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`        Consider the sequence: a1 = 101, a2 = 10101 , a3 = 1010101, and so on. Then ak is a composite number :(A) Iff k>=2 and 10k+1 +1 | 11(B) Iff k>=2 and 10k+1 -1 | 11(C) Iff k>=2 and k - 2 | 3(D) Iff k>=2`
one year ago

Arun
23506 Points
```							a) is wrong Consider k = 3. 10^(k+1) + 1 = 10^4 + 1 = 10001 Clearly, 11 does NOT divide 10001 since 10001/11 = 909.18 But, a3 = 1010101 is composite since it is divisible by 101. (b) is wrong Consider k = 2. 10^(k+1) - 1 = 10^3 - 1 = 999 Clearly, 11 does NOT divide 999 since 999/11 = 90.82 But, a2 = 10101 is composite since it is divisible by 3. (c) is wrong Consider k = 4. k - 2 = 4 - 2 = 2, which is NOT divisible by 3. But, a4 = 101010101 is composite since it is divisible by 41. So, (d) is your only choice left, even though it might be a bit hard to prove.
```
one year ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions