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Factoring 48 gives us - \(2^4*3\)

Looking at the multiplication, (k-1) * (k+1) * (k+3)

For any odd number in the given range, all these three values should be even and atleast one of them should be multiple of 4 as these three values would be 3 successive even integers and are always divisible by \(2^4\)

Also, for any odd number in the range, there are 2 cases -

Case 1 : k is divisible by 3

Then k+3 would also be divisible by 3

Case 2 : k is not divisible by 3

Then either (k-1) or (k+1) would be divisible by 3

In either case, given any odd number in the range, it would be always divisible by 48.

Range = 84 - 48 + 1 = 37
No. of odd numbers = Range/2 = 18

Probability = 18/37

Answer: D
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12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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48 = 2^4*3

If k = 48 => 47*49*51 (Not divisible by 2)
If k = 49 => 48*50*52 (48 divisible by 48)
If k = 50 => 49*51*53 (Not divisible by 2)
If k = 51 => 50*52*54 (50 have 2*1, 52 have 2*2, 54 have 2*1 and 3*1, divisible by 48)

Number between 48 to 84 is 84-48+1 = 37
Only if k = odd between 48 to 84 is divisible by 48
There are 19 even number and 18 odd number (because start from even), therefore the probability divisible by 48 is 18/37
D is the answer
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We need to check divisibility by 48 => 16 * 3

Therefore, (k - 1), (k + 1), and (k + 3) must be divisible by 16 and 3

This implies k cannot be an even integer, as that will lead to consecutive odd integers that won't have 16 as a factor.

We know that three consecutive numbers will always be divisible by 3, and when k is odd one of the numbers will be divisible by 4 and the other two by 2 => (k - 1) * (k + 1) * (k + 3) is divisible by 16 and 3 when k is odd

Number of odd integers between 48 and 84 = [(83 - 49)/2] + 1 = 18

Probability = Number of odd integers between 48 and 84 / Total no. of integers from 18 to 84 = 18/37

Answer D.
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12 Days of Christmas 2024 - 2025 Competition with $40,000 of Prizes

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

 


This question was provided by GMAT Club
for the 12 Days of Christmas Competition

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Number of terms = (84 - 48) + 1 = 37

If k = even,

(k - 1)(k + 1)(k + 3) = odd * odd * odd = odd

48 is even, hence no term will be divisible by 48.

If k = odd

(k - 1)(k + 1)(k + 3) = even * even * even = even

These are three consecutive even terms. In three consecutive even terms, on term is divisible by 4, and the one term is divisible by 6. Each term is divisible by 2.

Hence, we need to find the number of odd values of k.

First Term = 49
Last Term = 83

Number of terms = 83 - 49 / 2 + 1 = 17 + 1 = 18

Probability = 18 / 37

Option D
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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
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Solve this step by step:

1) First, let's understand what makes a number divisible by 48:
  • 48 = 16 × 3
  • So a number must be divisible by both 16 and 3

2) (k - 1)(k + 1)(k + 3) is a product of three consecutive odd numbers (because k is in the middle, so k±1 and k+3 are all odd)

3) For any three consecutive odd numbers:
  • One must be divisible by 3 (as every third odd number is divisible by 3)
  • Among three consecutive odd numbers, their product will always be divisible by 3

4) Now for divisibility by 16:
  • Among three consecutive odd numbers, one must be divisible by 8
  • Another must be divisible by 2
  • Together this makes the product divisible by 16

5) Therefore (k - 1)(k + 1)(k + 3) is ALWAYS divisible by 48!

6) Number of integers from 48 to 84 inclusive:
84 - 48 + 1 = 37 integers

7) Since it works for all values of k in this range:
Probability = 37/37 = 1 = 18/37

The answer is D. 18/37
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