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by Rahul@gurome » Mon Dec 13, 2010 10:41 pm
So n = 2^3 * y^1 * z^1.
So total number of divisors it will have will be 4*2*2 = 16.
Number of odd divisors is 2*2 = 4.
So number of even divisors is 16 - 4 = 12.
This includes n.
So excluding n, we have 12 - 1 = 11 even divisors.
The correct answer is (C).
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by junegmat221 » Tue Dec 14, 2010 12:08 am
@Rahul,
this method takes odd factors namely
2 to the power 0 ,
y to the power 0 (since y is odd and 2 is the only even prime)
y to the power 1
z to the power 0 (since z is odd and 2 is the only even prime)
z to the power 1

I thought these five would negate the Total number of factors(16) and result in 11...
Am i doing anything wrong here..????


I had doubts on this and Rather plugged in the smalles values in y and z namely 3 and 5 and computed the even divisors of
8 * 3 * 5 which are 2,4,6,8,10,12,20,30,40,50,60 which comes to eleven..But this method really kills my time..
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by Rahul@gurome » Tue Dec 14, 2010 12:25 am
junegmat221 wrote:@Rahul,
this method takes odd factors namely
2 to the power 0 ,
y to the power 0 (since y is odd and 2 is the only even prime)
y to the power 1
z to the power 0 (since z is odd and 2 is the only even prime)
z to the power 1

I thought these five would negate the Total number of factors(16) and result in 11...
Am i doing anything wrong here..????


I had doubts on this and Rather plugged in the smalles values in y and z namely 3 and 5 and computed the even divisors of
8 * 3 * 5 which are 2,4,6,8,10,12,20,30,40,50,60 which comes to eleven..But this method really kills my time..
2^0, y^0 and z^0 are all giving the same value 1.
So they cannot be taken as distinct odd factors.
Distinct odd factors will be 1, y, z and y*z.
They will be 4 in number.
Rahul Lakhani
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
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