If k is an integer greater than 6, all of the following must be divisi

Question

If k is an integer greater than 6, all of the following must be divisible by 3 EXCEPT

A.  k(k + 3)(k - 1)

B.  3k^3

C.  (k+1)(k+5)(k+6)

D.  (k+2)(k-2)(k+3)

E.  k(k+1)(k+2)

chetan2u · Accepted Answer

Since we are working with VARIABLE, We can be sure if the product can be shown as 3 consecutive numbers, by adding /subtracting 3 from any of the terms..

lets see the choices

a) k(k+3)(k-1) 
(k-1)*k*(k+1) are consecutives BUT instead of k+1, we have k+3 in k(k+3)(k-1) ..
and if k+1 is div by 3, k+3 will not and k and k-1 will also be not div by 3..
so this is NOT necessary to be div by 3
 
 b)3k^3 
3 is there in the term, so TRUE

c)(k+1)(k+5)(k+6) 
if K+1 is div by 3, k+1+3=k+4 will also be div by 3
so we have consecutive terms by adding 3 to k+1
k+4, k+5 and k+6..
TRUE

d)(k+2)(k-2)(k+3) 
add 3 to k-2. it becomes k-2+3=k+1..
again 3 consecutive terms k+1, k+2 and k+3
TRUE

e)k(k+1)(k+2) 
3 consecutive terms already
TRUE

Bunuel · Answer

Similar questions to practice: if-n-is-an-integer-greater-than-6-which-of-the-following-56233.html if-x-is-an-integer-then-x-x-1-x-k-must-be-evenly-divisible-126853.html

HKD1710 · Answer

I looked at this question and thought why would somebody make this question and post it here. then I saw that it is from GMATPREP. So now i assume this is one of those simple questions that make you think about your performance while doing the CAT. Algorithm's tricks.

devctg · Answer

I figured it out by taking K=7 and k=8
and option A is not divisible by 3 in case of k=8
so A is a answer.

Thanks !

kimbercs · Answer

I still don't understand this one why it would be choice A.

Bunuel · Answer

The question asks: If k is an integer greater than 6, then all the following  must  be divisible by 3  EXCEPT :

k(k+3)(k-1) (option A) is not necessarily divisible by 6. For example, if k = 8, then k(k+3)(k-1)=8*11*7, which is not divisible by 6. So, the answer is A.

kimbercs · Answer

Got it now. Thank you!

gracie · Answer

k(k+3)(k-1) will not be divisible by 3 when k=3n-1, where n is any positive integer
A

dcummins · Answer

Just test k=11.

Each answer choice but A will lead to a set of numbers divisible by 3.

ScottTargetTestPrep · Answer

FIrst, we eliminate choices B and E because both of those are divisible by 3 (3k^3 clearly has a factor of 3, and k(k+1)(k+2) is a product of 3 consecutive integers).

Now, let’s examine the other choices by letting k = 7.

A) k(k + 3)(k - 1) = 7(10)(6) is divisible by 3.
 
C) (k+1)(k+5)(k+6) = 8(12)(13) is divisible by 3.

D) (k+2)(k-2)(k+3) = 9(5)(10) is divisible by 3.

We see that all these choices are divisible by 3 when k = 7. We need to change k to a different number, so let k = 8.

A) k(k + 3)(k - 1) = 8(11)(7) is NOT divisible by 3.

We see that we’ve found the correct choice. However, let’s show that the other two indeed are divisible by 3.

C) (k+1)(k+5)(k+6) = 9(13)(14) is divisible by 3.

D) (k+2)(k-2)(k+3) = 10(6)(11) is divisible by 3.

Answer: A

whatisthematrix · Answer

I am confused by this question. I used 10 as my "plug in" number since it said greater than 6, and 10 seemed like an easy number to keep track of. For answer A it works out to (10)(13)(9) which equals 1170, which is divisible by 3, unless my math is wrong.

What did I do wrong here?

What did I do wrong here?

chetan2u · Answer

This is a MUST be question. 
Take k=11, then 11*14*10 is not divisible by 3

Paras96 · Answer

A. k(k + 3)(k - 1)

Take k=8,

Then, k(k + 3)(k - 1)=8x11X7 which is not divisible by 3, hence answer A

EnglishAgast · Answer

Hello  Bunuel

Option (A) will be divisible by 3 whenever k = 7, 10, 13, ... (3n+1) form.. and since all of these numbers are greater than 6, there is a possibility that 1/3rd of all numbers greater than 6 are all divisible by 3.

Pls could you help explain how a must-be-true question work here as only 2/3rd of the cases are valid and 1/3rd are not?

Bunuel · Answer

"MUST BE TRUE" ("ALWAYS TRUE"/"IS TRUE") questions:
 
These questions ask which statement is always true for every valid set of numbers. If you can find just one valid set of numbers where a statement is not true, it means the statement is not always true and therefore not the correct answer. So, for 'MUST BE TRUE' questions, the plug-in method is good for eliminating an option, but it does not provide a 100% guarantee that an option is always true. For "MUST BE TRUE" questions, when using the plug-in method, if you find that more than one option appears to be correct for a particular number or set of numbers, try using different numbers to double-check. Reevaluate only those options that were previously considered correct.

"COULD BE TRUE" questions:
 
The questions that ask which of the following statements could be true are different. If you can demonstrate that a statement is true for a specific set of numbers, it implies that the statement could be true and therefore is a correct answer.

You can practice more questions of this type  HERE .

For this particular problem, the question asks: 'All of the following MUST be divisible by 3 EXCEPT:' This means that all the options will be divisible by 3 for  every  value of k, except one option, which will NOT be divisible by 3 for  every  value of k. That correct option may be divisible by 3 for  some  values of k, but not for all.

Hope it helps.

MalachiKeti · Answer

Lets go by Elimination

B is out because it has 3 as a term

E is out because its clearly a product of 3 consecutive integers so it is must that one of them will be a factor

Now lets look at C and D 
Lets test three nos quickly for each option - an odd number (preferably prime and non 3 multiple), an even number (not multiple of 3) and a multiple of 3. This way you will test all possible cases. If it fits for this - it will be applicable for all.

C) (k+1)(k+5)(k+6) Test 7,8,9

D) (k+2)(k-2)(k+3) Test 7,8,9

In each case one term will be factor of 3.

A option is left behind so thats the answer.

Aboyhasnoname · Answer

Every 3rd integer is divisible by 3....0,3,6,9.....
This means...There are 3 cases 
Either K is divisible by 3 or then k+1 is divisible by 3 or then k +2 is divisible by 3.... 
Let's start with option B...Its clearly divisible by 3...so remove it.. 
Then E...It will also be clearly divisible by 3 ....because..we need either k or k+1 or k+2 to make it divisible....by 3... So E also out... 
Let's go to D
(k+2)(k-2)(k+3) 
Now if K is divisible by 3 then k+3 is also divisble by 3. 
if K+1 is divisible by 3 then K+1-3 is also divisible by 3 means k-2 is divisble by 3 
if k+2 is divisible by 3...so not matter what the number is ...one of them is a multiple of 3..we dont care which one is it..

Now C... 
if K is divisble by 3 then k+6 is divisible...
if k +1 is divisible by 3 then its there... 
if k+2 is divisble by 3 then ...k +5 is also divisible by 3... 
hence no matter what the number is C will be divisible by 3... Eliminate

Left with A ... 
if k is divisbile by 3 ...then fine.. 
if k +1 is divisble by 3...then...k+3 can't be divisible...and k-1 can't be divisible... 
So..its not must be true..it could be ...but question asks MUST..so A is the answer

If k is an integer greater than 6, all of the following must be divisi

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If k is an integer greater than 6, all of the following must be divisi

Prep Toolkit

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My Rewards