1
Question
If Ci,i=1,2,....9 are perfect odd squares, then ∣∣
∣∣C1C2C3C4C5C6C7C8C9∣∣
∣∣ is always a multiple of
If Ci,i=1,2,....9 are perfect odd squares, then ∣∣
∣∣C1C2C3C4C5C6C7C8C9∣∣
∣∣ is always a multiple of
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Solution
The correct option is C 16
C1=(2a1+1)2C2=(2a2+1)2..So on
∣∣
∣
∣∣(2a1+1)2(2a2+1)2(2a3+1)2(2a4+1)2(2a5+1)2(2a6+1)2(2a7+1)2(2a8+1)2(2a9+1)2∣∣
∣
∣∣
C1→C1−C2,C2→C2−C3
∣∣
∣
∣∣4(a1−a2)(a1+a2+1)4(a2−a3)(a2+a3+1)(2a3+1)24(a4−a5)(a4+a5+1)4(a5−a6)(a5+a6+1)(2a6+1)24(a7−a8)(a7+a8+1)4(a8−a9)(a8+a9+1)(2a9+1)2∣∣
∣
∣∣
=16(K)
C1=(2a1+1)2C2=(2a2+1)2..So on
∣∣ ∣ ∣∣(2a1+1)2(2a2+1)2(2a3+1)2(2a4+1)2(2a5+1)2(2a6+1)2(2a7+1)2(2a8+1)2(2a9+1)2∣∣ ∣ ∣∣
C1→C1−C2,C2→C2−C3
∣∣ ∣ ∣∣4(a1−a2)(a1+a2+1)4(a2−a3)(a2+a3+1)(2a3+1)24(a4−a5)(a4+a5+1)4(a5−a6)(a5+a6+1)(2a6+1)24(a7−a8)(a7+a8+1)4(a8−a9)(a8+a9+1)(2a9+1)2∣∣ ∣ ∣∣
=16(K)
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