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KC 1 Q30: Prime Fact / Combo

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by Night reader » Sun Feb 20, 2011 7:36 am
first imagine this band crowd coming up not in horizontal BUT vertical rows, s rows by t musicians
so 240=s*t, next put conditions 8=<t=<30 since s*t=240 it can be also formulated as 8=<s=<30. Basically not combo IOM counting 30-8=22 and 22*22 is huge so already the biggest answer is our choice. Also, prime factorization of 240=2^4 * 3^1 * 5^1, note here odd and even prime factors, if you take only odd then four factors are possible OR four rectangles. Together with one even factor not repeating many times to avoid the same rectangular formations we get 4*2=8. IOM [spoiler]comb/counting gives not concrete answer [/spoiler]
yellowho wrote:How do you do this using combination?

Possible or do you have to just manage a list?
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by gmatmachoman » Sun Feb 20, 2011 12:12 pm
Given condition : 8<=t <=30.

s : Number of rows

t: Number of musicians per row.

s*t = 240.

1. ( 10,24)
2. (24,10)
3. 12,20
4. 20,12

5. (8,30)

6. (30,8)

7 .(10,24)

8. ( 24,10)

Here we are restricted with t less than or equal to 30 & greater than/equal to 8, totalling 8 combinations.

Shortcut : 240 = 8 * 30 ( accounting for the restrictions)
2 ^3 . ( 2*3*5)

Number of factors with this given restriction will be : (3+1) * (1+1)
: 4*2 = 8
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by garuhape » Sun Feb 20, 2011 3:11 pm
gmatmachoman wrote:
Shortcut : 240 = 8 * 30 ( accounting for the restrictions)
2 ^3 . ( 2*3*5)

Number of factors with this given restriction will be : (3+1) * (1+1)
: 4*2 = 8
I do not understand the shortcut, can you please explain it?

Thx
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by mk101 » Sun Feb 20, 2011 8:57 pm
garuhape wrote:
gmatmachoman wrote:
Shortcut : 240 = 8 * 30 ( accounting for the restrictions)
2 ^3 . ( 2*3*5)

Number of factors with this given restriction will be : (3+1) * (1+1)
: 4*2 = 8
I do not understand the shortcut, can you please explain it?

Thx


I dont think the short cut suggested is correct. THE suggested shortcut merely provides the value of the number of factors

of 240 which are also multiples of 8. these 8 are 8,16,24,40,48,60,120,240.

The correct solution to the question according to me is as below -

In this particular case , all you need to do is count the number of factors of 240 from 8 to 30 inclusive. These factors are

8,10,12,15,16,20,24,30 which are 8 in number.

If you dont want to make a mistake in finding the number of factors you could prime factorise 240 and arrive at these factors[/b]
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by gmatmachoman » Sun Feb 20, 2011 9:00 pm
garuhape wrote:
gmatmachoman wrote:
Shortcut : 240 = 8 * 30 ( accounting for the restrictions)
2 ^3 . ( 2*3*5)

Number of factors with this given restriction will be : (3+1) * (1+1)
: 4*2 = 8
I do not understand the shortcut, can you please explain it?

Thx
Generally, to find the number of factors of a given number N we follow this following approach :




X= a ^n * b*m......

No of factors : (n+1) * (m+1) { this is a formula which u can google it for finding the number of factors for a number)


For ex:X = 10

10 = 2 * 5

we have (1+1) *(1+1)
= 4
We have 4 factors for 10.

Same technique was followed above with the given restrictions of 8 <=t<=30
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