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Please clarify me this doubts - number properties

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Please clarify me this doubts - number properties

by mjjking » Tue Apr 07, 2009 6:24 am
1. Is the sum or the difference of numbers divisibles by a certain number always divisible by that number? (like will always the sum of 2 or more multiples of 7 be divisible by 7? Will always be the same with difference?).
2. If I know that x<y<0 and I know that x^2=(y2-1), will I have to take the positive or negative value for the result? How do I know that?

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by moutar » Tue Apr 07, 2009 6:31 am
1. Say we have the addition of many numbers divisible by a.

ax1 + ax2 + ax3 + ax4 + ... + axn = a(x1 + x2 + x3 + ... + xn)

So the sum of lots of multiples of the same number is still a multiple of that number.

2. I don't really know what you're getting at. Is there a question this came from?
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by mjjking » Tue Apr 07, 2009 7:19 am
yes, the question was: if x^2-y^2 = 7 and x<y<0, then x=?
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by gmat740 » Tue Apr 07, 2009 7:28 am
yes, the question was: if x^2-y^2 = 7 and x<y<0, then x=?
(x+y)(x-y) = 7*1

here two case arise
Case-1
x+y= 7
and x-y = 1

Solve both equations
x= 4 and y = 3

Case-2:


(x+y)(x-y) =1*7

x+y = 1
x-y =7

solve both the equations

we again get x=4 but this time y= -3

since the question asks only about the value of x then clearly x=4


But I am not sure why x<0



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by moutar » Tue Apr 07, 2009 8:36 am
The solutions you need are x = -4 and y = -3.

7 is also -7*-1 so you need to analyse 4 cases.

Case 1:

(x+y)(x-y) =7*1
x+y= 7
x-y = 1

x = 4 and y = 3

Case 2:

(x+y)(x-y) =1*7
x+y = 1
x-y = 7

x = 4 and y = -3

Case 3:

(x+y)(x-y) = -7*-1
x+y=-7
x-y =-1

x = -4 and y = -3

Case 4:

(x+y)(x-y) = -1*-7
x+y=-1
x-y=-7

x = -4 and y = 3

Only the solutions in case 3 satisfy x<y<0 so x=-4 and y=-3
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by gmat740 » Tue Apr 07, 2009 1:37 pm
Thanks Moutar,

I missed the two other cases
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