Divisibility and converse statements

Consider the statement "if $M^2$ be divisible by $d$ , then $M$ is divisible by $d$ ".

Case I: $d$ is the multiple of a perfect square, that is, $d = kn^2$ with $n > 1$ .

Let $M = kn$ . Then $M^2 = k^2n^2 = kd$ is divisible by $d$ , but $M = d/n$ is not divisible by $d$ . Therefore the statement is false in this case.

Case II: $d > 1$ is square-free, that is, $d = p_1p_2\cdots p_N$ for distinct prime numbers $p_i$ .

Suppose $M^2$ is divisible by $d$ . Then $M^2$ is divisible by any of the prime numbers $p_i$ . But that is only possible if $M$ itself is divisible by each of those prime numbers. Therefore the statement is true in this case.

Conclusion : Since $11,13$ are prime, they are automatically square-free; since $14 = 2\cdot 7$ , it is square-free. Therefore the statement is true for $d = 11, 13, 14$ . But $12 = 2^2\cdot 3$ is the multiple of a square, so that case I applied: the statement is false, as is obvious when we take $M = 6$ .

The statement "If $M \times M$ is divisible by $12$ , then $M$ must be divisible by $12$ " is not true for $M = 6$ .

In order for the converse to be true, $N$ must not be square free. That means that the correct answer must be $\boxed{12}$ .

Example: If 36 is divisible by 12, then 6 is divisible by 12. False! Since 12=2 * 2 * 3 at least one of its prime divisors squares

M is specified to be an integer. Thus, M^2 is a square number, M^2 has a prime factorization with an even number of each prime, and M has all the same primes in its prime factorization, but half as many of each.

In answer A, 11 is prime. M^2 is stated to be divisible by 11, thus 11 is in its prime factorization more than zero times, thus likewise in that of M. Answer C is similar, 13 also being prime.

Answer D is a bit trickier, but only a bit. 14 breaks down to 2*7, thus to be divisible by 14, M^2 must include both 2 and 7 in its prime factorization, thus M must do so as well, thus M must also be divisible by 14.

So what is different with answer B? The prime factorization of 12 includes the primes 2 and 3, but it has the 2 twice. Thus, establishing that all the same primes are present in the prime factorization of M as in that of 12 is insufficient to establish divisibility, because two copies of 2 are needed, not just one. Thus, M^2 could be divisible, having two 2s, one from each of its constituent Ms, each of which only has one 2, and is this indivisible by 12.

For example, 18's prime factorization contains exactly one 2, and at least one 3, thus 18 squared is divisible by 12, but 18 is not.

Because 36 square of 6 is / by 12 but 6 is not/ by 12

If 11/m^2 then m^2=11a . If m= 11k+u , 0<u<11 , then m^2= 121k^2 +22ku + u^2 = 11a . So 11/u^2 impossible because u^2 = 1 ,4,9,16,25,36,49,64,81,100 . So u=0 , m=11k . The same for 14 , 13 . But for 12 , 12/36 then u=6 . Example k=2 u=6 , m=28 , m^2 = 784 =12.62 but 28=12.2+6 .

There may be a no. M=6k which is not divisible by 12 but M square will be divisible by 12. There is no such a case in 11,13& 14 .