When the positive integer x is divided by 6, the remainder is 4. Each of the following could also be an integer EXCEPT

- x/2
- x/3
- x/7
- x/11
- x/17

When the positive integer x is divided by 6, the remainder is 4. Each of the following could also be an integer EXCEPT

- x/2
- x/3
- x/7
- x/11
- x/17

is the answer B?

So the number x is of the following form → x=6k+4

- For option 1, we can see any number x is perfectly divisible.
- For option 2, we can see any number x will always leave a reminder of 1
- For option 3, we can find out that for some value of k the number x will be divisible by 7. Just try it with a few numbers and check the pattern to confirm.

First, take k = 0 => x = 4. x/7 leaves a reminder of 4.

Second, take k = 1 => x = 10. x/7 leaves a reminder of 3.

Third, take k = 2 => x = 16. x/7 leaves a reminder 2.

So if you go on, you can eventually find a number divisible by 7 for some k.

Try the same for other options and you’ll find the reminder keeps changing in a particular order implying that it will hit zero at some value of k.

This is how I solved it. Lengthy I know. There might be other methods, we’ll see what our gre mates come up with

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Hey!

@C.Koushik’s method is right. I took a little different approach.

Since, the remainder is 4, I need to look at all the numbers which when divided with 6 would leave a remainder 4. The set would be something like this:

S = {4, 10, 16, 22, 28, 34, 40…}

Now, I just check which option doesn’t fit.

- Fits (every num is divisible by 2)
- Looks like the answer
- Fits
- Fits
- Fits (17 x 2 = 34)

The answer would be **B**.

Good Luck!