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DS on OG

This topic has 6 member replies
magical cook Master | Next Rank: 500 Posts
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30 Jul 2006
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DS on OG

Thu Aug 03, 2006 12:47 am
I would appreciate it if someone could explain this why.... t

Jane

If X and Y are positive intergers such that X = 8Y + 12, what is the common divisor of X and Y?

1) X=12u where u is an integer
2) y=12z where z is an integer

The answer is B and only question 2 is correct.

gdhiman Junior | Next Rank: 30 Posts
Joined
31 Jul 2006
Posted:
12 messages
Sat Aug 05, 2006 10:50 pm
dblazquez wrote:
sorry for the delay
hmm, 12u = 8y + 12, with y = 1 is 20 therefore 2 is common divisor, thats why i was confused... i cant see the mistake

daniel
OK.. so i will try to detail it out.:

If X and Y are positive intergers such that X = 8Y + 12, what is the common divisor of X and Y?

1) X=12u where u is an integer
2) y=12z where z is an integer

In the first option, we have 2 equations:
X=12u & X=8Y + 12
Lets make Y=1, we get x=20.
So there is no common divisor of 1 and 20. (2 is not a common divisor).

In the second option, we have 2 equations:
Y=12z & X=96z + 12.
Lets make z=1, We get Y=12 and X=108.
(So now we have a common divisor for every value of z).

dblazquez Senior | Next Rank: 100 Posts
Joined
29 Apr 2006
Posted:
37 messages
1
Sun Aug 06, 2006 6:51 am
Hey thanks a lot for the explanation, now i see it much clearer... i hate this number properties problem... defintely is B

gdhiman Junior | Next Rank: 30 Posts
Joined
31 Jul 2006
Posted:
12 messages
Thu Aug 03, 2006 10:40 am
Good one.

putting in first data to the above equation gets us: 12u = 8y + 12 (u cannot get a common divisor from this).

putting in second data provided we get:
x = 8*12z + 12. So we know 12 divides X for sure, and we also know y is a multiple of 12. So 12 is the common divisor.

Hope i made sense. let me know if u need more clarification.

dblazquez Senior | Next Rank: 100 Posts
Joined
29 Apr 2006
Posted:
37 messages
1
Thu Aug 03, 2006 11:03 am
Hey, i was also trying to solve this good one

Why on the statement I we can state that you cannot get a common divisor from 12u = 8y + 12, what about 2 and four?

gdhiman Junior | Next Rank: 30 Posts
Joined
31 Jul 2006
Posted:
12 messages
Thu Aug 03, 2006 7:58 pm
dblazquez wrote:
Hey, i was also trying to solve this good one

Why on the statement I we can state that you cannot get a common divisor from 12u = 8y + 12, what about 2 and four?
Coz what if y = 1. then 2 is not a common divisor or even 4.

dblazquez Senior | Next Rank: 100 Posts
Joined
29 Apr 2006
Posted:
37 messages
1
Sat Aug 05, 2006 12:59 pm
sorry for the delay
hmm, 12u = 8y + 12, with y = 1 is 20 therefore 2 is common divisor, thats why i was confused... i cant see the mistake

daniel

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