Two distinct numbers a and b are chosen randomly from the set of first

Question

Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190

Bunuel · Accepted Answer

VERITAS PREP OFFICIAL SOLUTION:

Any number could be written in the form of one among the following:

(i) 3k

(ii) 3k + 1

(iii) 3k + 2

Only if the 2 numbers are of different types would |a – b| not be divisible by 3:

Numbers of type 3k --> 3, 6, 9, ......... 18 --> Total 6 numbers

Numbers of type 3k + 1 --> 1, 4, 7, .... 19 --> Total 7 numbers

Numbers of type 3k + 2 --> 2, 5, 8, 11 ...20 -->Total 7 numbers

Total possible outcomes = 20 * 19 = 380

Favorable Cases:

1) Picking from Type 1 and Type 2: 2 * (6 * 7) = 84

2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84

3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98

Hence, the required probability = (84 + 84 + 98) / 380, and E is the correct answer.

crazu4 · Answer

Here is what i thought:

From 20 first positive integers, we get that 6 of them are divisible by 3. For the equation not to be divisible by 3, we have the following possible combinations:

both of the numbers not being divisible by 3:

14/20 * 13/19 = 91/190

one of them is divisible by 3 and the other isn't:

14/20 * 6/19 = 42/190

because we need combination 1 OR combination 2 to achieve the requested, we need to sum both fractions:

91/190 + 42/190 = 133/190

For me, answer E. i guess i forgot something and made it completly wrong.

shriramvelamuri · Answer

Firstly,

The pattern is

1 4 7 10 13 16 19 = Pick two numbers from 7 -> 7p2= 42

2 5 8 11 14 17 20 = pick two number from 7 -> 7p2= 42

3 6 9 12 15 18= Pick two numbers from 6-> 6P2= 30.

Total permuations that are divisible by 3 are 114.

Total permutions are 20p2= 190.

1- 114/190.

This is exactly where I am struck. I am getting a strange answer. I guess, I have to wait for someone to crack this one. \

Bunuel please guide me what I am doing wrong.

zerosleep · Answer

I think you just made the mistake that 20P2 will be 20 * 19 = 380 and NOT 190..

If you consider this, that would make 114/380 (probability that the no is divisible by 3)
Subtract this probability from 1,
1- 114/380 = 266/380 = 133/190

Ans (E)

decadecaf · Answer

I started to write the numbers out. Eventually, I thought that since a little less than a third of the numbers are divisible by three, the probability of the absolute value of any number minus another number not being divisible by three would be a little more than 2/3, so I went with answer E.

bhatiavai · Answer

Hi Bunuel,

Why exactly are you multiplying by 2 when you are calculating the favourable cases.

Many thanks

Bunuel · Answer

1) a is from Type 1 and b is from Type 2 + b is from Type 1 and a is from Type 2... The same for (2) and (3).

cherryli2015 · Answer

1. Pick 2 from {1,2, ..... 20} 20*19 = 380 
2. figure out the probability that |a–b| would be divisible by 3
 1) {3, 6, 9, 12, 15, 18 } 6*5 = 30 
 2) {1,4,7,10,13, 16, 19} 7*6 = 42 
 3) {2, 5, 8, 11, 14, 17, 20} 7*6 = 42 
3. the probability that |a–b| would not be divisible by 3 = (380 - 30-42-42) / 380 = 133/190

So the correct answer is E

ravindra88 · Answer

when you select 2 numbers from a set of numbers, you use combination and not permeutation (one of the very common mistakes), so you should do =  1 - (7C2 + 7C2 + 6C2)/ 20C2

mike34170 · Answer

I have handled this one with no real mathematics involved.

For each number A selected you have 19 possible results for |a-b|, because when you select A you have 19 numbers remainding.

You know that only 6 results of |a-b| will give you a number divisible by 6 ( 3, 6, 9, ...18) . Then, 14 results out of 19 won't be divisible by 3.

Look, we've just said that 14 won't be divisible by 3 out of 19 possibles outcomes... 133/190 is approximatively this number.

ashokk138 · Answer

Let find the probability that |a-b| is divisible by 3.

Then |a-b| has to be 3, 6, 9, 12, 15, 18 (cannot be 0 since a and b are distinct)

To get |a-b| as 3, |a-b| could be |4-1|, |5-2|.....|20-17| => 17 Times. 
But since this is modulus, |4-1| = |1-4|.

Hence to get 3, there are 34 ways.
Similarly to get 6 => 14*2 = 28
9 => 11*2 = 22
12 => 8*2 = 16
15 => 5*2 = 10
18 => 2*2 = 4

Hence total ways to get |a-b| divisible by 3 = 114

Total Outcomes = 20 x 19 = 380

Probability to get |a-b| divisible by 3 is 114 / 380 => 57/190

Probability that |a-b| is not divisible by 3 is = 1 - (57/190) = 133/190

Option E

Dat Le · Answer

I have a better way to do this problem:
number 1-20 we have 19*20= 380 ways to choose a pair.
from 1-20 each number have 6 ways to choose a number that a-b divisible by 3: EX 20 must subtract by 17,14,11,8,5,3 to divisible by 3:
so from 1-20 we have 20*6= 120 ways to choose, but we didnt want a same number, hence 120-6= 114
so the percentage is 114/380 to choose a pair that divisible by 3, 
===> hence the chance to pick a pair that not divisible by 3 is 1- (114/380)= 266/380 =133/190 and this is final answer.

EMPOWERgmatRichC · Answer

Hi All,

The answer choices to this question provide a great 'logic shortcut' that you can take advantage of.

The calculation that we're asked to consider (|A-B|) and the limited possible values for the two distinct variables (1-20, inclusive) means that we're dealing with possible answers that are consecutive integers: 1 through 19, inclusive.

When dealing with consecutive integers, only 1 out of every "set" of 3 will be divisible by 3. This question asks for the probability that the calculation will NOT be divisible by 3. Thus, the answer MUST be greater than 1/2 (closer to about 2/3). There's ONLY one answer that's greater than one half....

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

ankitsood5 · Answer

|a-b| can have 19 values from 1 to 19 and out of these 19 values only 6 i.e. 3,6,9,12,15 and 18 are divisible by 3. Thus, 13 values are "not" divisible by 3. The probability can is thus 13/19 or 130/190. A close approximate to 133/190.
In any case, the probability should be more than 1/2. Only option E is >1/2

pappal · Answer

A B
1 4,7,10,13,17,19
2 5,8,11,14,17,20
3 6,9,12,15,18
.
20 ..................

pattern suggests that, 
1. for A not a multiple of 3 we can have 6 possibilities each i.e. 14*6=84 
2. for A multiple of 3 we have 5 possibilities each. i.e. 6*5=30
total of (1)+(2)=114 . possible combinations where absolute value of the difference is a multiple of 3.
Now, no. of ways to select 2 digits from 1-20 digits =20C2
however since (4,6) and (6,4) are two different combinations of A & B.
so, total no. of cases =>20C2 *fact(2)=20P2= fact(20) / fact(18)=20*19=380
 prob.(A,B are divisible by 3)=114/380
so, probability that (A,B are not divisible by 3)= 1-(114/380)=266/380=133/190.
option (E)

Wadree · Answer

Let....a>b..... we'll first figure out how how many | a - b | ARE divisible by 3....so....(a - b)/3 = n .... where n is an integer.....so.....a = 3n + b ..... But highest possible value of | a - b | in this case is 19..... Highest multiple of 3 here is 18 ..... So....all possible multiple of 3 are 18, 15, 12, 9, 6, 3....Now, 18 can be made by | a - b | in.... 20 - 18 = 2 ways....18 = 20 - 2 = 19 - 1....... 15 can be made by | a - b | in .....20 - 15 = 5 ways....15 = 20 - 5 = 19 - 4 = 18 - 3 = 17 - 2 = 16 - 1.....
So 12 can be made by | a - b | in 8 ways
9 can be made by | a - b | in 11 ways
6 can be made by | a - b | in 14 ways
3 can be made by | a - b | in 17 ways
So.... | a - b | can be divisible by 3 in 57 possible ways....
Total ways possible to choose | a - b | and get different results in this condition is 19 + 18 + 17 ...... 3 + 2 + 1 = 190 ways...
So..... | a - b | cant be divisible by 3 in 190 - 57 = 133 ways....
So... Answer is E = 133/190.....

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