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avatar+893 

Can anyone solve this? Answer: 5/16

 Mar 14, 2019
edited by dgfgrafgdfge111  Mar 15, 2019
edited by dgfgrafgdfge111  Mar 15, 2019
 #1
avatar+4299 
+2

Hint: A number is divisible by 77 when it is divisible by 7 and 11. There are a total of 2^6=64 total outcomes. 

 Mar 15, 2019
 #2
avatar+893 
0

Well, since there's no divisibility rule for 7, here's what I did:

 

77 can only divide the number if there are an even number of 7s. Also, the number has to start with 77 because if it only started with 7, then how would it be divisible by 77? So, we could have the numbers:

 

770000

770770

770077

777700

777777

 

There are 5 numbers. Each has a probability of (1/2)^5 of occuring, because the 7 in the hundred thousands place is already determined. So, I got 5/32, but the answer is actually 5/16? So what did I do wrong?

dgfgrafgdfge111  Mar 15, 2019
 #3
avatar+4299 
+2

Wait, so the second "7" is fixed, too? So, 1/16*5=5/16?

tertre  Mar 15, 2019
 #4
avatar+893 
0

No, actually I was trying to find the cases that would work. You could have the number be 700000.

dgfgrafgdfge111  Mar 15, 2019
 #5
avatar+4299 
+1

700000 is not divisible by 77, by the way.

tertre  Mar 15, 2019
 #6
avatar+893 
0

Yeah, but it's one of the possible cases, so it should be in the denom and counted.

dgfgrafgdfge111  Mar 15, 2019
 #8
avatar+4299 
+1

No, it's not. Try again. Look at Melody's solution. 

tertre  Mar 16, 2019
 #7
avatar+102763 
+3

Yep

I got 10 desirable outcomes out of a total of 32 possible options which is  10/32  =   5/16     laugh

 

Method:

The first digit must be 7 then there are 5 more digits with 2 posibilities each so there are    2^5 = 32 possible numbers.

If a number is divisable by 77 then it must be divisable by 7 and 11.

Since all the digits are either 0 or 7 all the numbers will be divisable by 7 so I need to worry about if it is divisable by 11.

 

Now I looked up the internet fo find tricks for deciding if a number is divisable by 11.

This is what I found.

https://www.math.hmc.edu/funfacts/ffiles/10013.5.shtml

 

So if I talk about odd and even digit places then I have

 

7(even), odd,even, odd, even, odd

I have to have the same number of 7s in even numbers as I do in odd ones so that they will cancel each other out.

so I can have

6 sevens      one way to get this   777777

4 sevens      the first even seven is set, 2 choices for second even one, only one odd one gest left out so that is 3 choices,

                     so I have 1*2*3 = 6 ways to have 4 sevens.

2 sevens      The even one is set and there are 3 odd ones to chose from so that is  1*3 = 3

So altogether that makes     1+6+3 =10 numbers that are divisable by 77

 

So that is a probablility of   10/32   = 5/16             laugh

 Mar 15, 2019
 #9
avatar+893 
+1

Thanks!!!

dgfgrafgdfge111  Mar 16, 2019

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