zhanbo wrote:
A positive integer M that divides k^2 but does not divide k must satisfy
M = k * f where f is a positive factor (other than 1）of k.
(1) Only
A factor of 13 can only be obtained if k^2 = P^12 where P is a prime number.
So, k = P^6
The following six numbers can divide K^2 but not k: P^7, P^8, P^9, P^10, P^11, P^12.
Sufficient.
(2) Only
For √k to have a total of 4 factors, it is possible that √k = P^3 where P is a prime number, which is the same case as in (1). There should be six number.
It is, however, also possible to have a total of 4 factors if √k = P * Q where P and Q are two different prime numbers.
In this case, k=P^2*Q^2, k^2=P^4*Q^4.
The following eight numbers can divide K^2 but not k: P^2*Q^3, P^2*Q^4, P^3*Q^2, P^3*Q^3, P^3*Q^4, P^4*Q^2, P^4*Q^3, P^4*Q4.
(The eight can be calculated via 3*3-1 because we can select one from P^2, P^3, P^4 and another one from Q^2, Q^3, Q^4. Then we remove P^2*Q^2)
Not sufficient.
The answer is (A).
Hi,
The solution posted above is almost correct but I think you missed some other nos. which can divide P^4 * Q^4 for eg. P^3, P^4, Q^3 and Q^4 these nos. shall divide y^2 but not y. Please let me know if you agree.