Question 80803
what are the factors of the number 2400? and how many are there in total?

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Break 2400 down into prime factors with a factor tree:

2400 = 2·2·2·2·2·3·5·5

It's easy to answer the last question.  There are only
3 prime factors of 2400. These are 2,3,and 5.  So
every factor is of the form

2<sup>P</sup>3<sup>Q</sup>5<sup>R</sup>

where P can take on values 0,1,2,3,4,5, that's 6 factors
      Q can take on values 0,1, that's 2 factors
      R can take on values 0,1,2, that's 3 factors

So that's a total of 6·2·3 = 36 factors.

But to get all those 36 factors is a lot of busy work:

We make all the combinations of those factors:

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Combinations of one factor:

1, 2, 3, and 5

This accounts for 4 factors

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Combinations of two factors:

2·2, 2·3, 2·5, 3·5. 5·5 or

This accounts for 5 factors

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Combination of three factors:

2·2·2, 2·2·3, 2·2·5, 2·3·5, 2·5·5, 3·5·5

This accounts for 6 factors

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Combinations of four factors:

2·2·2·2, 2·2·2·3, 2·2·2·5, 2·2·3·5, 2·2·5·5, 2·3·5·5 

This accounts for 6 factors

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Combinations of five factors:

2·2·2·2·2, 2·2·2·2·3, 2·2·2·2·5, 2·2·2·3·3, 2·2·2·3·5, 2·2·2·5·5, 2·2·3·5·5

This accounts for 7 factors

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Combinations of six factors:

2·2·2·2·2·3, 2·2·2·2·2·5, 2·2·2·2·3·5, 2·2·2·3·5·5

This accounts for 4 factors

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Combinations of seven factors:

2·2·2·2·2·3·5, 2·2·2·2·2·5·5, 2·2·2·2·3·5·5

This accounts for 3 factors

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Combinations of eight factors:

2·2·2·2·2·3·5·5

This accounts for 1 factor.

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That's  4 + 5 + 6 + 6 + 7 + 4 + 3 + 1 = 36 factors, so we have them
all. These are, in order:

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 32, 40, 48, 50,
60, 75, 80, 96, 100, 120, 150, 160, 200, 240, 300, 400, 480, 600, 800,
1200, 2400

Edwin</pre></b>