TheCATExam
Sign in

CAT 2023 Slot 3 — QA Question 7

IndicesEasy

Let n be any natural number such that 5n-1 < 3n+1. Then, the least integer value of m that satisfies 3n+1 < 2n+m for each such n, is?

Answer & solution

Answer: 5

Solution

Easy

First find every natural number nn that satisfies 5n−1<3n+15^{n-1}\lt 3^{n+1} — there are only finitely many. Then mm must make 3n+1<2n+m3^{n+1}\lt 2^{n+m} hold for all of those nn, so test the worst case (largest nn).

1

Find which nn satisfy 5n−1<3n+15^{n-1}\lt 3^{n+1}. Test successive nn:

n=5: 54=625<36=729 ✓n=6: 55=3125>37=2187 ×\begin{aligned} &n=5:\ 5^{4}=625\lt 3^{6}=729\ \checkmark\\ &n=6:\ 5^{5}=3125\gt 3^{7}=2187\ \times \end{aligned}

The left side grows faster (base 55 vs 33), so it holds only for n=1,2,3,4,5n=1,2,3,4,5.

2

The binding case is the largest nn, namely n=5n=5 (it makes 3n+13^{n+1} biggest, so it is hardest to beat). The required inequality becomes

36<25+m ⇒ 729<25+m3^{6}\lt 2^{5+m}\ \Rightarrow\ 729\lt 2^{5+m}
3

Find the least power of 2 above 729:

29=512<729<1024=2102^{9}=512\lt 729\lt 1024=2^{10}

So we need 5+m≥105+m\ge 10, i.e. m≥5m\ge 5.

4

Check m=5m=5 clears every nn. Since n=5n=5 is the hardest case and m=5m=5 satisfies it, all smaller nn are automatically fine. Hence the least integer mm is

m=5m=5

Least value of m=5m=\mathbf{5}.

"For each such nn" means cover the worst case. Only the largest valid nn matters — solve the inequality just once at n=5n=5 instead of for all five values.

CAT 2023 Slot 3 QA Q7: Let n be any natural number such that 5 n-1 < 3 n+1 . Then, the least integer value of m that satisfies 3 n+1 — Solution | TheCATExam