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From S1:We can infer that a^3-a=(a-1)*a*(a+1) which are 3 consecutive integers.So they're always divisible by 3 in the range of 'a' given.We just need to prove that any one or more of these will have 5 as a factor too.
S1 gives that 'a' ends with 9 so (a+1) will end with 0.always div by 5.Hence suff.

From S2:we get that 'a' will be in the range of 30s.We could consider 2 simple egs. (30,31,32)-Ans Yes.
While (31,32,33)-And No.Not sufficient.

Option A.

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Here If we write doen A^3-A as A (A+1)(A-1) => we can say that A is sufficient
remember we need to check only 3 values for the pattern of 3
hence 39=> yes
49=> yes
59=> yes
hence A
statement 2 can be discarded as for 39 => yes
else for 32 => NO
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