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Factor

by aakash82 » Tue Feb 10, 2009 7:57 pm

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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

Can somebody please explain??
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by sureshbala » Tue Feb 10, 2009 9:23 pm
The answer must be 35.

Given x = 5p+3 and y = 5q+3.

Hence x-y = 5(p-q). So x-y is divisible by 5

Similarly x = 7m+4 and y = 7n+4

Hence x-y = 7(m-n). So x-y is divisible by 7.

Since x-y is divisible by 5 and 7, it must be divisible by LCM(5,7) = 35
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Re: Factor

by x2suresh » Tue Feb 10, 2009 9:28 pm
aakash82 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

Can somebody please explain??
x=5k+3
x=7l+4
y=5m+3
y=7n+4

x>y

x-y= 5 (k-m)
x-y = 7 (l-n)

from the above it is clear that x-y= 35 (constant)

E
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by GMATGuruNY » Wed Jul 05, 2017 4:10 am
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by Brent@GMATPrepNow » Fri Jul 07, 2017 4:54 am
aakash82 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
x2suresh's solution is probably the fastest. However, here's another approach to consider...

When it comes to remainders, we have a nice rule that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
------------------ONTO THE QUESTION!!!----------------
When positive integer x is divided by 5, the remainder is 3
Possible values of x: 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58...

When positive integer x is divided by 7, the remainder is 4.
Possible values of x: 4, 11, 18, 25, 32, 39, 46, 53, 60, 67, ...
So, x can equal 18 or 53 or....

Since y has the same properties as x, y can equal 18 or 53 or....

If x > y, which of the following must be a factor of x - y?
Let x = 53 and let y = 18
So, x - y = 53 - 18 = 35
Only E is a factor of 35

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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Re: Factor

by Scott@TargetTestPrep » Sun Mar 15, 2020 7:48 am
aakash82 wrote: ↑
Tue Feb 10, 2009 7:57 pm
 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

Can somebody please explain??
Solution:

Let’s list some integers that have a remainder of 3 when divided by 5:

3, 8, 13, 18, 23, …

Let’s list some integers that have a remainder of 4 when divided by 7:

4, 11, 18, 25, …

We see that 18 is common to both lists and is the smallest positive integer that satisfies both conditions. Thus, y could be 18. To get a value for x, we can simply add the LCM of 5 and 7, i.e, 35, to 18. So, x could be 53 (notice that 53 also satisfies both conditions). We see that in this case, x - y = 35, and of all the choices, only 35 is factor of x - y.

Alternate Solution:

Since x produces a remainder of 3 when divided by 5, we can write x = 5p + 3 for some integer p.

Since x produces a remainder of 4 when divided by 7, we can write x = 7q + 4 for some integer q.

Since y produces a remainder of 3 when divided by 5, we can write y = 5s + 3 for some integer s.

Since y produces a remainder of 4 when divided by 7, we can write y = 7r + 4 for some integer r.

If we subtract the third equality from the first, we obtain: x - y = 5p - 5s = 5(p - s). Thus, x - y is a multiple of 5.

If we subtract the fourth equality from the second, we obtain: x - y = 7q - 7r = 7(q - r). Thus, x - y is a multiple of 7.

Since x - y is a multiple of both 5 and 7, it is a multiple of the LCM of 5 and 7 as well, namely 35.

Answer: E

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