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Factors of n.

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by neelgandham » Wed Oct 26, 2011 4:21 am
8,10,12,15,16,20,24,30 !

Answer : [spoiler]A)8[/spoiler]
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by user123321 » Wed Oct 26, 2011 4:26 am
is it 8?
quickly checking
with just single different factors
1) 2^3,2^4
with two
2) 2^2*3,2^3*3,2*5,2^2*5,5*3
with three
3) 2*3*5

on the whole 8

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by lenagmat » Wed Oct 26, 2011 11:54 pm
Please, could you explain how you did it quickly?

neelgandham wrote:8,10,12,15,16,20,24,30 !

Answer : [spoiler]A)8[/spoiler]
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by GmatKiss » Thu Oct 27, 2011 12:46 am
lenagmat wrote:If n=2^4*3*5, how many factors of n are greater than or equal to 8 and less than or equal to 30?

a). 8
b). 9
c). 10
d). 12
e). 15
too bad post, is it n=(2^4)*3*5 or 2^(4*3*5)???
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by neelgandham » Thu Oct 27, 2011 1:33 am
lenagmat wrote:Please, could you explain how you did it quickly?
neelgandham wrote:8,10,12,15,16,20,24,30 !

Answer : [spoiler]A)8[/spoiler]
2^4*3*5 = 16*3*5 (Assuming 2^4*3*5 = (2^4)*3*5)= 16*15 = 12*20 = 10*24 = 8*30 = 6*40 (I stopped here) = 5*48 = 4*60 = 3*80 = 2*120 = 1*240 - This is how put on paper.

If you see the first term in the expression 16*15 = 12*20, it is following a decremental pattern. If you could write the number as product of two equal or almost equal numbers then your job is half done(16*15). From there, you just got to decrease the value of the first term by 1 and see if it is a factor of 240(If it can divide 240, without leaving a reminder).

i.e
16*15 = 240, 15 and 16 are factors.Decrease the value of the first term by 1.
15*x =240 => 15*16 = 240.So, 15 and 16 are factors. Now, decrease the value of the first term by 1
14*x =240 Nopes as 14 (7*2) has got one 7 in it and 240 has none.
13*x =240 Nopes as 13 (Prime number) has got one 13 in it and 240 has none.
12*x =240 => 12*20 = 240.So, 12 and 20 are factors. Now, decrease the value of the first term by 1
and so on...
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