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factors and multiples

by alltimeacheiver » Wed May 11, 2011 5:20 am
When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7,
the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when
y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of
x - y?
A. 12
B. 15
C. 20
D. 28
E. 35

I got 20 ans but the ans is 35.
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by taneja.niks » Wed May 11, 2011 6:28 am
If u try and find out the numbers that saisfy the conditions

18 and 53 are the two numbers

X>y so, X=53, Y=18

X-Y = 35

So the number that must be the factor of 35 are 1,5,7,35 but out of these we have only 35 available in the answer choices:-)

Hope this helps
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by Brent@GMATPrepNow » Wed May 11, 2011 6:43 am
alltimeacheiver wrote:When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7,
the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when
y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of
x - y?
A. 12
B. 15
C. 20
D. 28
E. 35

I got 20 ans but the ans is 35.
When dealing with questions involving remainders, it's often useful to be able to list possible values of the dividend (aside: here, the dividends are x and y).

So, if x is positive (you won't see negative numbers used in GMAT remainder questions) and if we get a remainder of 3 when x is divided by 5, then possible values of x include:
x : 3, 8, 13, 18, 23, 28, 33, 38, 42, 48, 53, 58, 63, etc
So, x can be any number in this infinite list.

Now the second piece of information restricts possible values of x. We're told that when x is divided by 7, the remainder is 4

So, let's examine our list of possible values for x, and see which ones are such that we get a remainder of 4 when divided by 7.

Of the possible values for x (3, 8, 13, 18, 23, 28, 33, 38, 42, 48, 53, 58, 63, etc), we see that 18 satisfies this condition, 53 satisfies this condition, and so on.

So, x could equal 18, 53, as well as other numbers.

Aside: we could continue to list numbers but two numbers will suffice for this question.

Since y has the same two conditions, we know that y can equal 18, 53, and so on.

Now if x > y, then one possible set of values is x=53 and y=18

This means that x - y = 35, in which case only answer choice E (35) is a factor/divisor of x-y.

Aside: Now there are other possible values for x and y that we could have chosen, but since the numbers we used yielded a correct answer (and since there cannot be two correct answers on the GMAT), E must be correct.

Cheers,
Brent
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by MAAJ » Wed May 11, 2011 10:23 am
Hi, is there other way to find two values of x, other than plugging numbers? I could only find one value with my crappy logic:

x = 5q + 3 -> So the MIN value of x is 8
x = 7q + 4 -> So the MIN value of x is 11

So x must be divisible by 8 and also divisible by 11, hence x must be 88, which satisfy both conditions. But how can I get another number for x using this approach...?
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by MAAJ » Wed May 11, 2011 10:39 am
NVM I should have done the following:

x = 5q + 3
x = 7k + 4

5q + 3 = 7k + 4
5q - 7k = 1

Find values that satisfy this condition:
5(3) - 7(2) = 1
5(10) - 7(7) = 1

x = 5(3) + 3 = 18; x = 7(2) + 4 = 18
x = 5(10) + 3 = 53; x = 7(7) + 4 = 53
MAAJ wrote:Hi, is there other way to find two values of x, other than plugging numbers? I could only find one value with my crappy logic:

x = 5q + 3 -> So the MIN value of x is 8
x = 7q + 4 -> So the MIN value of x is 11

So x must be divisible by 8 and also divisible by 11, hence x must be 88, which satisfy both conditions. But how can I get another number for x using this approach...?
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by Whitney Garner » Wed May 11, 2011 10:57 am
MAAJ wrote:Hi, is there other way to find two values of x, other than plugging numbers? I could only find one value with my crappy logic:

x = 5q + 3 -> So the MIN value of x is 8
x = 7q + 4 -> So the MIN value of x is 11

So x must be divisible by 8 and also divisible by 11, hence x must be 88, which satisfy both conditions. But how can I get another number for x using this approach...?
Be very careful with your logic here:

If:
x = 5q + 3 -> So the MIN value of x is 3
**q can = 0**

x = 7p + 4 -> So the MIN value of x is 4
**again p could = 0, also, avoid using the same variable to represent the constant because we do not know that the value of the constant multiplier is the same for both expressions**

But just because these are the Min potential positive values for x, it does not mean that x is a multiple of their product. Instead it only means that x must be something that satisfies BOTH conditions.

The issue with a "clean" method to product values for x is the presence of a remainder. If x was a clean multiple of 4 and 7, then yes - all possible values for x must have 4 and 7 as factors. BUT, the introduction of a remainder means that we have introduced a linear relationship into the variable, not just a multiplicative one.

For example:
The number 18, when divided by 5 leaves a remainder of 3, but when divided by 7 leaves a remainder of 4. Notice that while 3 happens to be a factor of 18, 4 is not. The next number that satisfies the condition is 53 (remainder when divided by 5 is 3, and remainder when divided by 7 is 4). But 53's only factors are itself and 1 because it is prime.

Unfortunately the only reasonable way to solve this problem is Brute Force...start listing numbers until you find a pattern (if one exists).

Listing the values for "division by 5 leaves a remainder of 3" lets us see that we are going to pick up all the numbers with a 3 or 8 as the units digit. (3,8,13,18,23,...). So we can stop fairly quickly and just look for values where "division by 7 leaves a remainder of 4" and where the units digit is a 3 or an 8. We first hit 18, then the next is 53.

Given that the difference x - y MUST be divisible by only one of the choices in the list, we can try these. Let X=53 and Y=18, then 53-18=35. Checking the answer choices, 35 is not divisible by A, B, C, or D so the only possible answer is E.

:)
Whit
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by pankajks2010 » Sun May 15, 2011 10:35 pm
alltimeacheiver wrote:When positive integer x is divided by 5, the remainder is 3; and when x is divided by 7,
the remainder is 4. When positive integer y is divided by 5, the remainder is 3; and when
y is divided by 7, the remainder is 4. If x > y, which of the following must be a factor of
x - y?
A. 12
B. 15
C. 20
D. 28
E. 35

I got 20 ans but the ans is 35.
As per the question, x-3 is divisible by 5 and x-4 is divisible by 7
Similarly, y-3 is divisible by 5 and y-4 is divisible by 7

Now to obtain x-y we do the following substraction:
A=(x-3)-(y-3) & B=(x-4)-(y-4)

As the term A is derived from two multiples of 5, the difference is also a multiple of 5
similary, B is a multiple of 7

However, A=B, thus, A & B both should be multiples of 5 & 7 which is technically a multiple of LCM of 5 & 7. Thus, the minimum value is 35. Thus, D is the answer.
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by kareem.firoz » Mon May 16, 2011 3:30 am
Is this a good approach?

When positive integer x is divided by 5, the remainder is 3. => x+2 is divisible by 5--------(1)
When x is divided by 7,the remainder is 4 => x+4 is divisible by 7--------------------(2)
When positive integer y is divided by 5, the remainder is 3 => y+2 is divisible by 5-------(3)
When y is divided by 7, the remainder is 4=> y+4 is divisible by 7----------------(4)

Now, (1)-(3)=> x-y is divisible by 5
& (2)-(4) => x-y is divisible by 7

From the above, conclude that x-y should be divisible by 7*5 ie 35
E
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by Brent@GMATPrepNow » Mon May 16, 2011 6:02 am
kareem.firoz wrote:Is this a good approach?

When positive integer x is divided by 5, the remainder is 3. => x+2 is divisible by 5--------(1)
When x is divided by 7,the remainder is 4 => x+4 is divisible by 7--------------------(2)
When positive integer y is divided by 5, the remainder is 3 => y+2 is divisible by 5-------(3)
When y is divided by 7, the remainder is 4=> y+4 is divisible by 7----------------(4)

Now, (1)-(3)=> x-y is divisible by 5
& (2)-(4) => x-y is divisible by 7

From the above, conclude that x-y should be divisible by 7*5 ie 35
E
Hi kareem.firoz,

That's an AWESOME solution - very inventive!
Great work.

Brent
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by Sarran » Mon May 16, 2011 3:52 pm
Kareem,

your first line about x divided by 5 is correct, but not the second one.
When x is divided by 7, the remainder is 4 => how can x+4 will be divisible by 7? Should it not be x-4? Or x+3?

I thought it is a Typo, but found the same with Y divided by 7 as well.


Thanks,
Sarran
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by kareem.firoz » Mon May 16, 2011 5:06 pm
Sarran wrote:Kareem,

your first line about x divided by 5 is correct, but not the second one.
When x is divided by 7, the remainder is 4 => how can x+4 will be divisible by 7? Should it not be x-4? Or x+3?

I thought it is a Typo, but found the same with Y divided by 7 as well.


Thanks,
Sarran
Sarran,
Thanks for pointing it out. First a typo, followed by copy/ paste/ edit error; thats what it was!
Its x+3 and y+3 divisible by 7. Makes no difference in the end result, though.

thanks once again.
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