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Divisibility of integers, remainder. HELP!!!

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by shankar.ashwin » Fri Dec 02, 2011 1:55 am
Consider a number which when divided by 3 leaves a remainder of 2 - say 5.

5 when divided by 4 leaves a remainder of 1 (satisfies the other condition as well)

So, when 5 is divided by 12 - the remainder will be 5.
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by Mr.Hollywood » Fri Dec 02, 2011 2:26 am
shankar.ashwin wrote:Consider a number which when divided by 3 leaves a remainder of 2 - say 5.

5 when divided by 4 leaves a remainder of 1 (satisfies the other condition as well)

So, when 5 is divided by 12 - the remainder will be 5.
Oh wonderful! thanks. So besides plug in a number to try, is there any other method for this?
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by pemdas » Fri Dec 02, 2011 2:55 am
n=3a+2 and n=4b+1, find 'r' for n=12c+r
c<b<a, Always set c=0, you should have n=r, so find n?
to find n, find a or b (these two must be fitting each other)

3a+2=4b+1 OR a=(4b-1)/3 and a>b
b=1, a=(4*1-1)/3=1, condition doesn't work a=b
b=2, a=(4*2-1)/3=7/3, a must be integer
b=3, a=(4*3-1)/3=11/3, doesn't work
b=4, a=(4*4-1)/3=5, good choice

a=5,b=4 and n=3a+2=4b+1=17. Since we set n=r and r=17 which is greater than 12, we have to deduct and receive final remainder 5 (17/12 = 1*12 + 5)

I would use this method for many more numbers like n divided by x. Only we need to have two expressions with known remainders to find 'n' as in our case.
Mr.Hollywood wrote:Integer n is divided by 3 the remainder is 2, when integer n is divided by 4 the remainder is 1, what is the remainder when n is divided by 12?[/b]
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by Mr.Hollywood » Fri Dec 02, 2011 3:13 am
pemdas wrote:n=3a+2 and n=4b+1, find 'r' for n=12c+r
c<b<a, Always set c=0, you should have n=r, so find n?
to find n, find a or b (these two must be fitting each other)

3a+2=4b+1 OR a=(4b-1)/3 and a>b
b=1, a=(4*1-1)/3=1, condition doesn't work a=b
b=2, a=(4*2-1)/3=7/3, a must be integer
b=3, a=(4*3-1)/3=11/3, doesn't work
b=4, a=(4*4-1)/3=5, good choice

a=5,b=4 and n=3a+2=4b+1=17. Since we set n=r and r=17 which is greater than 12, we have to deduct and receive final remainder 5 (17/12 = 1*12 + 5)

I would use this method for many more numbers like n divided by x. Only we need to have two expressions with known remainders to find 'n' as in our case.
Mr.Hollywood wrote:Integer n is divided by 3 the remainder is 2, when integer n is divided by 4 the remainder is 1, what is the remainder when n is divided by 12?[/b]
Now it's much more clear to me! Thank you.
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by LalaB » Fri Dec 02, 2011 4:16 am
n/3=a+2 or n=3a+2 if a =0 then n=2;a=1 n=5

n/4=b+1 or n=4b+1 if b=0 then n=1; if b=1 then n=5

we have 5 in common. then n=5

n/12 is 5/12 , and 5 is remainder

sometimes it is easier to plug in answer choices.u can just plug in numbers to find out a number that will work in both cases.that's all
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by GmatMathPro » Fri Dec 02, 2011 9:18 am
Mr.Hollywood wrote:The question is:

Integer n is divided by 3 the remainder is 2, when integer n is divided by 4 the remainder is 1, what is the remainder when n is divided by 12?

Help guys, and when it comes to this kind of question how to we approach first?

Thanks a million!
In my opinion, listing numbers is clearly the most efficient way to solve these types of problems on the GMAT. As shown above, these situations can be modeled by using equations, but finding the actual solutions still requires a bit of guessing a checking. So I would recommend bypassing the equations and jumping straight to listing numbers.

An integer that has a remainder of 2 when divided by 3: Make a list by starting with the remainder (2) and then repeatedly adding what you're dividing by (3):

2, 5, 8, 11, 14, 17, 20,...

Do the same with the other statement: An integer that has a remainder of 1 when divided by 4:

1, 5, 9, 13, 17,...

5 is the smallest positive number that satisfies both statements. Other numbers that satisfy both statements will be of the form 5+12n, because 12 is the LCM of 4 and 3. All of these have a remainder of 5 when divided by 12.

This approach is fast and efficient because the numbers given are small and easy to work with, and we only have two conditions to consider. As things get more complicated, listing numbers becomes less practical. For more complex situations, mathematicians use something called the Chinese Remainder Theorem. You can read about it more on Wikipedia if you're interested, but if you haven't had at least a course in elementary number theory, you probably won't be able to follow it. And you definitely shouldn't worry about it for the purposes of the GMAT. The point is just that the "more mathematical" approach to these types of problems is fairly advanced and complex, and thus would not be required or expected on the GMAT. Just stick with solving these types of problems by inspection or by listing numbers and be confident that nothing better is expected of you.
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by Mr.Hollywood » Fri Dec 02, 2011 3:31 pm
LalaB wrote:n/3=a+2 or n=3a+2 if a =0 then n=2;a=1 n=5

n/4=b+1 or n=4b+1 if b=0 then n=1; if b=1 then n=5

we have 5 in common. then n=5

n/12 is 5/12 , and 5 is remainder

sometimes it is easier to plug in answer choices.u can just plug in numbers to find out a number that will work in both cases.that's all
Wow, I definitely see it now! Thank you so much!
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