If b is the product of three consecutive positive integers c, c + 1

b = c(c+1)(c+2)
now if c = even, then c and (c+2) are even and b = 8k as the smallest positive even integer is 2 so c+2 = 4. and c+1 = 3.
and if c = odd, then c and (c+2) are odd and (c+1) = even.
So for c = even it will be a multiple of 24.
We need to check for the case c=odd.

Statement 1 -
b is a multiple of 8.
now, for c= even we already know that b = 8k, and it will be a multiple of 24.
we can check as well by taking a few cases (2,3,4), (4,5,6),(6,7,8) and so on.

and if c = odd and b is a multiple of 8 so it implies that c+1 = 8k.
The cases possible are (7,8,9), (15,16,17) and so on and we can see that all will be multiple of 24.

Sufficient.

Statement 2 -
As seen from above just knowing that c is odd doesn't make sure that it will be a multiple of 24.
For cases such as (7,8,9) b will be a multiple of 24
BUT
for cases such as (1,2,3), b will not be a multiple of 24.

Not sufficient.

IMO, Answer is A.

Prime factorization of 24: 2^3 * 3

(1) b is a multiple of 8.

Prime factorization of 8 is 2^3. If b is a multiple of 8, then b has to have 3 2's in it. This means that the absolute lowest possible number that b can be is is (2)(3)(4), which has 3 2's in it. All the numbers that satisfy the criteria of being the product of 3 consecutive numbers and are also multiples of 8, will also be multiples of 24. For instance: 4*5*6, 6*7*8, 7*8*9, etc. So statement 1 is sufficient.

(2) c is odd.

This doesn't tell us anything. If c = 1, then b = 1*2*3, which is not a multiple of 24. If c = 3, then b = 3*4*5, which is a multiple of 24. Not sufficient.

Answer is A.

Any three positive consecutive integer is multiple of 3 and there is one 2 for sure in three consecutive integers. All we need to know is b divisible by 8( 24 = 8 * 3)
Hence A is suff
We do not need value of c. From B value of B is not deductible

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In 3 consecutive integers there will be one integer that is a multiple of 3. Therefore c(c+1)(c+2) = multiple of 3.

The question rephrased: Is b a multiple of 8?

(1) Directly answers our question. SUFFICIENT.

(2) Knowing c is odd is not enough to conclude b = multiple of 8. INSUFFICIENT.

Answer is A.

1) SUFF since if it is a multiple of 8, the product will definitely be a multiple of 3 (because 3 consecutive integers)
Thus, ,3*8 = 24, so will always be of 24n format

2) INSUFF, just plug in numbers
if c is odd, means middle number is multiple of 2. But we need at least 2^3 for it to be 24n
Not all sequences satisfy that so no.
e.g. 15,16,17 will work
BUT 11,12,13 will NOT

B = C(C+1)(C+2), and we want to know if B is a multiple of 24. B will only be a multiple of 24 if the product of C(C+1)(C+2) contains at least three factors of 2 (2^3) and at least one fact of 3 (3^1).

Statement one:

Since we're told that B is a multiple of 8 we can say that B = 8K in which K is integer. This means that 8K = C(C+1)(C+2) which we can further restate as K = C(C+1)(C+2) DIVIDED by 8.

Since K must be an integer then that means that 8 will divide into either of C, C+1, or C+2 evenly, and furthermore since the product of the three numbers is three consecutive integers there must also be a factor of 3 in K. So that means that K contains at least one factor of 3 and three factors of 2, sufficient.

Statement two:
C could be equal to 9 which would be not sufficient, or C could be equal to 7 which would be sufficient. So not sufficient alone.

A is the answer.

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