Problem 21 · 2019 Math Kangaroo
Stretch
Number Theory
factorizationperfect-square
Let \(a\) be the sum of all positive factors of 1024 and \(b\) be the product of all positive factors of 1024. (Note that 1 and 1024 are also factors of 1024.) Then which statement holds?
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Answer: B — \((a+1)^{5} = b\)
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Hint 1 of 2
\(1024 = 2^{10}\) has eleven factors: \(2^{0}\) through \(2^{10}\).
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Hint 2 of 2
Their product is 2 raised to \(0 + 1 + \cdots + 10\); compare it with a power of the sum \(a\).
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Approach: use that \(1024 = 2^{10}\) has 11 factors
- The factors of \(1024 = 2^{10}\) are \(2^{0}, \ldots, 2^{10}\), so their product is \(b = 2^{0+1+\cdots+10} = 2^{55}\).
- The sum of all factors is \(a = 1 + 2 + \cdots + 1024 = 2^{11} - 1 = 2047\), so \(a + 1 = 2^{11}\).
- Then \((a+1)^{5} = (2^{11})^{5} = 2^{55} = b\).
- Answer (B) \((a+1)^{5} = b\).
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