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[#permalink]
I think the answer is E.

i) Students can be divided evenly only if 3n/m is a positive integer. If n is odd then 3 n = odd x odd = odd. For m=even nos. statement 1 is not possible but for some m odd nos it is. Hence (i) is insufficient.

ii) Same as (i). 13n will be odd for n odd nos. and maybe possible for some m odd nos. Hence (ii) is also insufficient.

(i) & (ii) will also be insufficient.

Thus E.
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[#permalink]
I would say B

1) 3xn / m = k with k integer
and we cannot conclude anything at all.

2) 13xn / m = j with j integer
As we know that m < 13, it means that 13 cannot be a prime factor of m

which means all prime factors from m are prime factors of n, in other words n/m is an integer

>>> Sufficient
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[#permalink]
it is B

Statement 1 is insufficient: Though 3n is divisible by m, that doesn’t mean that n is also divisible by m. Consider n=15, m=9 or n=15, m=5.

Statement 2 is sufficient: Because 13 is prime, and 13n is divisible by m, than n must also be divisible by m.
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[#permalink]
I agree with B Also!

Nice question... Makes me feel good.
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Re: A school administrator will assign each student in a group [#permalink]
Can any one provide a much elaborate solution, did not understand the above's....
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Re: A school administrator will assign each student in a group [#permalink]
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