CAT 2023 Slot 1 · Quant · Question 22

The rule is {a}_{n}=2{a}_{n-1}+3, starting from {a}_{1}=2.

A recurrence of the form {a}_{n}=2{a}_{n-1}+c becomes a clean geometric one after shifting by a constant.

Adding 3 to both sides, {a}_{n}+3=2{a}_{n-1}+6=2\left({a}_{n-1}+3 \right).

So the shifted quantity {a}_{n}+3 doubles every day.

On the first day, {a}_{1}+3=2+3=5.

Therefore {a}_{n}+3=5\times{2}^{n-1}, which gives {a}_{n}=5\times{2}^{n-1}-3.

The requirement is 5\times{2}^{n-1}-3>1000000, that is {2}^{n-1}>\frac{1000003}{5}=200000.6.

Now {2}^{17}=131072, which is too small, while {2}^{18}=262144, which clears it.

So n-1=18, giving n=19.

Checking, {a}_{19}=5\times 262144-3=1310717, comfortably above one million, while {a}_{18}=5\times 131072-3=655357 falls short.

Shortcut. The fixed point of a↦2a+3 is a=-3, so measuring from -3 turns the rule into plain doubling. After that it is only a matter of knowing that {2}^{10}\approx{10}^{3}, so {2}^{18} is roughly 260000 — the first power to push 5\times it past a million.