12 Days of Christmas GMAT Competition - Day 12: If the integer k lies

Question

12 Days of Christmas GMAT Competition with Lots of Fun

If the integer k lies between 100 and 137, then what is the probability that k^3 – k is divisible by 72?

A. 1/9
B. 1/6
C. 7/36
D. 2/9
E. 1/4

Bunuel · Accepted Answer

The correct answer is option C Solution: Step 1: Understand Question Stem: k is an integer between 100 and 137. o So, we cannot consider 100 and 137. Thus, 100 < k < 137. We have a total of 36 numbers, from 101 to 136, inclusive. We are asked the probability that k^3-k will be divisible by 72 i.e., 2^3*3^2=8*9 Step 2: Define Methodology: We can simplify k^3-k =k(k^2-1) =(k-1)k(k+1) It is important to logically deduce that k – 1, k, and k + 1 are three consecutive integers. o This needs to be divisible by 72. o 72 = 2^3*3^2 Thus, the term (k-1)k(k+1) can be divisible by 72 if it is divisible by 2^3 and 3^2 separately. o Divisibility by 2^3 : This is possible in two ways. If we have one multiple of 8. (Condition 1.a) If we have k – 1 as even. (Condition 1.b) o In such a case k + 1 will also be even and one of them must a multiple of 4. For example, if k – 1 = 2, k + 1 = 4. If k – 1 = 12, k + 1 = 14. o Divisibility by 3^2 : One of the numbers must be a multiple of 9. (Condition 2) Please note, we cannot have 2 multiples of 3 because three consecutive integers will have only one multiple of 3. o For example, 5, 6, 7 has 6 or 9, 10, 11 has 9. We will have the term (k-1)k(k+1) as a multiple of 72 If Condition 2 is satisfied with 1.a or 1.b It is obviously easier for us to consider multiples of 9 (Condition 2) first. o The multiples of 9 in these ranges are 108, 117, and, 126 and 135. Step 3: Calculate Final Answer: We can make the following table using these options and check if it is a multiple of 8 or not. https://gmatclub.com/forum/download/file.php?id=69731 Total number of values of k for which it is multiple of 72 = 7 Total possible values of k = 36. o The required probability = (The number of favourable case)/(The number of total case) = 7/36 The answer is option C. 2022-12-26_17-56-22.png

Oppenheimer1945 · Answer

k^3-k= (k-1)k(k+1)
72= 9x8=2^3 3^2

Total Nos=36
we need to check the divisbility in the neighbourhood of numbers div by 9

Such div nos= 108, 117, 126, 135.
We need to check div at 106, 107, 108, 109,110, 115, 116, 117, 118, 119, 124, 125, 126, 127, 128, 133,134,135,136,137

No of such numbers that are div by both 9 & 8 are= 107, 109, 117, 125,127,135,136 (7 nos)

Required probability= 7/36

mysterymanrog · Answer

k^3-k can be rewritten as (k-1)(k)(k+1).

72 has prime factors 2^3*3^2.

Note that for our target number (k^3-k) can only contain one multiple of 3 and no more. Therefore, that multiple of 3 must also be a multiple of 9.

For our range of k, we have the following multiples of 9: 108,117,126,135. 
Now we can test each case, using the multiples of 9 as our basis. We will want to check for 1 number less and 1 number ahead of our multiple of 9. We only need to ensure that there are enough 2s, because we already have enough 3s.

k=107
(106)(107)(108)->108 is m4 as well, so we have 2 twos, and 106 has a 2 as well. Possible.
k=108
(107)(108)(109)->missing a two. Out.
k=109
(108)(109)(110): enough 2s. Possible.

k=116
(115)(116)(117)->lacking a two. out. 
k=117
(116)(117)(118)->enough 2s. keep.
k=118
(117)(118)(119)->not enough 2s. out.

k=125
(124)(125)(126)->24 is a m4, so we have 2 2s, and 126 has a 2 itself. keep. 
k=126
(125)(126)(127)->not enough 2s
k=127
(126)(126)(128)->128 is a m4, and 126 has a 2. keep.

k=134
(133)(134)(135)->not enough 2s. 
k=135
(134)(135)(136)->enough 2s (136 is m8)
k=136
(135)(136)(137)->136 has enough 2s on its own (136 is a m8). keep.

Total, we have k {107,109,117,125,127,135,136}

Our range has 36 numbers (136-101+1=36)
7/36. Choice C.

Lizaza · Answer

First, let's transform the condition a little:
 k^3 – k = k(k^2-1)=k(k-1)(k+1) 
In other words, we need to know whether  (k-1)k(k+1) = 72*n , where N is a random integer.

As we already see, we are dealing with three consecutive numbers here. Let's also take a look at our divisor - what factors must we have?
 72 = 8*9 = 2*2*2*3*3  
So basically, we need to check, in how many cases the possible K numbers - three consecutive ones - are divisible by 8 and 9 at the same time.

Starting with 9 is easiest, because from three consecutive numbers, obviously, only one will be divisible by three, because they will always be:  3m, 3m+1, 3m+2 .
Therefore, our K-combinations will be centered around the integers divisible by 9, because those are the only ones that will provide the necessary 3*3 factors.
Such numbers are: 108 | 117 | 126 | 135.
Now, we need to 'gather' 2*2*2 factors within those numbers and around them, to find the right combinations of K.

108 = 2*2*27 
Therefore, we need only one more 2 for the combination to be divisible by 72.
It means that we need to include at least one more even number - which may be 106 or 110.
So, there are two divisible combinations:  106*107*108  and  108*109*110 
So,   two possible K's are 107 and 109.  
 
  117=9*13 
Which means we need to have all three 2's for the divisibility.
From the neighbours, 116 gives us 2*2*29, and 118 = 2 * 59
Therefore, there's only one combination -  116*117*118  - and   one possibility for K: 117  
 
  126=2*63 
Therefore, we need two more 2's for the combination to be divisible by 72.
From the even neighbours, both 124 and 128 are divisible by 4.
So, there are two divisible combinations:  124*125*126  and  126*127*128 
So,   two possible K's are 125 and 127 . 
 
  135=9*15 
Which means we need to have all three 2's for the divisibility.
From the neighbours, 134 gives us 2*67, and 136 = 17*8. This means that only 136 by itself is sufficient to provide all 2*2*2!
Therefore, there are again two divisible combinations:  134*135*136  and  135*136*137 .
So,   two possible K's are 135 and 136.

Now it's time to solve the task. We have 7 suitable options for K - and in total, there's a range of 36 numbers available for it (from 101 to 136, inclusive).
Therefore, the probability of getting a number that is divisible by 72 is  \frac{Suitable-K's }{ Total-range-of-K's }=\frac{7}{36} .

The correct answer is C.

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12 Days of Christmas GMAT Competition - Day 12: If the integer k lies

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12 Days of Christmas GMAT Competition - Day 12: If the integer k lies

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