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?

Guest Apr 23, 2019

#1**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

ElectricPavlov Apr 23, 2019

edited by
Guest
Apr 23, 2019

edited by Guest Apr 23, 2019

edited by Guest Apr 23, 2019

#2**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

RandiLaine98 Apr 23, 2019

#3**+1 **

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

.Rom Apr 23, 2019

#6**-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

doorknoob Apr 23, 2019