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How many numbers can be a FACTOR of 90

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How many numbers can be a FACTOR of 90 [#permalink] New post 04 Apr 2006, 21:21
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How many numbers can be a FACTOR of 90?
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 [#permalink] New post 04 Apr 2006, 21:44
Any easier way to approach such problems??
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 [#permalink] New post 04 Apr 2006, 22:08
I would also appreacite an easier way...anybody!
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 [#permalink] New post 05 Apr 2006, 06:20
Unfortunately there is no quick way around these sort of problems. The most efficient way to solve these is to list two columns and progressively count the factors:

1 90
2 45
3 30
5 18
6 15
9 10

Once the numbers on the left column are about to become larger than the ones on the right column you have determined all factors. Division rules are helpful here.
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 [#permalink] New post 05 Apr 2006, 07:07
Charlie45 wrote:
Unfortunately there is no quick way around these sort of problems. The most efficient way to solve these is to list two columns and progressively count the factors:

1 90
2 45
3 30
5 18
6 15
9 10

Once the numbers on the left column are about to become larger than the ones on the right column you have determined all factors. Division rules are helpful here.


Pretty neat.
This is a more "traditional" way.
get the prime factorization of the number. Then multply every combination of those numbers.
so for 90:
2*3*3*5

so,
2*3 = 6
2*5 = 10
3*3 = 9
3*5 = 15
2*3*3 = 18
2*3*5 = 30
3*3*5 = 45
2*3*3*5 = 90

so, you have 6,9,10,18,15,30,45, and 90.
plus the original 3 prime factors: 2,3 and 5
plus 1 (duh)

answer is 12
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 [#permalink] New post 05 Apr 2006, 07:40
Well there is a neat and short cut way of doing this type of problem

take the prime factor of the number and from that take powers (exponent) of those prime factors add one to each of them and multiply .. the result you get is the factors u r looking for

for example 90 = 2^1 * 3 ^2 * 5^1

so from above exponents are 1 ,2 and 1 respectively
add one to each so it becomes 2,3 and 2

multiply them 2*3*2 = 12 hence 12 factors for 90 ..

the above will work for all number .

I hope i explained it clearly .
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 [#permalink] New post 05 Apr 2006, 08:23
ipc302 wrote:
the above will work for all number .


Well, I have to admit that I was a bit sceptical at first.
But I checked it and you're absolutely right :)

Very good, ipc302 :!:
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 [#permalink] New post 06 Apr 2006, 12:58
12.

90 = 2*45 =2*3*15= 2*3*5*3 =2^1 *3^2 *5^1

add up one to each power and multiply

2*3*2 =12
  [#permalink] 06 Apr 2006, 12:58
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