Number of integer values of expression

CAT 2021 Slot 2 · QA · Hard · Number Theory

For all possible integers nn satisfying 2.252+2n+22022.25 \le 2 + 2^{n + 2} \le 202, the number of integer values of 3+3n+13 + 3^{n + 1} is

Answer

7

Explanation

Given inequality: 2.252+2n+22022.25 \le 2 + 2^{n + 2} \le 202 0.252n+22000.25 \le 2^{n + 2} \le 200 222n+22002^{-2} \le 2^{n + 2} \le 200

Taking logarithm base 2: 2n+2log2200-2 \le n + 2 \le \log_2 200 Since 27=128<200<256=282^7 = 128 < 200 < 256 = 2^8, we have 7<log2200<87 < \log_2 200 < 8.

For nn to be an integer: 4n5-4 \le n \le 5

So n{4,3,2,1,0,1,2,3,4,5}n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}.

We need 3+3n+13 + 3^{n + 1} to be an integer. 3+3n+13 + 3^{n + 1} is an integer if n+10    n1n + 1 \ge 0 \implies n \ge -1.

Integer values of n1n \ge -1 are n{1,0,1,2,3,4,5}n \in \{-1, 0, 1, 2, 3, 4, 5\}, which gives 7 values.

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