dave13 wrote:

santorasantu wrote:

I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3

k = -2 => sum = 3x+1 --> not divisible by 3

k = -1 => sum = 3x --> divisible by 3

k = 2 => sum = 3x-3 --> divisible by 3

k = 5 ==> sum = 3x-6 --> divisible by 3

so answer choice is (b)

hi

pushpitkc in the solution above the product of x*(x-1)*(x-k) is converted into SUM x+x-1+x-k how can that be possible, that we convert PRODUCT of integers into the SUM of integers ?

can you please explain.

thank you and have a good day

Hey

dave13Look at the highlighted portion!

This says - If a number is divisible by 3, the sum of the integers must be a factor of 3

Eg. If the number be 123, the sum of the digits is a factor of 3.

Similarly, if the number is 241 - the sum of the digits is not a factor of 3.

This problem has been explained properly

here.

For clarification regarding the approach, you can see this link

https://gmatclub.com/forum/if-x-is-an-i ... l#p1539020Hope this helps you!

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