12 Days of Christmas GMAT Competition - Day 5: If k is an integer from

I would first calculate the denominator, which would be the total number of possible outcomes from 48 to 84, both inclusive. Nonetheless this is not much of an advantage since the different solutions have common denominators, which is 84 - 48 +1 = 37

Denominator = 37
Numerator = Total desired outcomes, now we want to know the total number of ways in which (k - 1)(k + 1)(k + 3) is divisible by 48.
48 = 2^4 * 3 = 16 * 3

Taking it in steps:

(k−1),(k+1),(k+3) are consecutive odd numbers. Their product will always include:

At least one multiple of 3 (since every third odd number is divisible by 3)

At least one multiple of 22 (as odd numbers are surrounded by even numbers)

[*]The additional condition for divisibility by 16=2^4 is that one of the terms must provide at least four factors of 2.
Multiples of 16 in the range are: 48, 64 and 80. We calculate the number of possible values so that at least one of k-1, k+1 or k+3 includes one of these.

For example:
k = 49 is one option
k -1 = 48
k+1 = 50
k+3 = 52
another option
k = 61
k + 3 = 64
k -1 = 60
k + 1 = 62
or
k = 65
k -1 = 64
k+1 = 66
k + 3 = 69

All these contain multiples of 16 and a multiple of 3, always.

counting we have 19 total outcomes
thus P((k-1)(k+1)(k+3) to be a multiple of 48) = 19/37 answer (E)

If k is an integer from 48 to 84, inclusive, what is the probability that (k - 1)(k + 1)(k + 3) is divisible by 48?

A. 9/37
B. 12/37
C. 15/37
D. 18/37
E. 19/37

The integers are consecutive even or odd integers. We cannot have k=even as all integers will be add and hence wont be divisible by 48. Thus k=odd. Odd integers in the range = (83-49)/2+1= 18

Since n consecutive even integers are divisible by 2^n (n!) hence all the combinations of even integers will be divisible by 48 since 48=2^4+3 and 2n *(2n+2)*(2n+4) will be divisible by 2^3 *(3!)

Ans D

k must be odd as:
(odd - odd) * (odd + odd) * (odd + odd) = even * even * even = even

and only an even number is divisible by 48.

Total numbers between 48 and 84 are 84 - 48 = 36 + 1 = 37
Total odd numbers between 48 and 84 are 83 - 49 = 34/2 = 17 + 1 = 18

Probability that a number is divisible by 48 is 18/37

[D] is the correct answer.

48 =< k =< 84

Total number of possinbilities of k = 37

(k - 1)(k + 1)(k + 3) is divisible by 48

Assume A = (k - 1)(k + 1)(k + 3)

If k was 48, A = 47 * 49 * 51 -> not divisible by 48

If k was 49, A = 48 * 50 * 51 -> divisible by 48

If k was 50, A = 49 * 51 * 53 -> not divisible by 48

If k was 51, A = 50 * 52 * 54 ->

48 = 2^4 * 3

So A needs to be a multiple of 16 and 3, which is not possible if A is odd
A is odd when k is even as seen above.

(k-1), k, (k+1), (k+2), (k+3)

Out of (k-1), (k+1), (k+3) -> one of them will be a multiple of 3

If (k-1) is a multiple of 4 -> k+3 is also a multiple of 4 -> A will be divisible by 16

If k is multiple of 4 -> k+4 is multiple of 4, A will not be divisble by 16

If k+1 is multiple of 4 - > (k-1) and (k+3) are multiples of 2, A will be divisible by 4 * 2 * 2 = 16

If k+2 is a multiple of 4 -> A will not be divisible by 16

If k+3 is a multiple of 4 -> k-1 will also be a multiple of 4, A will be divisible by 16


(k-1) or (k+1) or (k+3) should be a multiple of 4

49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83 (all odd numbers)

18 numbers

Probability = 18/37 (OPTION D)

84-48+1= 37 integers

if k is divisible by 16, then k could be 49,65,81,47,63,79,45,61,77

Prob = 9/37 Option A

We have 37 numbers to choose from (48 to 84), and we want (k - 1)(k + 1)(k + 3) to be divisible by 48, which is 2^4 * 3.

First, consider divisibility by 3.

Since the product involves three consecutive numbers, one of them is always divisible by 3, so we don't need to worry about divisibility by 3.

Next, we need to check divisibility by 2^4 = 16.

1. When k is even
   
   • k - 1, k + 1, and k + 3 are all odd, so their product can never be divisible by 16.
   • Therefore, we exclude all even numbers, which are 19 out of the 37 numbers, leaving 18 odd numbers.
2. When k is odd
   
   • Let k = 2x + 1. The product becomes 2x * (2x + 2) * (2x + 4) = 8x(x + 1)(x + 2).
   • Since x(x + 1)(x + 2) consists of three consecutive integers, one of them is always divisible by 2, ensuring divisibility by 16.
   • Thus, all 18 odd numbers will make the product divisible by 48.

So the probability is 18/37 (Option D)

If k is an integer from 48 to 84, inclusive, what is the probability that (k - 1)(k + 1)(k + 3) is divisible by 48?

A. 9/37
B. 12/37
C. 15/37
D. 18/37
E. 19/37

The denominator is the total number of possible outcomes, thus 84 - 48 + 1 = 37
Now numerator, desired outcomes -> given that (k-1)*(k+1)*(k+3) can be 3 consective odd or even numbers, depending wether k is odd or even.
For example, if k is 48 then -> 47*49*51 all are odd
if k is 49, which is odd then -> 48*50*52 all evens

For this product to be a multiple of 48 it has to be a multiple of 2^4*3^1 = which is the prime factorization of 48

now given that every 3rd consecutive number is a multiple of 3, odd or even; for example above, 51 is a multiple of 3 and 48 too, we know that every third number we have a multiple of 3 which coincides with our condition.

on the other hand, a multiple of 2^4 = 16 must be found. multiples of 16 are 16,32,48,64,80,96,... in the range only the bold ones.

(k - 1)(k + 1)(k + 3) is a sequence of 3 consecutive even or odd numbers, and for the purpose of finding multiples of 2^4 it k must be odd, so that (k - 1)(k + 1)(k + 3) contains at least 2 multiples of 4 and one multiple of 2 making it 2^5 and thus a multiple of 48.

Now for k to be odd in the range of 48 to 84, the common difference is 2 and the first and last odd numbers are 49 and 83, thus 83-49/ 2 +1 = 18 numbers

Answer (D) = 18/37

48 = 2 * 2 * 2 * 2 * 3

Now for Solving For 48<=K<=84

Every alternate no. has at least four 2's and one 3's.

Hence 18/37.

48=2^4*3

There are 84-48+1=37 numbers between 48 and 84. 19 are even numbers and 18 are odd numbers.

If k is and even number, (k - 1)(k + 1)(k + 3) is the product of three odd numbers and it is an odd number too, so it is not divisible by 48 (which is an even number).

So we have 18 candidates when k is an odd number and the product is the product of three even numbers.

The product of 3 consecutive even numbers is:
(2n-2)2n(2n+2)=8(n-1)n(n+1)

Which is always divisible by 48 because it is divisible by 16 (there is an 8 and in the other three consecutive numbers there is at least one even number) and it is divisible by 3 (the product of three consecutive integers is always divisible by 3).

probability=18/37

IMO D

nIMO D

The Only way I can think of is do is case by case but will see a pattern very quicky

So total number is fine 84-36+1= 37

now we can start with
48- No
49- Yes
50 - No
51- Yes
52- No
53-Yes

so we find the pattern till 84 every second number is multiple of 48 so there will be 36/2 = 18

hence 18/37

48 = 2^4 x 3

In order for (k - 1)(k + 1)(k + 3) to be divisible by 48 (or 2^4 x 3), k must be odd. Then all terms will be divisible by 2, and at least 1 term will be divisible by 4.

Looking at consecutive even numbers (2,4,6,8,10,12, etc.), we found that 1 in 3 consecutive even numbers will be always divisible by 3.

So we need to fund the total odd numbers within the range of 48 to 84 (37 consecutive numbers). That would be (83 - 49)/2 + 1 = 18.

Final answer: 18/37 (D)

We want to find the probability that (k-1)(k+1)(k+3) is divisible by 48 (3 * 16) for integers k from 48 to 84 inclusive.

• Divisibility by 3: The expression is always divisible by 3. This is because k-1, k, and k+1 are three consecutive integers and thus one of them is divisible by 3. Since k+3 can also be divisible by 3, the expression is always divisible by 3.
• Divisibility by 16: We need to check when (k2 - 1)(k + 3) is divisible by 16. Since we are looking for divisibility by 16, k must be odd. If k is even, k2-1 is odd and thus the whole expression cannot be divisible by 16.

Let k = 2n+1.
Then (k-1)(k+1)(k+3) = (2n)(2n+2)(2n+4) = 8n(n+1)(n+2).
We need 8n(n+1)(n+2) to be divisible by 16, which means n(n+1)(n+2) must be divisible by 2.This is always true as n, n+1, and n+2 are three consecutive integers.

So, we need to find when the expression is divisible by 16. When k is odd, k can be written as 2n+1.

Then the expression becomes (2n)(2n+2)(2n+4) = 8n(n+1)(n+2).
Since n(n+1)(n+2) is the product of three consecutive integers, it is always divisible by 2. Thus the expression is always divisible by 16 when k is odd.

Since k must be odd, we count the number of odd integers between 48 and 84 inclusive. The odd integers are 49, 51, 53, ..., 83. The number of such integers is (83 - 49)/2 + 1 = 34/2 + 1 = 17 + 1 = 18.

The total number of integers between 48 and 84 inclusive is 84 - 48 + 1 = 37.

Therefore, the probability is 18/37.

On analyzing this question we find that 48-84 inclusive would have around 37 numbers in the group with 19 even numbers & 18 odd numbers.

For (k-1)(k+1)(k+3) to be divisible by 48 we cannot have k as an even number as the resulting equation would be odd & 48 cannot divide an odd number. Hence we can eliminate all the even numbers from the list of numbers that can yield a number divisible by 48.

That would leave us with 18 odd numbers. 48 can be factorized as 16*3 or 2^4*3. Basically 4 powers of 2 & one power of 3. On putting odd values starting from 49 & going on till 83, we find that (k-1)(k+1)(k+3) will always generate 4 powers of 2 & one power of 3 which should be sufficient for the whole equation to be divisible by 48.

Hence the correct answer to this given question is (D) 18/37

Hi All,

Divisibility by 48 means \(2^4\) x 3

for divisibility by \(2^4\) the number should be divisble by 16

and for divisibility by 3 the all the 3 numbers should have atleast one 3 as a factor.

But as in [color=#0f0f0f](k - 1)(k + 1)(k + 3), atleast on number will be divisible by 3.[/color]


Multiple of 16 in the range are 48,64,80.

We need to calculate of the values for k so that we can get 48 in [color=#000000](k - 1)(k + 1)(k + 3).[/color]

[color=#000000]Therefore the values for K will be be 45,47,49,61,63,77,81 i.e. 15 values[/color]

[color=#000000]And the total number in 48<=x<=84 are 84-48+1 =37[/color]

[color=#000000]Therefore probability = 15/37[/color]

8The question is about even and odd divisibility .
Lets consider k even (i.e 48) 47*49*51 is not divisible by 48 . The number should be divisible by 8*6 (2^4 * 3) . Any three consecutive even numbers product will definitely be divisible by 8 and 6. (Because even numbers have common multiple 2 and if we take that common we have 2^3 * consecutive numbers . These consecutive numbers are definitely divisible by 6) .Consider an example ,let k=51 then product is 50*52*54 = 2*25*2*26*2*27 = 8* 5^2 * 13* 2 * 3^3 . So any three consecutive numbers product is definitely divisible by 6 and any three cosecutive even numbers product is divisible by 8 and 6. So we need to pick odd numbers from the list of 48-84 (37 numbers in total) . 18 odd and 19 even . So answer is D.

when (k - 1)(k + 1)(k + 3) is divisible by 48. Since 48 = 2^4 * 3, the expression must be divisible by both 16 and 3.


Notice that (k - 1), k, and (k + 1) are three consecutive integers. Therefore, at least one of them must be divisible by 3.
If k is divisible by 3, then (k-1)(k+1) is not, however k+3 is then also divisible by 3.
This means the product (k - 1)(k + 1)(k + 3) will always be divisible by 3.


case 1: k is even

If k is even, then (k - 1), (k + 1), and (k + 3) are all odd.
Therefore, the product (k - 1)(k + 1)(k + 3) will be odd and cannot be divisible by 16.
Case 2: k is odd

If k is odd, then (k - 1), (k + 1), and (k + 3) are all even.
Let k = 2n + 1 (where n is an integer). Then:
k - 1 = 2n
k + 1 = 2n + 2 = 2(n + 1)
k + 3 = 2n + 4 = 2(n + 2)
The product becomes: 2n * 2(n + 1) * 2(n + 2) = 8n(n + 1)(n + 2)
For this to be divisible by 16, n(n + 1)(n + 2) must be divisible by 2. Since the product of three consecutive integers is divisible by 2, this is always true when k is odd.
Now we need to consider when there is another factor of 2.
If n is even, then n and n+2 are both even. That means n(n+1)(n+2) is divisible by 4 and therefore the original term is divisible by 16.
If n is odd, then n+1 is even, which means that k-1 and k+3 are only divisible by 2, and therefore the original term is not divisible by 16.
So we need n to be even, which means that n+1 is odd. Thus k=4m+1.
Since k is between 48 and 84, the values for k that satisfy this are 49, 53, 57, 61, 65, 69, 73, 77, 81. That is 9 values.


Total possible values of k: 84 - 48 + 1 = 37
Values of k that work: 9
Probability: 9/37

48 = 4 * 4 * 3

Out of (k - 1)(k + 1)(k + 3) any one number is divisible by 3 that is for sure

1) k is even:
In this case all numbers are odd and odd/3 = odd so again all 3 numbers are odd so even k is not possible.

2) k is odd:
In this case all 3 numbers are even and even/3 = even so again all numbers are even
Now, in these 3 numbers 1 number is definitely divisible by 4 and other 2 are divisible by 2 for sure
So, (k - 1)(k + 1)(k + 3) is divisible is 48 for odd k only.
Total k = (84 - 48 + 1) = 37
Total odd k = (37 - 1)/2 = 18
Probability = Total odd k / Total k = 18/37

Answer: D. 18/37