## Problem Of The Week 12

**Let us have a series of numbers 16,1156,111556, …. . The series is formed by k ones, (k-1) fives and one 6 for k>0. Then each number of the series is a**

**A) Prime **

**B) Perfect Square **

**C) Prime Square **

**D) Atleast one of the foregoing**

**E) None of the foregoing**

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Tk = (11….k)*10^k + (55…k-1)*10 +6

= [(10^k-1)*10^k + 5(10^k-1 – 1)*10 + 54]/9

= [10^2k +4*10^k +4]/9

=(10^k+2)^2/9

=[(10^k+2)/3]^2

(10^k+2) will always be divisible by 3. Hence each term of the series is perfect square.

Apart from that tk = [(3….k-1times)*10+4]^2. Hence only B could be the answer

AmitSeptember 12, 2008 at 8:11 pm