Is the integer n odd ?

[accenture]

(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

I need more explanations on (2) especially.

[sason]

Is the integer n odd ?
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n

To me it looks like statement 2 is alone sufficient to answer the question but not 1

(1) a number divisible by 3 may be even
(2) let suppose the number n is divisible by m positive integers [1 ...... n ]
now if n is odd it will have [ 1 . ...... n] positive integers or have it in powers of prime
2^03^b5^c.......
now the number 2n will have twice integers divisible as we already have 2^0[3^a.5^b....]
numbers now we will also have 2^1[3^a.5^b....] so the total numbers would be exactly
double
2^0[3^a5^b....] & the set consisting of 2^1[3^a.5^b....]
however in case of even n this wont happen as we won get new number on doubling n
( think of it as we already have a combination of atleast 2^0.3^a... and 2^1.3^b... so on multiplying former by 2 would result in the latter that is already present in our set).

do write if there is any better solution or any correction needed to this ...

regards
sason

[ag240000]

we can do it by putting number too:

1. 1,3,6, : 6 is not
2. 3 and 6

for 3: 1,3
6: 1,2,3,6

try for 4 and 8....