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STEP I 1997 question 3 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1997 question 3 solution

(i) a_1 = 3, a_2 = 27 \equiv 7 \mod 10. So a_3 \equiv 7^3 \equiv 3 \mod 10. So a_4,a_6 \equiv 7 \mod 10, a_5,a_7 \equiv 3 \mod 10. So the last digit of a_7 is 3.

(ii) a_2 = 27 > 10^1 \implies a_3 > (10^1)^3 = 10^3 \implies a_4 > 10^9 \implies a_5 > 10^27 \\ \implies a_6 > 10^{81} \implies a_7 > 10^{343} > 10^{100}.

(iii) \displaystyle \frac{a_7+1}{2a_7+1} = \frac{1}{2}\frac{(2a_7+1)+1}{2a_7+1} = 1/2 + \frac{1}{2}\frac{1}{2a_7+1} < 1/2 + 10^{-100}.

Thus the fraction = 0.50 to 2d.p. (and rather a lot more!).

Solution by DFranklin

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