enigma123 wrote:
Is x divisible by 30?
(1) x = k*(m^3 - m), where m and k are both integers > 9
(2) x = n^5 - n, where n is an integer > 9
(1) x = k*(m^3 - m), where m and k are both integers > 9
(2) x = n^5 - n, where n is an integer > 9
Bunuel has already explained the second part. Just another approach :
From F.S 1, we know that x = k*(m-1)*m*(m+1). Now, product of any 3 consecutive integers is ALWAYS divisible by 3.Also, the product of any two consecutive integers is ALWAYS divisible by 2. Thus, x is divisibe by 2 and 3--> x is divisible by atleast 6. Now we have to find out whether x is divisible by 5 or not.For k=m=10, it is. For k=m=13, it isn't.Insufficient.
From F.S 2, we know that x =n^5-n = n(n^2-1)(n^2+1) = (n-1)(n)(n+1)(n^2+1). Just as above, we know that x is atleast divisibly by 6.Again, we have to find out whether it is divisibly by 5 or not-->If there was some way in which I could represent (n^5-n) as a product of 5 consecutive integers, then I would be succesfull in proving that n^5-n is divisibly by 5 also.
Now, assuming n as the median,to get a product of 5 consecutive integers, we would need (n-2) and (n+2)-->(n^2-4)-->(n-1)n(n+1)[(n^2-4)+(4+1)] = (n-2)(n-1)n(n+1)(n+2) + 5(n-1)n(n+1)--> This will be always be divisible by 30.
Note: We can also prove thatn^7-n will ALWAYS be divisible by 7, in a similar way.
n^7-n = n[(n^2)^{3}-1^3] = n(n-1)(n+1)(n^4+n^2+1). We would like to get a product of 7 consecutive integers. Thus, (n-3),(n-2),(n+2) and (n+3)-->(n^2-4)*(n^2-9)
= n^4-13n^2+36 -->(n-1)(n)(n+1)[(n^4-13n^2+36)-36+13n^2+n^2+1] =(n-3)(n-2)(n-1)n(n+1)(n+2)(n+3) + 7*(2n^2-5)*(n-1)(n)(n+1).
There is actually a very famous theorem which states that x^p-x is always divisible by p, provided p is prime and x an integer. It is a different thing that it is beyond the scope of GMAT.