1st question1) 2 is in K

Initially: K = {2,-2, ...}

Then, pick any two numbers: K = {2,-2,4,-4, ...}

Again: K = {2,-2,4,-4,8,-8, ...}

Not sufficient

2) 3 is in K

K = {3,-3, ...}

K = {3,-3,9,-9, ...}

Not sufficient

1 + 2)

K = {2,-2,3,-3,4,-4, ...}

K contains 3 and 4, hence, 12 is in K.

C.

2nd question1) Clearly knowing x is not sufficient e.g:

x = 1

10 + y could equal 10,11,12,13,14 all of which have different remainders

2) y = 2

Possible numbers are 12,102,1002

Just divide these by three at this stage and see if we can get an answer.

R(12 / 3) = 0

R(102 / 3) = 0

R(1002 / 3) = 0

Sufficient

I guess the question is, why? In the exam I think the above solution is quick and reliable, but let's think about it a bit more. The number will always be of the form:

2*5 + 2

2*2*5*5 + 2

When 2 is divided by 3, the remainder is 2

When 5 is divided by 3 the remainder is 2

We can multiple and add remainders together. The

Manhattan number properties book covers this well, and will explain this better than me. After doing a few practice questions, these become quite easy.

We just have to 'adjust', for example:

Lets think about 2*5 / 3. What is the remainder? Well clearly, the remainder when 10 / 3 is 1. But we know the remainder when 2 / 3 and when 5 / 3 is 2. Yet, when we multiply the remainders together, we get 4, not 1!. We need to divide by three again.

When we multiply remainders together, we must then divide them by the original divisor.

e.g R(2/3) * R(5/3) = R( 2*2/3 ) = 1

So now for a number which has a less obvious remainder:

R(2*2*5*5 / 3) = R(2/3) * R(2/3) * R(5/3) * R(5/3) = R( 2*2*2*2/3 ) = 1

Therefore: R((2*2*5*5 + 2) / 3)) = R((1 + 2) / 3) = 0.