Schachfreizeit wrote:
"Also it is a multiple of 3." What is it?
Hi
Schachfreizeit,
Thanks for your query.
Let me first quote the statement 2 analysis from zerosleep’s solution. I will then explain it in detail:
zerosleep wrote:
2) k = 3m +2. So, n=K(K+7) => n= (3m+2)(3m+9)= 3(3m+2)(m+3) => multiple of 3.
If m is odd, m+3 is even. Hence , multiple of 2. Also it is a multiple of 3 => multiple of 6
If m is even, 3m+2 is even. Hence , multiple of 2. Also it is a multiple of 3 => multiple of 6
So B is sufficient.
The analysis shows that if we put k = 3m + 2 in n = k (k + 7), we get n = (3m + 2) (3m + 2 + 7).
- That is, n = (3m + 2) (3m + 9) = (3m + 2) (3) (m + 3)
- From this final expression, we can say that n has ‘3’ as one of its factors. So, n MUST be a multiple of 3. ---(I)
Now, recall that we want to check divisibility by 6, which is 2 × 3. So, even though 3 is set, we still need to check the divisibility of n by 2. This will depend on the nature of ‘m’ - we have two possibilities:
- m is ODD:
Then (m + 3), a factor of n, will be (odd + odd) even. And every even number is divisible by 2. So, n MUST be divisible by 2. ----(II)
From (I) and (II), n is divisible by both 2 and 3, and hence, by 6.
Your point of confusion: ”Also, it is a multiple of 3” refers to ‘n’ being divisible by 3 (We got this in (I)).
- m is EVEN:
Then (3m + 2), a factor of n, will be (even + even) even. So, n MUST be divisible by 2. ----(III)
Again, from (I) and (III), n is divisible by both 2 and 3, and hence, by 6.
Overall, from both cases, we get that
n is a multiple of 6. This makes statement 2 SUFFICIENT.
Hope this helps!
Best,
Aditi Gupta
Quant expert,
e-GMAT