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

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

Grade:11

2 Answers

Arun
25750 Points
5 years ago
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.
Deepanjan Das
13 Points
2 years ago
Observe that a_n=100^n+100^(n-1)+...+1
Therefore a_n reduces to (100^(n+1)-1)/99
So a_n=(10^(n+1)-1)(10^(n+1)-1)/99
Observe for n>=2,both the factors in the numerator are greater than 99...do dividing by 99 won't cancel any factor by whole...so a_n is composite for all n>=2

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