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Prime factors

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Prime factors

by msd_2008 » Wed Aug 20, 2008 10:33 pm
What is the number of prime factors in

(6^12 x 35^28 x 15^16)/ 14^12 x 21^11

A.58 B.66 C.112 D. None of these

The OA is 66.....the powers of respective prime factors have been added to arrive at this answer. But why addition? I think we always take product of the powers after adding 1 to each factor.
for eg:- a^2 x b^2 x c^2 where a, b and c are prime numbers, then the total prime factors are (2+1) x (2+1) x (2+1) = 3 x 3 x 3 = 27.
However in this question, the answer has been arrived at after adding 2+2+2
Can anyone please explain what is the right mathematical rule while finding out total prime factors?

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by parallel_chase » Wed Aug 20, 2008 11:14 pm
(6^12 x 35^28 x 15^16)/ 14^12 x 21^11


take the numerator first

6^12 = 2^12 * 3^12
35^28 = 7^28 * 5^28
15^16 = 3^16 * 5^16

denominator

14^12 = 2^12 * 7^12
21^11 = 3^11 * 7 *11

we can cancel 2^12, 7^23 and 3^11

we are left with
3* 5^28 *5^16 * 7^5 * 3^16

1+28+16+5+16 =66
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by msd_2008 » Wed Aug 20, 2008 11:27 pm
I still have a doubt......why did we add the powers of the prime factors to find out the total number of prime factors.....why didwe say that total prime factors is 1+28+16+5+16 =66....why cant it be (1+1)x(28+1)x(5+1) x (16+1)....in several previous postings we actually took product of the powers after adding 1 to each power. I am confused as to what is the exact mathematical rule here?


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by parallel_chase » Wed Aug 20, 2008 11:33 pm
i think you are confused between number of factors and total number of prime factors.

If you need to calculate number of factors for 20

20 = 2*2*5

no. of factors of 20 = 2^2 * 5

2+1 = 3
1+1 = 2

3*2 = 6 factors

How many prime factor does 20 has i.e. 2*2*5. 2^2 * 5 = 2+1 =3

Like the question you posted.

Hope this helps.
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by msd_2008 » Thu Aug 21, 2008 1:16 am
OK...can you answer the following question for me......this will help me understand the concept better. Eventhough this is a DS example, I am still posting it here because this example is related to the same mathematical rule. Please help me understand the reason for your answer choice and if possible, plug in some numbers to highlight.

Q. If the square root of p^2 is an integer, which of the following must
be true?
I. p^2 has an odd number of factors
II. p^2 can be expressed as the product of an even number of prime
factors
III. p has an even number of factors
· I
· II
· III
· I and II
· II and III


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by parallel_chase » Fri Aug 22, 2008 4:23 pm
Sorry for replying so late, almost missed it!!!

I. p^2 has an odd number of factors

p^2 = 4 = 2^2
no. of factors = 2+1 = 3.
TRUE

II. p^2 can be expressed as the product of an even number of prime
factors

p^2=100=2*2*5*5=2^2 * 5^2
TRUE.

III. p has an even number of factors
I think you wanted to say p^2, I'll give you the answer for both.

p=10 = 2*5 = 2^1 * 5^1
no. of factors = (1+1)*(1+1) = 4
TRUE.

P^2=144=2^4 * 3^2
no. of factors= (4+1)*(2+1)=15
FALSE.


If third statement is p then answer would be I, II & III.
If third statement is p^2 then answer would be I, II.

Let me know if you still have any doubts.
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by stop@800 » Fri Aug 22, 2008 9:20 pm
I differ, please see inline



II. p^2 can be expressed as the product of an even number of prime factors

p^2=100=2*2*5*5=2^2 * 5^2
TRUE

lets say p^2 = 4
sp [^2 = 4 = 2^2
hence can't be expressed as the product of an even number of prime factors







III. p has an even number of factors
I think you wanted to say p^2, I'll give you the answer for both.

p=10 = 2*5 = 2^1 * 5^1
no. of factors = (1+1)*(1+1) = 4
TRUE.


lets take p = 4
no of factors = 3
so FALSE
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by parallel_chase » Fri Aug 22, 2008 10:15 pm
stop@800 wrote:I differ, please see inline



II. p^2 can be expressed as the product of an even number of prime factors

p^2=100=2*2*5*5=2^2 * 5^2
TRUE

lets say p^2 = 4
sp [^2 = 4 = 2^2
hence can't be expressed as the product of an even number of prime factors
p^2=4=2*2
product of even number of prime factors.
Therefore TRUE.
III. p has an even number of factors
I think you wanted to say p^2, I'll give you the answer for both.

p=10 = 2*5 = 2^1 * 5^1
no. of factors = (1+1)*(1+1) = 4
TRUE.


lets take p = 4
no of factors = 3
so FALSE
Agree!!

Hence the answer is I & II
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by stop@800 » Sat Aug 23, 2008 4:35 am
II. p^2 can be expressed as the product of an even number of prime factors

Agree
P^2 will always be a product of even number of prime factors [as it is p*P so prime factors will always be even :)]

Thanks for correcting me :)
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