What you should know in the back of your head is that any three consecutive numbers multiplied give us a number that is divisible by three. This means that x(x+1)(x+2) will always be divisible by three. Try using some numeric examples to be sure this is true. Anything works here!
So you have x(x-1)(x-k). In order for the numbers to be consecutive, k could be:
a. 2, since x(x - 1)(x - 2) are three consecutive numbers, with x - 2 being the smallest.
b. -1, since x(x - 1)(x + 1) are again 3 cons numbers, with x - 1 the smallest of the batch
So for now we've eliminated C and D. Let's see what we can do about the rest.
1. Say k = -4. This makes it x(x-1)(x + 4) = x(x - 1)[(x + 1) + 3] = x(x - 1)(x + 1) + 3x(x - 1). Since we've already established that x(x - 1)(x + 1) is divisible by 3 no matter the value of x and 3x(x - 1) is obviously divisible by 3, this means that when k = -4, x(x – 1)(x – k) is divisible by 3. This means that we've eliminated A.
2. Say k = -2. So we get x(x - 1)(x + 2), but since x + 2 = x - 2 + 4 = x + 1 + 1, we can conclude that x(x – 1)(x + 2) is not divisible by 3. So the correct answer will be B.
I would advise just stopping here and picking B, but let's test k = 5 just for some extra practice.
We get that x(x – 1)(x – 5) = x(x - 1)(x - 2 - 3) = x(x - 1)(x - 2) - 3x(x - 1). Since both (x - 1)(x - 2) and 3x(x - 1) are divisible by 3, then the number is divisible by 3.
number prop
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Source: Beat The GMAT — Problem Solving |
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sanjay_dce
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well explained DanaJ your answers are as articulate as u areDanaJ wrote:What you should know in the back of your head is that any three consecutive numbers multiplied give us a number that is divisible by three. This means that x(x+1)(x+2) will always be divisible by three. Try using some numeric examples to be sure this is true. Anything works here!
So you have x(x-1)(x-k). In order for the numbers to be consecutive, k could be:
a. 2, since x(x - 1)(x - 2) are three consecutive numbers, with x - 2 being the smallest.
b. -1, since x(x - 1)(x + 1) are again 3 cons numbers, with x - 1 the smallest of the batch
So for now we've eliminated C and D. Let's see what we can do about the rest.
1. Say k = -4. This makes it x(x-1)(x + 4) = x(x - 1)[(x + 1) + 3] = x(x - 1)(x + 1) + 3x(x - 1). Since we've already established that x(x - 1)(x + 1) is divisible by 3 no matter the value of x and 3x(x - 1) is obviously divisible by 3, this means that when k = -4, x(x – 1)(x – k) is divisible by 3. This means that we've eliminated A.
2. Say k = -2. So we get x(x - 1)(x + 2), but since x + 2 = x - 2 + 4 = x + 1 + 1, we can conclude that x(x – 1)(x + 2) is not divisible by 3. So the correct answer will be B.
I would advise just stopping here and picking B, but let's test k = 5 just for some extra practice.
We get that x(x – 1)(x – 5) = x(x - 1)(x - 2 - 3) = x(x - 1)(x - 2) - 3x(x - 1). Since both (x - 1)(x - 2) and 3x(x - 1) are divisible by 3, then the number is divisible by 3.
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ontopofit
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for this we should keep the value of x such that both x and x-1 are not divisible by 3.
say x=5,x-1=4
now from the answer choices we can check whether x-k is divisible by 3 or not
A x-(-4) = 9
B x-(-2) = 7; hence correct.its better that u leave it here and move on if u have less time rem. in the test.
answer B
say x=5,x-1=4
now from the answer choices we can check whether x-k is divisible by 3 or not
A x-(-4) = 9
B x-(-2) = 7; hence correct.its better that u leave it here and move on if u have less time rem. in the test.
answer B
