What is the probability that a three-digit integer corresponding with

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}

sukoon9334 · Answer

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.

sujoykrdatta · Answer

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: 
 
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

MathRevolution · Answer

=>

In order for a three-digit integer  abc  to be a multiple of  11, a – b + c  must be a multiple of  11  since we have  100a + 10b + c = 99a + a + 11b – b + c = (99a + 11b) + a – b + c = 11(9a+b) + a – b +c.

If  a – b + c = 0 , then the possible cases for  (a, b, c)  are

(1, 2, 1), (1, 3, 2), (1, 4, 3), (1, 5, 4), (1, 6, 5), (2, 3, 1), (2, 4, 2), (2, 5, 3), (2, 6, 4), (3, 4, 1), (3, 5, 2), (3, 6, 3), (4, 5, 1), (4, 6, 2)  and  (5, 6, 1),  and we have  15  cases.

If  a – b + c = 11 , then we have only the case  (a, b, c) = (6, 1, 6).

Thus the probability is  \frac{16}{(6^3)} = \frac{16}{216} = \frac{2}{27}.

Therefore, the answer is B.
Answer: B

What is the probability that a three-digit integer corresponding with

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What is the probability that a three-digit integer corresponding with

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