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K is a set of numbers such that: i. if x is in K, then -x is

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K is a set of numbers such that: i. if x is in K, then -x is [#permalink] New post 16 Jun 2011, 13:01
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K is a set of numbers such that:

i. if x is in K, then -x is in K, and
ii. if each of x and y is in K, then xy is in K

is 12 in K?

1) 2 is in K
2) 3 is in K


if x and y are positive integers, what is the remainder when 10^x + y is divided by 3?

1) x = 5
2) y = 2


the answers to these two problems are giving me a headache, can somebody explain them to me in simpler terms? Thanks!
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Re: I need a better explanation, some help? [#permalink] New post 16 Jun 2011, 13:59
1st question
1) 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 question
1) 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.
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Re: I need a better explanation, some help? [#permalink] New post 16 Jun 2011, 18:34
1 a+b

set will have 2,3 hence 6 and 12 too.

C it is.


2 10^n for n>0 the units digit will always be 0.
hence
10^2 = 100, 10^3 = 1000 or so.

however, 10^2 + 2 = 102/3 = integer
similarly, 10^3 + 2 = 1002/3 = integer again.
and 10^3 + 3 = 1003/3 gives remainder = 1.

thus you can see its the values 2 and 3 determining the remainder 0 and 1 respectively.

applying this logic,you will get B.
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Re: I need a better explanation, some help? [#permalink] New post 16 Jun 2011, 19:40
Expert's post
iheathcliff wrote:
K is a set of numbers such that:

i. if x is in K, then -x is in K, and
ii. if each of x and y is in K, then xy is in K

is 12 in K?

1) 2 is in K
2) 3 is in K


Let's see what the given data implies:
i. if x is in K, then -x is in K, and
So if 4 is in K, -4 is in K too. If -5 is in K, 5 is in K too. Ok.

ii. if each of x and y is in K, then xy is in K
If 2 and 3 are in K, 6 is in K too. Then 2*6 = 12 is in K too. Then 3*6 = 18 is in K too. Then 2*12 = 24 is in K too... etc

Question: Is 12 in K?

Statement 1: 2 is in K
So -2 is in K too. So 2*-2 = -4 is in K too. So -8 is in K too. So 8 is in K too etc. Notice here that all numbers that we see in K are powers of 2. We have no information about 12 yet.

Statement 2: 3 is in K.
Exactly same as statement 1 above.

Using both together: 2 and 3 are in K. So -2 and -3 should be in K too. Also, 2*3 = 6 will be in K. Also, 2*6 = 12 will be in K.
Sufficient.
Answer C
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Re: I need a better explanation, some help? [#permalink] New post 16 Jun 2011, 19:45
psethi wrote:
pike wrote:
1st question
...


I have a doubt in this


Psethi: If you feel there is an error in my solution or comments, then please let me know. I'm not sure quoting my post in its entirety, and saying you have doubts about it is particularly helpful.
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Re: I need a better explanation, some help? [#permalink] New post 16 Jun 2011, 19:47
Expert's post
if x and y are positive integers, what is the remainder when 10^x + y is divided by 3?

1) x = 5
2) y = 2

First thing that comes to mind: What is the divisibility rule of 3?
Sum of all digits should be divisible by 3. What is the remainder when a number is divided by 3? It is same as the remainder you get when you divide the sum of the digits by 3. So, to get the remainder, all you need is the sum of the digits of the given number.

Given number: 10^x + y

If x is a positive integer, what will 10^x look like?
10 or 100 or 1000 or 10000 or 100000 etc. What is the sum of digits in each one of these cases? Of course 1.
Whatever y is, it gets added to this number. The sum of the digits of 10^x + y will depend on the value of y.
Say y = 20. Sum of digits of 10^x + y will be 1 (from before) + 2 = 3
Say y = 7. Sum of digits of 10^x + y will be 1 (from before) + 7 = 8

You can find the sum of the digits the moment you get the value of y. Hence statement 2 is sufficient alone.
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Re: I need a better explanation, some help? [#permalink] New post 16 Jun 2011, 20:28
I didn't even think of the divisibility rules for the second one, nice approach Karishma
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Re: I need a better explanation, some help? [#permalink] New post 17 Jun 2011, 02:54
Thanks to the three of you for taking some time to explain this to me, especially to Karishma for being so detailed with it.

In the first one the use of multiples of 2 and 3 still bugs me a little, but I am going to take it as a rule of thumb and look around for similar problems. The second problem is crystal clear, actually now it looks pretty simple, so I am kind of ashamed.

Again, thank you all for your time. I'll see you around the forum.

-R
Re: I need a better explanation, some help?   [#permalink] 17 Jun 2011, 02:54
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