If integer C is randomly selected from 20 to 99, inclusive.

We know C^3-C=(C-1)C(C+1), product of 3 consecutive integers.

Now,
19,20,21
20,21,22
21,22,23
22,23,24
23,24,25
...
96,97,98
97,98,99
98,99,100

These are the entire set. Note; the middle value represents "C" and we know 20<=C<=99; (C-1) AND (C+1) are left and right values based on C.

Total count=99-20+1=80

We need to see how many of these sets will be divisible by 4;

If C=odd; left value=C-1=even; C+1=even; Multiplication of 2 Even numbers will always be divisible by 4 because it will contain at least two 2's in its factors.
e.g.
48,49,50
49=odd
48=even
50=even;
So, 48*49*50 must be a multiple of 4 as it has 2 even numbers. Thus, we count all odds C's, for it will make the left and right values even.

Or,
23, 24, 25
Since, 24 is a multiple of 4, it will also be divisible by 4. Thus, we count that too.

But, sets such as
21,22,23: will not be divisible by 4 as 21 AND 23 are odds and don't contain any 2 in their factors. and 22 is not divisible by 4.

forgot to mention:
the product of three consecutive integers will always be divisible by 3, so we don't need to mind that. We just need to make sure that there are at least 2 2's.

C^3-C = C(C-1)(C+1)

Divisibility by 12 requires divisibility by both 4 and 3.

given C-1, C and C+1 are 3 consec. integers, divisibility by 3 is guaranteed. The question remains whether it is divisible by 4.

2 cases:
a) C is odd -> C-1 and C+1 are even which guarantees at least 2 factors of 2 and hence divisible by 4
b) C is even -> only C is even -> divisibility by 4 requires C to be divisible by 4

20-99 inclusive is 80 integers, 40 odd and 40 even. Of those that are even every other one is divisible by 4, that is half of the 40. Therefore there are 20 divisible by 4

Thus the probability is (40+20)/80 = 3/4

How many of the integers from 20 to 99 are either ODD or Divisible by 4.
could you explain why have u take "either ODD"

Thank you

Would you please explain (1) The formulas you used for finding the # of ODD and "Divisible by 4" outcomes?
AND: (2) Why are we looking for the number of ODD integers within 20-99? I'm confused b/c 23 for example is not divisible by 12....

Thank you for the great post

My approach was just two steps:
1) As we have (C-1)*C*(C+1) we can forget about devisibility by 3 and think only of div. by 4;
2) Then I didn't care about the total number of sets but considered that moving through number line and taking triples of sequent numbers we have only one option of four possible not to cover multiple of 4.
So, the answer must be 1-1/4 = 3/4.

I realize that it could be less straightforward if number of possible options was not 80, but in that case denominator of the answer would be smth less simple than 2,3,4,5 as we see in answers.

Bunuel : I am finding these type of question difficult to understand ..
Can you please simplify a bit more ..explaining how we determine the cases in which the given number is a multiple of the given divisor... please

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Ah! I had to take some time to figure out this 1. This working might be helpful to you!

C^3 - C= (c-1)(c)(c+1) --> this is some rule

Any integer (c-1)(c)(c+1) will be divisible by 12, if any of these numbers are factors of 4 and 3, either alone or together.
Now, in a set of any 3 consecutive integers, one will automatically be divisible by 3. So the Q is simplied as whether the integer is a multiple of 4!

Now, there are 2 possibilities - C is even or odd.
a) C is even. In this case 2 numbers (c-1 and c+1) will be odd. This leads to 2 sub possibilities!
(i) C is a multiple of 4 - This means the integer will be divisible by 4. (We will have: odd X multiple of 4 X odd)
(ii) C is not a multiple of 4- This means the integer will NOT be divisible by 4. (We will have: odd X non multiple of 4 X odd)

b) C is odd. two numbers (c-1 and c+1) are divisible by 2 and therefore the set is divisble by 4.

Now the total numbers in the set are 80 (not 79, since 20 and 99 are included). The number of possibilities for C are
(a) (i) --> 99/4 - 19/4 = 24 -4 = 20
(a) (ii) --> 99/2 - 19/2 = 49 -9 = 40; reduce 20 from (a)(i) ==> 20
(b) --> 100/2 - 20/2 = 50 -10 = 40

Sum of (a) (i) + (a) (ii) + (b) --> 20+20+40 = 80 (80 is the total number of numbers between 20 and 99, this means we have considered all possibilities )

Favourable outcomes above (a) (i) + (b) = 60

Therefore probability = 60/80 or 3/4

Answer = C

The answer is C
My solution is longer, time consuming.
The given expression is (c-1) c (c+1), product of three consecutive integer and it is always divisible by 6. However, we have to find whether this is divisible by 12. we need an additional factor of 2.
I used the induction method. When c is 22, 26, 30 and so on up to 98, it is not possible.
There are 20 such values. Therefore, the answer 60/80, which is 3/4.
If anybody has a crisp solution please....

Where I say that induction method is the soul of Math's we should not forget that Induction methods with a little bit of Observation is the finest way to solve the question.

The key is "To question how and when will (c-1) c (c+1) be divisible by 12"

12 = 4*3

i.e. (c-1) c (c+1) will be divisible by 12 if it's divisible by 4 as well by 3

in order to make sure that this is divisible by 4

1) either (c-1) should be divisible by 2 so that (c+1) is also even and their product is divisible by 4 i.e. (c-1) can b any even number from 20 through 99 = (98/2)-(18/2) = 49-9 = 40 cases

2) or if (c-1) is odd then (c+1)will also be odd i.e. c must be the multiple of 4 itself i.e. c can be any multiple of 4 from 20 though 99 = (99/4)-(19/4) = 24-4 = 20 cases

Total favorable cases = 40 + 20 = 60

Hence, Probability = 60/80 = 3/4

I hope this helps!

Question: is What is the probability that\(\frac{C^3 - C}{12}\) ---->\(\frac{C(C^2-1)}{4*3}\) --->\(\frac{(C-1)C(C+1)}{4*3}\)?

(20,21,22,23),(24,25,26,27),(28,29,30,31),......,(96,97,98,99)

In above groups group of three numbers divisible by 12 and the group always excluded one number that is not divisible by 4.( any three consecutive numbers are always divisible by 3)

So the probability is \(\frac{3}{4}\)

Correct Answer C

Posting official solution of this problem.

We are given that an integer C is to be chosen at random from the integers 20 to 99 inclusive, and we need to determine the probability that C3 – C will be divisible by 12.

We should recall that when a number is divisible by 12, it is divisible by 4 and 3. We should also recognize that C3 – C = C(C2 – 1) = C(C – 1)(C + 1) = (C – 1)(C)(C + 1) is a product of three consecutive integers. Furthermore, we should recognize that any product of three consecutive integers is divisible by 3! = 6; thus, it’s divisible by 3. We have to make sure it’s also divisible by 4.

Case 1: C is odd

If C is odd, then both C – 1 and C + 1 will be even. Moreover, either C – 1 or C + 1 will be divisible by 4. Since C(C – 1)(C + 1) is already divisible by 3 and now we know it’s also divisible by 4, C(C – 1)(C + 1) will be divisible by 12.

Since there are 80 integers between 20 and 99 inclusive, and half of those integers are odd, there are 40 odd integers (i.e., 21, 23, 25, …, 99) from 20 to 99 inclusive. Thus, when C is odd, there are 40 instances in which C(C – 1)(C + 1) will be divisible by 12.

Case 2: C is even

If C is even, the both C – 1 and C + 1 will be odd. If C is a multiple of 4 but not a multiple of 3, then either C – 1 or C + 1 will be divisible by 3 (for example, if C = 28, then C – 1 = 27 is divisible by 3, and if C = 44, then C + 1 = 45 is divisible by 3). In this case, C(C – 1)(C + 1) will be divisible by 12.

If C is a multiple 4 and also a multiple of 3, then C is a multiple of 12 and of course C(C – 1)(C + 1) will be divisible by 12. Therefore, if C is even and a multiple of 4, then C(C – 1)(C + 1) will be divisible by 12. So, let’s determine the number of multiples of 4 between 20 and 99 inclusive.

Number of multiples of 4 = (96 – 20)/4 + 1 = 76/4 + 1 = 20. Thus, when C is even, there are 20 instances in which C(C – 1)(C + 1) will be divisible by 12.

In total, there are 40 + 20 = 60 outcomes in which C(C – 1)(C + 1) will be divisible by 12.

Thus, the probability that C(C – 1)(C + 1) will be divisible by 12 is: 60/80 = 6/8 = 3/4.

Answer: C

Hi Bunuel
Thanks for the explanation.
I have a doubt in this.

After creating groups, why are we checking for the even integers not divisible by 4.
Can't we check directly from the set that how many numbers are divisible by 4?

{20,21,22,23}=> Here only 1 number is divisible by 4
{24,25,26,27}=> Here only 1 number is divisible by 4
{28,29,30,31}=> Here only 1 number is divisible by 4
{32,33,34,35}=> Here only 1 number is divisible by 4

Similarly,
{96,97,98,99}=> Here only 1 number is divisible by 4

So p(E) = 1/4

Please tell me where I am wrong.
Thanks in advance

Yes, in {20,21,22,23} only one number is divisible by 4 but (C-1)*C*(C+1) is divisible by 12 also when C = 21 or 23. So, for three numbers out of {20,21,22,23} (C-1)*C*(C+1) IS divisible by 12, for 20, 21, and 23. Similarly for all other sets of four numbers there will be 3 out of 4 numbers for which (C-1)*C*(C+1) will be divisible by 12. P = 3/4.

Hope it's clear.

Hi Bunuel
Do you think this is a 740+ level GMAT question?
There are so many conditions to be thought of. Although your explanation is brilliant, I don't think I could do this under even 3 minutes.
Thanks.

ankitamundhra28
There is no difficulty level bucket like 740+ but if you are asking if this question is very difficult question then you are probably right...