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JMark

  • one year ago

Hi can you please help me to solve this prob? Find the number divisors of 240

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  1. AlexandervonHumboldt2
    • one year ago
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    the first divisor is 2: 240/2=120 120/2=60 60/2=30 30/2=15 15/3=5 5/5=1 each time you divide new number by the it's smallest divisor: thus your divisors are 2, 4, 8, 16, 6, 10, 48, 30, 15, 60, 5, 120, 240, 1, 3, 80 , 24, 12, 20, 40, that's all if i didn't forgot anything prime factorization is 2*2*2*2*3*5

  2. AlexandervonHumboldt2
    • one year ago
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    to find divisors of 240, first write it's prime factorization, then calculate all of the combinations between the prime factors

  3. AlexandervonHumboldt2
    • one year ago
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    @JMark

  4. mom.
    • one year ago
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    what kind of divisors? 2.4 is also a divisor :)

  5. mathmath333
    • one year ago
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    is the answer 20

  6. ganeshie8
    • one year ago
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    \(240 = 2^4*3*5\) So, any divisor of \(240\) will be of form \(2^a*3^b*5^c\), where \(0\le a\le 4\) and \(~0\le b,c\le 1\)

  7. AlexandervonHumboldt2
    • one year ago
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    yeah gane is right

  8. mathmath333
    • one year ago
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    mom=Mars Orbiter Mission

  9. mom.
    • one year ago
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    mom=mother of mankind (;

  10. ganeshie8
    • one year ago
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    Notice that we can choose the exponent \(a\) in \(5\) ways, \(b\) in \(2\) ways, \(c\) in \(2\) ways therefore total number of positive divisors = \(5*2*2=20\)

  11. mathmath333
    • one year ago
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    i learnt this stuff 1 year ago here now i forgot it

  12. mom.
    • one year ago
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    same

  13. AlexandervonHumboldt2
    • one year ago
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    looks like Mister.Mark is not in a hurry to see our explanations @JMark

  14. mathmath333
    • one year ago
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    i just remembered the pattern (a+1)(b+1)(c+1)....=no of divisors

  15. mom.
    • one year ago
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    looks like he did it himself (;

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