For positive integer a, is the product (a)(a + 1)(a + 2) divisible by

There was a typo error earlier. This one is corrected.

for (a)(a + 1)(a + 2) to be divisible by 48...it should also have the factors of 48 i.e. 2^3 & 3.

1) a is even...Insufficient
2) 4a is divisible by 32...that means 'a' is the multiple of 8, so considering 'a' as 8 or 16 or 32 or ..... , the expression (a)(a + 1)(a + 2) will have the required factors 2^3 & 3. Therefore (a)(a + 1)(a + 2) to be divisible by 48...Sufficient

Answer is 'B'.

Question : is the product (a)(a + 1)(a + 2) divisible by 48?

Inference : (a)(a + 1)(a + 2) is a product of three consecutive Positive Integer
48 = 2*3*8
Product of 3 consecutive Integers always include a multiple of 3 so all we have to find is whether product of these three consecutive Integers is divisible by 16 or not

Question REPHRASED : is the product (a)(a + 1)(a + 2) divisible by 16?

Statement 1: a is even.
@a=2, (a)(a + 1)(a + 2) = 24 i.e. NOT divisible by 16 or 48
@a=6, (a)(a + 1)(a + 2) = 6*7*8 i.e. Divisible by 16 and 48
NOT SUFFICIENT

Statement 2: 4a is divisible by 32.
i.e. a is divisible by 8
i.e. (a)(a + 1)(a + 2) will have two even integers a and (a+2) and one is multiple of 8 and other is definitely a multiple of 2
i.e. (a)(a + 1)(a + 2) will be divisible by 8*2*3=16 *3 = 48
SUFFICIENT

Answer: option B

VERITAS PREP OFFICIAL SOLUTION:

Question type: Yes/No. The question asks: “Is the product (a) (a + 1) (a +2) divisible by 48?” You can think ahead of time that in order to be divisible by 48 this product must have the prime factors of 48: 2^4 • 3. So the question is really: Does this product contain at least four 2s and one 3?

Given information in the question stem or diagram: “a is a positive integer.” Since a is a positive integer the rule that “in any three consecutive integers one of those integers will be a multiple of 3” applies. That means that before you even go to the statements you know that the product (a) (a + 1) (a +2) will be a multiple of 3. The question then can be simplified even more from above because you know that the factor of 3 will be present. The simplified question is: Does this product have 2^4 as a factor? It is essential that you always leverage all given information in the question before moving to statements. Also note that this problem (as is true for most arithmetic problems) is best solved with your conceptual understanding of factors and divisibility. While you could prove sufficiency/insufficiency with number picking, it would be cumbersome and risky in this example.

Statement 1: a is even. If a is an even number, it means that a will contain at least one 2 as a factor. It also means that a + 2 will be even and that one of those two even numbers will be a multiple of 4. For example, if x = 2 then (x +2) = 4. This means that you have at least 2^3 as a factor. However, this statement is not sufficient as it only guarantees three 2s in the product and not the required four 2s. Eliminate choices A and D.

Statement 2: If 4a is divisible by 32 then “a” must be divisible by 8. If a contains three 2’s as factors then this information is sufficient as you know that (a + 2) will have to contain at least one 2 as well. This statement is sufficient to prove that the product will contain 2^4 • 3 and the correct answer is B.

Hi Bunuel, did you forget me by the distribution of Kudos ))...

Hi,

A user cannot give more than 5 kudos points to another in a day.

Best regards,
Bunuel.

48= 2^4*3
Since a is a +ve integer and

a, a+1, a+2 are +ve successive integers eoe or oeo. And one of them is multiple of 3

From 1 a is even ... Multiple solutions ( 2*3*4 or 8*9*10)

From 2

A is divisible by 8

A = 8m , a+1 = 8m+1 , a+2= 8m+2

Multiplied together = 64m^2+8m* 8m+2 = 512m^3+ 192m^2+16m + 16m( 32m^2 + 12m+1) ... The expression between brackets whatever the values of m is is a multiple of 3 ... divisible by 48.... B








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If we have three consecutive integers, they must be divisible by 3!=6, hence we only need to find out whether a(a+1)(a+2) is divisible by 2^3, not 2^4, right??

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