FALSE.
Statement C: Mod-258 Counter
Mod-N counter ke liye, agar humein n flip-flops chahiye, toh formula hota hai:
Yahan N=258:
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Agar n=8, toh 28=256. Kyunki 256 < 258, isliye 8 flip-flops kaafi nahi hain.
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Agar n=9, toh 29=512. Kyunki 512≥258, isliye 9 flip-flops ki zaroorat hogi.
Statement C is FALSE. (8 nahi, 9 flip-flops chahiye).
Final Answer
Diye gaye options ke hisaab se:
Only Statement A is correct.
