challenge number properties ........ continued

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Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Mon Dec 13, 2010 10:29 pm

If n = 8zy, where z and y are distinct prime numbers greater than 2, how many different positive even divisors does n have, excluding n ?

(A) Nine
(B) Ten
(C) Eleven
(D) Twelve
(E) Thirteen

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Source: Beat The GMAT — Problem Solving | Mon Dec 13, 2010 10:29 pm

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

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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junegmat221: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Wed Apr 28, 2010 4:03 am

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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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

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
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)

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