Doomsday

Algebra Level 1

As some people prophesied it, today marks the end of the world! A deadly virus has been released and contaminated on Day 1 one human being! On the second day, this person will contaminate on extra person. On the third day, both people will infect 2 other and different people. Suppose that, in general, on each subsequent day, every infected person will contaminate a person, which has not yet been contaminated (so, to avoid ambiguities, no 2 infected people can contaminate the same person). Supposing that the world population is at 7 1 0 9 7 * 10^9 individual on Day 1, and that the births following this day are negligible, find X such that Day X marks the very day when every human being is infected!


The answer is 34.

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2 solutions

Allan Zeta
Feb 7, 2015

It can be expressed as a geometric sequence. Find The Xth.term(day). 7.9 × 1 0 9 7.9 \times 10^{9} = 1 × r X 1 1 \times r^{X-1} then convet to logarithmic form. Use chane of base. Tada! answer is 34

Patrick Bourg
Feb 6, 2015

The question can be reformulated as :

What is the smallest positive integer x x such that 2 x 1 7 1 0 9 2^{x-1} \geq 7*10^9 One easily finds that the answer is 34 \boxed{34}

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