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This is something I havent seen before, thank you. Are there other formulas for figuring out if something is divisible by a certain number. I think my problem is that i don't know the best way to approach the questions.
At frist split up the given stem
1) says that x is divisible by 7 - but what about y ? It could be 0, or 1, each yielding a different result. We still don't know so statement 1 is insufficient.
2) xy is divisible by 49, so do we know anything about the respective values? No, xy can be 0 or 49 ... it's not suffcient
1+2) Aren't sufficient as well, since the case that y is 0 can't be excluded
This is one of the easy divisibility questions, will try to find a difficult one to explain the method
Re: whats the best way to approach divisibility questions? [#permalink]
26 Jan 2006, 22:55
I saw this question in an earlier post and always have problems with these.... Thanks! Is x^2*y^4 divisible by 49?
(1) x is an integer divisible by 7 (2) x*y is an integer divisible by 49.
Is x^2*y^4 divisible by 49?
(1) x is an integer divisible by 7
(2) x*y is an integer divisible by 49.
1) If x is an integer divisible by 7, it must be a multiple of 7. x can take values 7, 14, 21, 28... etc. Notice that x will be of the form 7*1, 7*2, 7*3.....7*n
So x^2 = 49n, x^2*y^4 = 49n(y^4). This will always be divisible by 49 since the numerator and denominator cancel out leaving you with just y^4. Y can be a fraction, an integer, a whole number, but it doesn't matter.
2) x*y = 49 -> This will result in (x,y) = (1,49), (49,1), (7,7) and no other combinations. All 3 combinations will be able to cancel out the 49.
I'm not sure what you meant. But what I did previously was work a lot on these type of questions and then fine-tune my own method. There is no hard and fast way to solve such problems (probably the same thing with RC) but you just have to expose yourself to more of such questions and soon you'll feel much more comfortable facing them.