# Question: K Divided by 67

## Comment on K Divided by 67

### So the first equation is set

So the first equation is set up with

(1) K = 2345*n1 + 89

and the second is

(2) K = 67*n2 + R

with n1 and n2 being quotients that are not given, and R is what you're trying to find.

In the 2nd approach, you're setting n1=0, so K = 2345*0 + 89 = 89. But if you set K=89 in (2) and make n2=0, you get 89 = 67*(0) + R, which makes R= 89. Why do you set the second quotient, n2=1, in order for this to work?

### I'm not setting any quotients

I'm not setting any quotients here. The quotients vary, depending on what value of K you choose to use.

We're told that, when K is divided by 2345, the remainder is 89. So, K can equal 89 or 2434 or 4779 or....

If K = 89, then we're saying that the quotient is 0 when we divide K BY 2345. That's different from the quotient when we divide K BY 67

Likewise, when we divide 17 by 3, we get a quotient of 5 with remainder 2. When we divide 17 by 9, we get a quotient of 1 with remainder 8. The quotients are different in these two cases because we're dividing by a different number each time.

### Very tricky question. Will we

Very tricky question. Will we get this type on real exam? I'm scared :).

### Tricky indeed!

Tricky indeed!

The GRE is computer adaptive, so if you answer a bunch of difficult questions, then it's possible to encounter a question like this.

Cheers,
Brent

### OMG :) Ok, Thanks Brent.

OMG :) Ok, Thanks Brent.

### I think, finding the smallest

I think, finding the smallest value of possible dividend in these type of questions basically solves the question.

### That's certainly the case

That's certainly the case with this question.
More importantly, finding the smallest possible dividend will typically speed up your calculations.

### Hi Brent,

Hi Brent,

I tried for the latter approach but I am worried that each possible dividend might have a different remainder. So if I encountered this kind of question, plug into possible dividend from the smallest one will be always correct?

### I'm hesitant to say testing

I'm hesitant to say testing values will work for ALL questions of this nature, since it's hard to define which questions fall into this category. That said, testing values is often a useful approach.

### Hi Brent

Hi Brent
for these kind of questions, if there is only one condition like x is divided by 2345 and remainder is 89
I am just calculating the 2345 multiples and adding 89 to that taking that result as X

That is 2345*2 = 4690
Now add 89 to 4690 = 4779

If same 4779 is divided by 67 then we can easily find the remainder as 22

Is this the right approach

Thanks

### That approach certainly works

That approach certainly works. However the point of the question was to show that it is much easier to work with the smallest possible value of K.

For example, from the given information, it COULD be the case that K = (2345)(279) + 89
In this case, K = 654,344
From here, when we divide 654,344 by 67, we get a remainder of 22.

However, as you can imagine, it is MUCH easier to answer the question if we use the smallest possible value of K.
Since K divided by 2345 leaves a remainder of 89, it COULD be the case that K = 89
From here, when we divide 89 by 2345, we get a remainder of 22.

Also, it can take a while to determine the remainder when a number is divided by 67.
For example, you wrote "4779 is divided by 67 then we can easily find the remainder as 22"
What calculations are you performing to determine that the remainder is 22?

Cheers,
Brent