Question

A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the nth day exceeds one million, then the lowest possible value of n is

Answer 1

Given that, on day 1, there are 2 organisms.
On day-2, there are \(2\times2 + 3 = 7\)
and on day-3, there are \(2\times7 +3 = 17....\)
Let us try to form a pattern:
\(2=2+0\)      \((n=1)\)
\(7=4+3\)       \((n = 2)\)
\(17=8+9\ \ \ \  [8+3\times3]\)        \((n = 3)\)
\(37=16+21\ \ \ \ [16 + 3\times7]\)    \((n = 4)\)
\(T(n) = 2^n + 3 (2^{n-1} − 1)\)
We know that,  \(2^{20}= 2^{10} × 2^{10}\) \(=1024 \times 1024\), which is more than \(1\) million.
Now, check for \(n = 19\),
\(2^{19} +3 (2^{18}− 1) = 2^{19}+ 3 ⋅ 2^{18} − 3 = 2 · 2^{19} + 2^{18} − 3 = 2^{20} +2^{18} – 3\), which is more than 1 million.
Now, check for \(n = 18\),
\(2^{18} + 3 (2^{17} − 1) = 2^{18}+ 3 ⋅ 2^{17} − 3 = 2 · 2^{18} + 2^{17} − 3 = 2^{19} + 2^{17} – 3\), which is not more than a million.
\(⇒ n = 19\)

So, the answer is \(19\).