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geerky42

  • 3 years ago

How many numbers in the form \(a^4\), where \(a \in \mathbb{Z}^+\) divide \(3! \times 4! \times 7!\) ?

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  1. anonymous
    • 3 years ago
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    i don't think there are too many since you need primes to the power of 4

  2. anonymous
    • 3 years ago
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    included in \(3!4!7!\) is \(2^7\) and \(3^4\) all other primes are to lower powers

  3. anonymous
    • 3 years ago
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    i am not certain but on the basis of prime factorization i only see \[2^4,3^4,(2\times 3)^4\]

  4. anonymous
    • 3 years ago
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    oops i miscounted!! it is \(2^8\) and \(3^4\)

  5. anonymous
    • 3 years ago
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    so maybe there are 4 all together, \[2^4, 3^4, (2^2)^4,(2\times 3)^4, (2^2\times 3)^4\]

  6. anonymous
    • 3 years ago
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    well that is actually 5, not 4

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