Let n and m be two positive integers such that there are exactly 41 integers greater than 8ᵐ and less than 8ⁿ, which can be expressed as powers of 2. Then, the smallest possible value of n+m is:
Question:
Let n and m be two positive integers such that there are exactly 41 integers greater than 8ᵐ and less than 8ⁿ, which can be expressed as powers of 2. Then, the smallest possible value of n+m is:
Let n and m be two positive integers such that there are exactly 41 integers greater than 8ᵐ and less than 8ⁿ, which can be expressed as powers of 2. Then, the smallest possible value of n+m is:
Options
Answer: 16
Explanation:
8^k = (2^3)^k = 2^{3k}. Powers of 2 strictly between 8^m and 8^n are 2^t with 3m < t < 3n, i.e. t = 3m+1, 3m+2, …, 3n−1. Count of such t = (3n−1) − (3m+1) + 1 = 3(n − m) − 1. We are given 3(n − m) − 1 = 41 → 3(n − m) = 42 → n − m = 14. To minimize n + m with n − m fixed = 14, choose the smallest positive integer m = 1. Then n = 15 and n + m = 16
Explanation:
8^k = (2^3)^k = 2^{3k}. Powers of 2 strictly between 8^m and 8^n are 2^t with 3m < t < 3n, i.e. t = 3m+1, 3m+2, …, 3n−1. Count of such t = (3n−1) − (3m+1) + 1 = 3(n − m) − 1. We are given 3(n − m) − 1 = 41 → 3(n − m) = 42 → n − m = 14. To minimize n + m with n − m fixed = 14, choose the smallest positive integer m = 1. Then n = 15 and n + m = 16
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