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Integer A

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Integer A

by outreach » Fri Jul 02, 2010 10:11 am
Integer A is the product of 4 different prime numbers; integer B is the product of 5 different prime numbers. If the largest common factor of A and B is 30, how many factors does A*B have?
(A) 72
(B) 108
(C) 144
(D) 216
(E) 256
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by kvcpk » Fri Jul 02, 2010 10:26 am
outreach wrote:Integer A is the product of 4 different prime numbers; integer B is the product of 5 different prime numbers. If the largest common factor of A and B is 30, how many factors does A*B have?
(A) 72
(B) 108
(C) 144
(D) 216
(E) 256
A = product of 4 diferent primes
B = product of 5 different primes

Largest common factor of A and B is 30 = 2*3*5

So 2,3,5 are present in both A and B

so A = 2*3*5*P1
B = 2*3*5*P2*P3
So A*B = 2^2*3^2*5^2*P1*P2*P3

Number of divisors of a^n*b^m where a and b are primes is (n+1)(m+1)

So number of divisors = (2+1)(2+1)(2+1)(1+1)(1+1)(1+1)
=27*8 = 216

Hope this helps!!
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by sumanr84 » Fri Jul 02, 2010 11:00 am
kvcpk wrote:
Number of divisors of a^n*b^m where a and b are primes is (n+1)(m+1)

Hope this helps!!
Nice one..
I am on a break !!
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by SM2010 » Fri Jul 02, 2010 11:10 am
kvcpk wrote: Number of divisors of a^n*b^m where a and b are primes is (n+1)(m+1)
How did you deduce this?
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by kvcpk » Fri Jul 02, 2010 11:41 am
SM2010 wrote:
kvcpk wrote: Number of divisors of a^n*b^m where a and b are primes is (n+1)(m+1)
How did you deduce this?
I did not deduce this.. I remember using this property earlier.. So I used it here..

But since, you asked for it, I tried a proof of this concept, Here it comes:

We know that any number can be expressed as a product of primes..

So let us say, n = a^p * b^q * c^r * d^s * e^t * ...... [a,b,c.... are prime numbers]

what will be the divisors of n?

any product of these prime numbers will give a divisor of n. for example: a*b^q is a divisor, similarly a^0*b^q is also a divisor.

so how many sch numbers can i get? answer to this is similar to a P&C problem.

I have p a's, q b's, r c's ... ..

I need to choose any number of a's less than p, any number of b's less than q .....

so I can have 0 a's or 1 a's or 2 a's .. so on upto p

How many are these? (p+1)
on a similar note, for b we will get (q+1)

so the number of divisors is (p+1)(q+1)...

I might have been a bit shabby in my explanation.. Let me know in case you do not understand this..

Praveen
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by SM2010 » Fri Jul 02, 2010 12:25 pm
Ohh ok now I understand it! thanks alot!
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by kvcpk » Fri Jul 02, 2010 12:40 pm
SM2010 wrote:Ohh ok now I understand it! thanks alot!
Glad it helped :)
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by pradeepkaushal9518 » Sat Jul 03, 2010 2:00 am
really kvcpk its great

i never think of this property. u are really helpful to explain the problems. next time i will send u pm as soon as i put any new question to get explanations from u.

i re
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