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24 Jan 2020, 03:03
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[GMAT math practice question]
What is the probability that a three-digit integer corresponding with the three numbers thrown on three different dice is a multiple of \(11\)?
A. \(\frac{1}{3} \)
B. \(\frac{2}{27} \)
C.\( \frac{4}{27}\)
D. \(\frac{5}{27} \)
E.\( \frac{2}{9}\)
What is the probability that a three-digit integer corresponding with the three numbers thrown on three different dice is a multiple of \(11\)?
A. \(\frac{1}{3} \)
B. \(\frac{2}{27} \)
C.\( \frac{4}{27}\)
D. \(\frac{5}{27} \)
E.\( \frac{2}{9}\)
sukoon9334
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24 Jan 2020, 03:53
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Let's first find the different possible numbers on the three dice.
The first die can show six different results i.e. 1,2,3,4,5,6
Sly, die 2 and die 3.
So, 3 digit numbers possible are (6x6x6).
Now, let us find out the possible numbers divisible by 11. This will be interesting.
Let's suppose that first die shows 1.
Now, possible 3 digit multiples of 11 are 110, 121, 132, 143, 154, 165, 176, 187, 198
We cannot consider numbers that have digits other than the die can show i.e. 0, 7,8, 9.
So, acceptable solutions are 121, 132, 143, 154, 165.
Sly, when you start solving with 2,3,4,5,6 in the 100th place, you'll notice that the possible solutions are 5,4,3,2,1,1.
Total possible solutions = 5+4+3+2+1+1 = 16.
Our probability is 16/ (6x6x6) = 2/27.
The first die can show six different results i.e. 1,2,3,4,5,6
Sly, die 2 and die 3.
So, 3 digit numbers possible are (6x6x6).
Now, let us find out the possible numbers divisible by 11. This will be interesting.
Let's suppose that first die shows 1.
Now, possible 3 digit multiples of 11 are 110, 121, 132, 143, 154, 165, 176, 187, 198
We cannot consider numbers that have digits other than the die can show i.e. 0, 7,8, 9.
So, acceptable solutions are 121, 132, 143, 154, 165.
Sly, when you start solving with 2,3,4,5,6 in the 100th place, you'll notice that the possible solutions are 5,4,3,2,1,1.
Total possible solutions = 5+4+3+2+1+1 = 16.
Our probability is 16/ (6x6x6) = 2/27.
24 Jan 2020, 06:55
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Question:
What is the probability that a three-digit integer corresponding with the three numbers thrown on three different dice is a multiple of 11?
Solution:
The digits on a dice are 1, 2, 3, 4, 5, and 6
Let us use the divisibility rule for 11:
If a 3-digit number abc is divisible by 11, then
#1. a + c - b = 0 => a + c = b --- example: 121, 143, etc.
Or
#2. a + c - b = 11 => a + c = b + 11 --- example: 616, 506, etc.
We need to find out how many such cases exist for #1 and #2
For #1: a + c = b: Possible solutions for (a, b, c) are:
Starting with a=1: (1,2,1), (1,3,2), (1,4,3), (1,5,4), (1,6,5) => 5 cases
Starting with a=2: (2,3,1), (2,4,2), (2,5,3), (2,6,4) => 4 cases
Starting with a=3: (3,4,1), (3,5,2), (3,6,3) => 3 cases
Starting with a=4: (4,5,1), (4,6,2) => 2 cases
Starting with a=5: (5,6,1) => 1 case
For #2: a + c = b + 11: Possible solutions for (a, b, c) are:
[Note: a + c must be 11 or greater. Thus, the values of a and b must be at least 5 or 6]
Starting with a=6: (6,1,6) => 1 case
Starting with a=5: (5,0,6) ---- not possible
Thus, total number of solutions = 5 + 4 + 3 + 2 + 1 + 1 = 16
Total number of 3-digit numbers that can be formed = 6 x 6 x 6 (since each dice has numbers from 1 to 6)
Thus, required probability = 16/216 = 2/27
Answer B
What is the probability that a three-digit integer corresponding with the three numbers thrown on three different dice is a multiple of 11?
Solution:
The digits on a dice are 1, 2, 3, 4, 5, and 6
Let us use the divisibility rule for 11:
If a 3-digit number abc is divisible by 11, then
#1. a + c - b = 0 => a + c = b --- example: 121, 143, etc.
Or
#2. a + c - b = 11 => a + c = b + 11 --- example: 616, 506, etc.
We need to find out how many such cases exist for #1 and #2
For #1: a + c = b: Possible solutions for (a, b, c) are:
Starting with a=1: (1,2,1), (1,3,2), (1,4,3), (1,5,4), (1,6,5) => 5 cases
Starting with a=2: (2,3,1), (2,4,2), (2,5,3), (2,6,4) => 4 cases
Starting with a=3: (3,4,1), (3,5,2), (3,6,3) => 3 cases
Starting with a=4: (4,5,1), (4,6,2) => 2 cases
Starting with a=5: (5,6,1) => 1 case
For #2: a + c = b + 11: Possible solutions for (a, b, c) are:
[Note: a + c must be 11 or greater. Thus, the values of a and b must be at least 5 or 6]
Starting with a=6: (6,1,6) => 1 case
Starting with a=5: (5,0,6) ---- not possible
Thus, total number of solutions = 5 + 4 + 3 + 2 + 1 + 1 = 16
Total number of 3-digit numbers that can be formed = 6 x 6 x 6 (since each dice has numbers from 1 to 6)
Thus, required probability = 16/216 = 2/27
Answer B
