Guys,
I don`t understand how to solve this. I`m taking the exam this friday.
DS Remainder
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Statement (1) tells us that x is an odd number (has remainder of 1 when divided by 2) and that x is divisible by 3. If you list out the odd multiples of 3 you will see that they all have a remainder of 3 when divided by 6...
3,9,15,21,27,.....
Therefore, we know that x will have a remainder of 3 when divided by 6.
Statement (2) tells us that x leaves a remainder of 3 when divided by 12. In other words x = 12*k + 3 for some integer k. We can rewrite this expression for x as 6*2*k + 3. Dividing this expression by 6, the first term of the sum will be evenly divisible while the integer 3 will leave a remainder of 3 when divided by 6. Therefore, the number x will have a remainder of 3 when divided by 6. You could also list out numbers that satisfy Statement (2) and see if a pattern develops...
3, 15, 27, 39, ...
all leave a remainder of 3 when divided by 6, so you could probably guess that this will continue.
3,9,15,21,27,.....
Therefore, we know that x will have a remainder of 3 when divided by 6.
Statement (2) tells us that x leaves a remainder of 3 when divided by 12. In other words x = 12*k + 3 for some integer k. We can rewrite this expression for x as 6*2*k + 3. Dividing this expression by 6, the first term of the sum will be evenly divisible while the integer 3 will leave a remainder of 3 when divided by 6. Therefore, the number x will have a remainder of 3 when divided by 6. You could also list out numbers that satisfy Statement (2) and see if a pattern develops...
3, 15, 27, 39, ...
all leave a remainder of 3 when divided by 6, so you could probably guess that this will continue.