Problem Solving

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

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

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