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12 Jan 2008, 15:35
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Need more economical way to solve remainder problems.
I generally substitute values and try to figure but just wondering if anybody has even a better way and if somebody can explain step by step.
OA is D ( releasing cause I am not looking for answers )
I generally substitute values and try to figure but just wondering if anybody has even a better way and if somebody can explain step by step.
OA is D ( releasing cause I am not looking for answers )
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12 Jan 2008, 17:18
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Can anyone show me how to use the remainder formula to solve this?
12 Jan 2008, 17:52
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For #1, you have x/2 = q+1 and x/3 = q+0
From this you can get that x is a multiple of 3 (since if you divide x by 3 you get a whole number).
List all multiples of 3, you have 3,6,9,12,etc.
Out of this list, the even numbers when divided by 2 will not have a remainder, so they should be excluded. Example 6/2=3.
So in essence, statement 1 is telling you that x is odd number and is a multiple of 3.
So x can be any of the following 3,9,15,etc.
Try to divide any value of x (for example 3 or 9) by 6 and you will always get a remainder of 3.
So it is sufficient.
For #2, you have x/12 = q+3
From this you can identify possible values of x as 3,15,27 etc.
Since if you divide x by 12 you get a remainder of 3.
Example, divide 3/12. You get remainder of 3. Divide 15/12, you get a remainder of 3.
In all cases, your remainder is 3. So it sufficient.
From this you can get that x is a multiple of 3 (since if you divide x by 3 you get a whole number).
List all multiples of 3, you have 3,6,9,12,etc.
Out of this list, the even numbers when divided by 2 will not have a remainder, so they should be excluded. Example 6/2=3.
So in essence, statement 1 is telling you that x is odd number and is a multiple of 3.
So x can be any of the following 3,9,15,etc.
Try to divide any value of x (for example 3 or 9) by 6 and you will always get a remainder of 3.
So it is sufficient.
For #2, you have x/12 = q+3
From this you can identify possible values of x as 3,15,27 etc.
Since if you divide x by 12 you get a remainder of 3.
Example, divide 3/12. You get remainder of 3. Divide 15/12, you get a remainder of 3.
In all cases, your remainder is 3. So it sufficient.
