Let n and m be two positive integers such that there are exactly 41 integers greater than 8m and less than 8n,whichcanbeexpressedaspowersof2.Then,thesmallestpossiblevalueofn + m$ is
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Official Correct Answer: A. The question says: Let n and m be two positive integers such that there are exactly 41 integers greater than 8^m and less than 8^n which can be expressed as powers of 2. Then, the smallest possible value of n + m is... and there are options given.
Hmm, okay. So, We find n and m such that between 8^m and 8^n, there are exactly 41 integers that are powers of 2. Then, find the smallest n + m.
First, let me understand what the question is asking. We have two numbers, 8^m and 8^n, and between them, there are 41 integers that are powers of 2. So, these powers of 2 must lie strictly between 8^m and 8^n.
8 is 2^3, so 8^m is (2^3)^m = 2^{3m}, and similarly, 8^n = 2^{3n}. So, the numbers we're looking at are powers of 2, so they can be written as 2^k where k is an integer.
So, the integers greater than 8^m and less than 8^n that are powers of 2 are the numbers 2^k such that 2^{3m} < 2^k < 2^{3n}. Since 2^k is increasing with k, this inequality simplifies to 3m < k < 3n.
Therefore, the number of integers k satisfying 3m < k < 3n is equal to the number of integers between 3m and 3n, not including the endpoints. So, the count is (3n - 1) - (3m + 1) + 1? Wait, no, that's not quite right.
Wait, the number of integers strictly between a and b is b - a - 1. So, if we have k such that 3m < k < 3n, then the number of integers k is (3n - 1) - (3m + 1) + 1? Wait, no, let me think again.
Actually, the number of integers strictly between a and b is (b - a - 1). So, if a = 3m and b = 3n, then the number of integers k is (3n - 1) - (3m + 1) + 1? Wait, no, that's not correct.
Wait, no. Let me think of it differently. If we have integers from a to b inclusive, the count is b - a + 1. But here, we have integers strictly greater than a and strictly less than b. So, the number of integers is (b - 1) - (a + 1) + 1 = b - a - 1.
So, in this case, a = 3m and b = 3n. So, the number of integers k is (3n - 1) - (3m + 1) + 1 = 3n - 1 - 3m - 1 + 1 = 3n - 3m - 1.
Suppose a = 2 and b = 5. Then, integers strictly between 2 and 5 are 3 and 4, which is 2 numbers. Using the formula, (5 - 1) - (2 + 1) + 1 = 4 - 3 + 1 = 2. That works. So, yes, the formula is (b - 1) - (a + 1) + 1 = b - a - 1.
So, in our case, the number of integers k is 3n - 3m - 1. According to the problem, this is equal to 41.
So, 3n - 3m - 1 = 41.
Simplify that: 3(n - m) - 1 = 41.
So, 3(n - m) = 42.
Therefore, n - m = 14.
So, n = m + 14.
Our goal is to find the smallest possible value of n + m. Since n = m + 14, n + m = m + 14 + m = 2m + 14.
To minimize 2m + 14, we need to minimize m. Since m is a positive integer, the smallest possible m is 1.
So, if m = 1, then n = 1 + 14 = 15. Then, n + m = 15 + 1 = 16.
Wait, but let me check if this works. Let me plug m = 1 and n = 15 into the original condition.
Compute 8^m = 8^1 = 8, and 8^n = 8^15. The powers of 2 between 8 and 8^15 are 2^3, 2^4, ..., up to 2^{44} because 8^15 = 2^{45}. Wait, 8^m = 2^{3m}, so 8^1 = 2^3, and 8^15 = 2^{45}. So, the powers of 2 between 2^3 and 2^{45} are 2^4, 2^5, ..., 2^{44}. How many are these?
The exponents k satisfy 3 < k < 45, so k can be 4,5,...,44. The number of terms is 44 - 4 + 1 = 41. Wait, 44 - 4 = 40, plus 1 is 41. So, yes, exactly 41 integers.
So, m = 1 and n = 15 gives us n + m = 16, which is option A.
But Wait, m is a positive integer, so the smallest m is 1. So, n = 15, n + m = 16.
Is there a possibility of a smaller n + m? .
Suppose m = 2, then n = 16, n + m = 18, which is larger than 16.
If m = 0, but m has to be a positive integer, so m cannot be 0.
So, the minimal n + m is 16, which is option A.
Wait, but let me think again. The problem says "integers greater than 8^m and less than 8^n". So, 8^m is not included, and 8^n is not included. So, the