DS on OG

[magical cook]

I would appreciate it if someone could explain this why.... t

hanks in advance!!!
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]

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]

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]

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]

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

[gdhiman]

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]

Hey thanks a lot for the explanation, now i see it much clearer... i hate this number properties problem... defintely is B