even divisors

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even divisors

by pappueshwar » Sat Feb 04, 2012 11:28 pm
hi all,
pls assist :


If y is a prime number greater than 5 and x = 10y, how many different positive even divisors does x have, including x ?

Ans choices :

2
3
4
5
6

For me the answer is 3, but the original answer seems to be 4 how?


My solution :

A) prime number greater than 5 is : 7,11,13, 17
B) there fore y = 7,11,13,17 etc
Given : x = 10y

x = 10* 7 = 70
x=10*11= 110
x=10*13 = 130

even divisors for 70 are : 2,10,14,70 (odd divisors : 1,7,5,35)
even divisors for 130 are : 2,10,130 (odd divisors : 1,13,5)

is the question incorrect ? or have i missed any even divisor of 130?
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by African Dude » Sun Feb 05, 2012 1:19 am
pappueshwar wrote:hi all,
pls assist :


If y is a prime number greater than 5 and x = 10y, how many different positive even divisors does x have, including x ?

Ans choices :

2
3
4
5
6

For me the answer is 3, but the original answer seems to be 4 how?


My solution :

A) prime number greater than 5 is : 7,11,13, 17
B) there fore y = 7,11,13,17 etc
Given : x = 10y

x = 10* 7 = 70
x=10*11= 110
x=10*13 = 130

even divisors for 70 are : 2,10,14,70 (odd divisors : 1,7,5,35)
even divisors for 130 are : 2,10,130 (odd divisors : 1,13,5)

is the question incorrect ? or have i missed any even divisor of 130?
Let y =7. Hence x=10(7) = 70. Divisors of 70 (i.e. numbers that can divide through 70 without a remainder) are: 1, 2, and 5,7,10,14,35,70. However, please note that the question is asking about only EVEN divisors. From the list EVEN numbers are 2, 10, 14, and 70 = a total of 4

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by pappueshwar » Sun Feb 05, 2012 2:11 am
HI,
what about if x is 130 ?

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by African Dude » Sun Feb 05, 2012 4:21 am
Hi Pappueshwar,

The answer is still 4 because all factors of 130 are [1, 2, 5, 10, 13, 26, 65, 130] implying that the even divisors are 2, 10, 26 and 130.

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by GMATGuruNY » Sun Feb 05, 2012 4:25 am
pappueshwar wrote:hi all,
pls assist :


If y is a prime number greater than 5 and x = 10y, how many different positive even divisors does x have, including x ?

Ans choices :

2
3
4
5
6

For me the answer is 3, but the original answer seems to be 4 how?


My solution :

A) prime number greater than 5 is : 7,11,13, 17
B) there fore y = 7,11,13,17 etc
Given : x = 10y

x = 10* 7 = 70
x=10*11= 110
x=10*13 = 130

even divisors for 70 are : 2,10,14,70 (odd divisors : 1,7,5,35)
even divisors for 130 are : 2,10,130 (odd divisors : 1,13,5)

is the question incorrect ? or have i missed any even divisor of 130?
The even factors of 130 include 26, omitted from your list above.

x = 2*5*y

Any combination of factors that includes 2 will be even.
Thus, if y=13 -- so that x=2*5*13=130 -- x will have the following even factors:
2
2*5 = 10
2*13 = 26
2*5*13 = 130.
A total of 4 even factors.

The correct answer is D.

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by somsubhra86 » Sun Feb 05, 2012 4:47 am
ans is 4