The integer x is positive. What is the remainder when the x is divided by 14?
(1) The remainder when 4x is divided by 28 is 12.
(2) The remainder when x is divided by 21 is 3.
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The integer x is positive. What is the remainder when x is d
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Statement 1:eitijan wrote:The integer x is positive. What is the remainder when the x is divided by 14?
(1) The remainder when 4x is divided by 28 is 12.
(2) The remainder when x is divided by 21 is 3.
In other words, 4x is equal to 12 more than a multiple of 28.
In math terms:
4x = 28a + 12, whether a is a nonnegative integer.
Dividing the equation above by 4, we get:
x = 7a + 3.
Options for x:
3, 10, 17, 24, 31...
If x=3, then x/14 = 3/14 = 0 R3.
If x=10, then x/14 = 10/14 = 0 R10.
Since the remainder can be different values, INSUFFICIENT.
Statement 2:
In other words, x is equal to 3 more than a multiple of 21.
In math terms:
x = 21b + 3, whether b is a nonnegative integer.
Options for x:
3, 24, 45...
If x=3, then x/14 = 3/14 = 0 R3.
If x=24, then x/14 =24/14 = 1 R10.
Since the remainder can be different values, INSUFFICIENT.
Statements combined:
Option for x yielded by Statement 1: 3, 10, 17, 24, 31...
Option for x yielded by Statement 2: 3, 24, 45..
Values common to both lists: 3, 24...
If x=3, then x/14 = 3/14 = 0 R3.
If x=24, then x/14 = 24/14 = 1 R10.
Since the remainder can be different values, INSUFFICIENT.
The correct answer is E.
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If we solve above question as:
St1: 4x/28 or 2x/14 gives 6 as remainder. Now, if dividend decreases, remainder also decreases in same proportion. Eg: x= 20 => so (2 * 20)/14 gives 6 as remainder and 20/14 will give 3(12/2) as remainder. So as per this,statement1 can find out what will be the remainder.
Please tell me the error in my approach.
St1: 4x/28 or 2x/14 gives 6 as remainder. Now, if dividend decreases, remainder also decreases in same proportion. Eg: x= 20 => so (2 * 20)/14 gives 6 as remainder and 20/14 will give 3(12/2) as remainder. So as per this,statement1 can find out what will be the remainder.
Please tell me the error in my approach.
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Hi eitijian,
When a question deals with 'remainders', you CANNOT 'reduce' the given fraction because that changes what the remainders could be.
Here, with 4X/28, and the restriction that X is an positive integer, we have the following possible remainders:
X = 1 --> 4/28 = 0r4
X = 2 --> 8/28 = 0r8
X = 3 --> 12/28 = 0r12
X = 4 --> 16/28 = 0r16
X = 5 --> 20/28 = 0r20
X = 6 --> 24/28 = 0r24
X = 7 --> 28/28 = 1r0
X = 8 --> 32/28 = 1r4
Etc.
By reducing the fraction to 2X/14, you CHANGE the possible values of the remainder....
X = 1 --> 2/14 = 0r2
X = 2 --> 4/14 = 0r4
X = 3 --> 6/14 = 0r6
X = 4 --> 8/14 = 0r8
X = 5 --> 10/14 = 0r10
X = 6 --> 12/14 = 0r12
X = 7 --> 14/14 = 1r0
Etc.
In real basic terms, by doing that math, you are NOT answering the question that is asked.
GMAT assassins aren't born, they're made,
Rich.
When a question deals with 'remainders', you CANNOT 'reduce' the given fraction because that changes what the remainders could be.
Here, with 4X/28, and the restriction that X is an positive integer, we have the following possible remainders:
X = 1 --> 4/28 = 0r4
X = 2 --> 8/28 = 0r8
X = 3 --> 12/28 = 0r12
X = 4 --> 16/28 = 0r16
X = 5 --> 20/28 = 0r20
X = 6 --> 24/28 = 0r24
X = 7 --> 28/28 = 1r0
X = 8 --> 32/28 = 1r4
Etc.
By reducing the fraction to 2X/14, you CHANGE the possible values of the remainder....
X = 1 --> 2/14 = 0r2
X = 2 --> 4/14 = 0r4
X = 3 --> 6/14 = 0r6
X = 4 --> 8/14 = 0r8
X = 5 --> 10/14 = 0r10
X = 6 --> 12/14 = 0r12
X = 7 --> 14/14 = 1r0
Etc.
In real basic terms, by doing that math, you are NOT answering the question that is asked.
GMAT assassins aren't born, they're made,
Rich.
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The statement in red is not true in every case.eitijan wrote:If we solve above question as:
St1: 4x/28 or 2x/14 gives 6 as remainder. Now, if dividend decreases, remainder also decreases in same proportion. Eg: x= 20 => so (2 * 20)/14 gives 6 as remainder and 20/14 will give 3(12/2) as remainder. So as per this,statement1 can find out what will be the remainder.
Please tell me the error in my approach.
Consider the following:
If x=10, then dividing by 2 yields a remainder of 0:
10/2 = 5 R0.
If x=5, then dividing by 2 yields a remainder of 1:
5/2 = 2 R1.
Here, the dividend DECREASES from x=10 to x=5, while the remainder INCREASES from 0 to 1.
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You could also use:
S1:
4x = 28*something + 12
x = 7*something + 3
S2:
x = 21*something + 3
Taking the two together, we see that the second case is a subset of the first one: in the first case, x could be 3, 10, 17, 24, ...., but in the second, x can only be 3, 24, ...
Luckily for us, the first two values (3 and 24) give contradictory results, so we're done!
S1:
4x = 28*something + 12
x = 7*something + 3
S2:
x = 21*something + 3
Taking the two together, we see that the second case is a subset of the first one: in the first case, x could be 3, 10, 17, 24, ...., but in the second, x can only be 3, 24, ...
Luckily for us, the first two values (3 and 24) give contradictory results, so we're done!
