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The integer x is positive. What is the remainder when x is d

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by GMATGuruNY » Sat Jan 30, 2016 8:18 am
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.
Statement 1:
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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by eitijan » Sat Jan 30, 2016 8:24 am
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.
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by [email protected] » Sat Jan 30, 2016 9:06 am
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.

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by GMATGuruNY » Sat Jan 30, 2016 3:40 pm
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.
The statement in red is not true in every case.

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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by eitijan » Sun Jan 31, 2016 10:21 am
Thank you Rich and Mitch.
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by Matt@VeritasPrep » Sun Jan 31, 2016 7:09 pm
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!
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