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02 Dec 2021, 04:21
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If \(n\) and \(k\) are positive integers, then how many ordered pairs \((n, k)\) satisfy \(n! + 8 = 2^k\) ?
A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. Infinitely many
A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. Infinitely many
02 Dec 2021, 04:21
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Official Solution:
If \(n\) and \(k\) are positive integers, then how many ordered pairs \((n, k)\) satisfy \(n! + 8 = 2^k\) ?
A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. Infinitely many
\(n! + 8 = 2^k\);
\(n! = 2^k-8\);
\(n! = 2^k-2^3\);
\(n! = 2^3(2^{k-3}-1)= 2^3*odd\).
\(n! = 2^3*odd\) means that \(n! \geq 8\), thus \(n \geq 4\).
\(n! = 2^3*odd\) also means that \(2^3\) is the highest power of 2 that can divide \(n!\). This on the other hand means that \(n \leq 5\) (if \(n > 5\), then \(n!\) will have more than three 2's).
So, we get that \(4 \leq n \leq 5\).
If \(n = 4\), then \(k=5\).
If \(n = 5\), then \(k=7\).
Thus, \((n, k)\) can only be \((4, 5)\) and \((5, 7)\)
Answer: C
If \(n\) and \(k\) are positive integers, then how many ordered pairs \((n, k)\) satisfy \(n! + 8 = 2^k\) ?
A. \(0\)
B. \(1\)
C. \(2\)
D. \(3\)
E. Infinitely many
\(n! + 8 = 2^k\);
\(n! = 2^k-8\);
\(n! = 2^k-2^3\);
\(n! = 2^3(2^{k-3}-1)= 2^3*odd\).
\(n! = 2^3*odd\) means that \(n! \geq 8\), thus \(n \geq 4\).
\(n! = 2^3*odd\) also means that \(2^3\) is the highest power of 2 that can divide \(n!\). This on the other hand means that \(n \leq 5\) (if \(n > 5\), then \(n!\) will have more than three 2's).
So, we get that \(4 \leq n \leq 5\).
If \(n = 4\), then \(k=5\).
If \(n = 5\), then \(k=7\).
Thus, \((n, k)\) can only be \((4, 5)\) and \((5, 7)\)
Answer: C
13 Apr 2025, 00:58
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Hi Bunuel, I didn't understand this part
" This on the other hand means that n≤5 "
" This on the other hand means that n≤5 "
