Hmm. Indeed you are correct. My bad!!! How could I missed that, in some earlier post to engin, I already asked him to consider first 1/multiple of the given number as the factor of the given number. And now I missed myself .
lol
Okay then lets try to solve it again.
Lets say n = 18/5 and its not an integer. Then n/18 would not be an integer.
Hence, statement 1 is alone not sufficient.
For statement 2, we have already proved that in my earlier posts.
Lets try to take both statements together.
If n=18/5, then 5n/18 is an integer. OK
but putting n=18/5 in 3n/18 will contradict the 2nd statement. Hence n can't be equal to 18/5.
Statement 1, also reveals that n could be 18x/5 where x = 1, 2, 3, ....
and
Similarly Statement 2, reveals that n could be 18y/3 where y = 1, 2, 3....
Now we need to write a series of 18x/5 and 18y/3 and need to find 1st value where it satisfy both the statements.
The first value would be 270/15. (You can check that, as the denominator of both statements are different, so you need to make equal by multiplying and dividing 1st series by 3 and second series by 5)
The 2nd value would be 540/15...
So the common series would be 270/15, 540/18, 810/15.....
When we simplify it, then it would reduced to 18, 36, 54..... which are multiple of 18.
So, that is the thing that we need to prove that n is the multiple of 18.
Hence, n/18 is an integer, proved by both statements.
Now Enjoy
