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243
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How many three-digit whole numbers have no 9's and no 7's as digits?

 

Ugh getting different answers every time I redo this... Is it 228?

 Apr 23, 2019
 #1
avatar+19942 
0

You bsically have 8 digits that can go into each of the three positions (seven in the first one....no zero)

 

1-6 and 8    x  8   x  8    = 448 possibilities

 Apr 23, 2019
edited by Guest  Apr 23, 2019
edited by Guest  Apr 23, 2019
 #2
avatar+34 
0

https://answers.yahoo.com/question/index?qid=20130519114338AAziTPE

 

Start with 100-999 

eliminate 700-999, leaving 100-699 

eliminate anything ending in 70-79,80-89,90-99, leaving 100-169,200-269,300-369,400-469,500-569,...

eliminate anything ending in 7,8,9. leaving 100-106,110-116,120-126,130-136,140-146,... through 666...49x6=294

 Apr 23, 2019
 #4
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0

How many three-digit whole numbers have no 9's and no 7's as digit 

 

Not no 8

Guest Apr 23, 2019
 #3
avatar+6046 
+1

\(\text{There are 7 valid digits for the first number, and 8 valid digits for numbers 2 and 3}\\ \text{Thus there are }\\ n = 7\cdot 8^2 = 448 \text{ 3 digit numbers with no 7's or 9's}\)

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 Apr 23, 2019
 #5
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+1

Oh thanks a lot! Why didn't I think of that? ( time for bed... 😴 I think my brain malfunctioned...)

Guest Apr 23, 2019
 #6
avatar+81 
-6

So in order to solve this we can use two ways, complementary counting or how they had done it below (permutation)

Complementary counting starts with finding out the total amoutn of 3 digit whole numbers, in which we can use the Arithmatic Sequence.(1000-1/1)+1 = 999. 

 

We can then find how many numbers HAVE 9 or 7 as their digits. So we get 551 numbers that have 9 and 7 as digits. 

Using this information we can subtract 551 from 999, 999-551 = 448 Which is our answer.

 

ANSWER: 448

 Apr 23, 2019

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