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Grade: 11
        
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

Answers : (1)

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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