Remainder of positive integer x ?

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Remainder of positive integer x ?

by gander123 » Mon Dec 31, 2012 5:21 am
Got another one:

"What is the remainder when positive integer x is devided by 6?

(1) When x is devided by 2, the remainder is 1; and when x is devided by 3, the remainder is 0.
(2) When x is devided by 12, the remainder is 3."

Correct answer: D

I really have no clue whatsoever how to tackle this problem.
Can anyone help?

Kind regards,

Tobi

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by Param800 » Mon Dec 31, 2012 8:05 am
Hey gander123,

I did like this--

St 1: When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.

It just means that the number should be a multiple of 3 such that when we divide by 2 it leaves remainder 1. So.. such numbers could be 3,9,15..etc
And when we divide any such number with 6, we will always get 3 as remainder.
Thus, A is sufficient

St 2: When x is divided by 12, the remainder is 3.

It means our number should be 12k+3... where k is an integer. Ex`s 15, 27 etc...

When we divide this number by 6 we will always get a reminder of 3 as 12 k will be divided evenly and only 3 will be left.

Thus, B is sufficient

Since, both A and B are sufficient we choose D

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by Anurag@Gurome » Mon Dec 31, 2012 10:30 pm
gander123 wrote:What is the remainder when positive integer x is devided by 6?

(1) When x is devided by 2, the remainder is 1; and when x is devided by 3, the remainder is 0.
(2) When x is devided by 12, the remainder is 3.
Statement 1: This means x is an odd multiple of 3.
Hence, x is of the form 3*(2m + 1) where m is some non-negative integer.
Hence, x = 3*(2m + 1) = 6m + 3 = Some multiple of 6 + 3
Hence, x will leave a remainder of 3 when divided by 6.

Sufficient

Statement 2: x is of the form 12n + 3 where n is some non-negative integer.
Hence, x = 12n + 3 = 6*2n + 3 = Some multiple of 6 + 3
Hence, x will leave a remainder of 3 when divided by 6.

Sufficient

The correct answer is D.
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by GMATGuruNY » Mon Dec 31, 2012 11:47 pm
A quick lesson on remainders:
When x is divided by 5, the remainder is 3.
In other words, x is 3 more than a multiple of 5:
x = 5a + 3.

When x is divided by 7, the remainder is 4.
In other words, x is 4 more than a multiple of 7:
x = 7b + 4.

Combined, the statements above imply that when x is divided by both 5 and 7 -- in other words, when x is divided by 35 -- there will be a constant remainder R.
Put another way, x is R more than a multiple of 35:
x = 35c + R.

To determine the value of R:
Make a list of values that satisfy the first statement:
When x is divided by 5, the remainder is 3.
x = 5a + 3 = 3, 8, 13, 18...
Make a list of values that satisfy the second statement:
When x is divided by 7, the remainder is 4.
x = 7b + 4 = 4, 11, 18...
The value of R is the SMALLEST VALUE COMMON TO BOTH LISTS:
R = 18.

Putting it all together:
x = 35c + 18.

Another example:
When x is divided by 3, the remainder is 1.
x = 3a + 1 = 1, 4, 7, 10, 13...
When x is divided by 11, the remainder is 2.
x = 11b + 2 = 2, 13...

Thus, when x is divided by both 3 and 11 -- in other words, when x is divided by 33 -- the remainder will be 13 (the smallest value common to both lists).
x = 33c + 13 = 13, 46, 79...
Onto the problem at hand:
gander123 wrote:Got another one:

"What is the remainder when positive integer x is divided by 6?

(1) When x is divided by 2, the remainder is 1; and when x is divided by 3, the remainder is 0.
(2) When x is divided by 12, the remainder is 3."

Correct answer: D
Statement 1: When x is divided by 2, the remainder is 1, and when x is divided by 3, the remainder is 0.
List 1: x = 2a + 1 = 1, 3, 5, 7...
List 2: x = 3b = 3, 6, 9, 12...
The smallest value common to both lists is 3.
Thus:
x = 6c + 3 = 3, 9, 15, 21...
When any value here is divided by 6, the remainder in each case will be 3.
SUFFICIENT.

Statement 2: When x is divided by 12, the remainder is 3.
x = 12a + 3 = 3, 15, 27, 39...
When any value here is divided by 6, the remainder in each case will be 3.
SUFFICIENT.

The correct answer is D.
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by gander123 » Wed Jan 02, 2013 4:07 am
Hey,

LIKE! Thanked all your posts :) and most importantly: printed out the quick reminder on remainders...


cheers,

Tobi