How many integers from 1 to 10000 do not conain digit 8?

[4meonly]

How many integers from 1 to 10000, inclusive, do not conain digit 8?

I dont remember answers and OA.


what is the best approach to solve this question and this type of questions?

[sudhir3127]

How many integers from 1 to 10000, inclusive, do not conain digit 8?

I dont remember answers and OA.


what is the best approach to solve this question and this type of questions?

IMO 9^4-1 assuming we dont consider zero...
9^4 if we take zero into consideration.

[parallel_chase]

one digit = 8 ways = 8
two digit = 8*9 ways = 72
three digit = 8*9*9 ways = 648
four digit = 8*9*9*9 ways = 729*8 = 5832
five digit = 1*1*1*1*1 ways = 1

8+72+648+5832+1

[4meonly]

I fond the answer

it is 6562

parallel_chase,

could you please explain?

one digit = 8 ways = 8
two digit = 8*9 ways = 72
three digit = 8*9*9 ways = 648
four digit = 8*9*9*9 ways = 729*8 = 5832
five digit = 1*1*1*1*1 ways = 1

8+72+648+5832+1 =6561

You are on the right way but I don't the logic

[sudhir3127]

I fond the answer

it is 6562

parallel_chase,

could you please explain?

one digit = 8 ways = 8
two digit = 8*9 ways = 72
three digit = 8*9*9 ways = 648
four digit = 8*9*9*9 ways = 729*8 = 5832
five digit = 1*1*1*1*1 ways = 1

8+72+648+5832+1 =6561

You are on the right way but I don't the logic

are u sure abt the Answer because for me its 9^4 = 6561.. when u consider zero.

[Senator 153]

I wrote an algorithm to check. It's 6561.

[4meonly]

I wrote an algorithm to check. It's 6561.

Could you please specify an algorithm to me as to a kindergardener?

thank u!

[Senator 153]

It's just a way of checking; it's a computer script that counts the numbers that don't have eight in them.

I solved it using the same method as sudhir3127. 9 out of ten numbers will "survive" the test, per digit -- and four digits are tested, so it's 9^4.

[4meonly]

It's just a way of checking; it's a computer script that counts the numbers that don't have eight in them.

I solved it using the same method as sudhir3127. 9 out of ten numbers will "survive" the test, per digit -- and four digits are tested, so it's 9^4.

It means that for every place we have only 9 digits?
_ _ _ _
9*9*9*9
but this is for 4-digit number. What about 3, 2 and 1 digit numbers?
About 1-digit there is only one that will not survive. What about 3, 2 digit numbers?

[anju]

I fond the answer

it is 6562

parallel_chase,

could you please explain?

one digit = 8 ways = 8
two digit = 8*9 ways = 72
three digit = 8*9*9 ways = 648
four digit = 8*9*9*9 ways = 729*8 = 5832
five digit = 1*1*1*1*1 ways = 1

8+72+648+5832+1 =6561

You are on the right way but I don't the logic

are u sure abt the Answer because for me its 9^4 = 6561.. when u consider zero.

Sudhir, can you pls. elaborate why you did 9^4?

thanks

[parallel_chase]

Here is the process:

For example:

You have to find the number of integers between 1-999 that doesnt contain 3

formula : (9^3) - 1 = 728

Long method.
one digit number (1,2,4,5,6,7,8,9) = 8 ways
two digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9 ways
three digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9*9 ways

8 + 72 + 648 = 728

You have to find the number of integers between 1-1000 that doesnt contain 3

formula : 9^3 = 729

Long Method.
one digit number (1,2,4,5,6,7,8,9) = 8 ways
two digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9 ways
three digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9*9 ways
four digit number (0,1), first digit cannot be 0 and remaining digit cannot be greater than 0 = 1*1*1*1 = 1 ways

8 + 8*9 + 8*9*9 + 1 = 729

Try to solve the posted question with the same method.

"4meonly" i think you have got the wrong answer, the answer should 6561.

And the algorithm used by "Senator" is a computer program written probably in C++ or php, therefore i dont think he would be able to explain you the program unless you have any prior knowledge of computer programming.

Let me know if you still have any doubts.

[anju]

Here is the process:

For example:

You have to find the number of integers between 1-999 that doesnt contain 3

formula : (9^3) - 1 = 728

Long method.
one digit number (1,2,4,5,6,7,8,9) = 8 ways
two digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9 ways
three digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9*9 ways

8 + 72 + 648 = 728

You have to find the number of integers between 1-1000 that doesnt contain 3

formula : 9^3 = 729

Long Method.
one digit number (1,2,4,5,6,7,8,9) = 8 ways
two digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9 ways
three digit number (0,1,2,4,5,6,7,8,9), first digit cannot be 0 = 8*9*9 ways
four digit number (0,1), first digit cannot be 0 and remaining digit cannot be greater than 0 = 1*1*1*1 = 1 ways

8 + 8*9 + 8*9*9 + 1 = 729

Try to solve the posted question with the same method.

"4meonly" i think you have got the wrong answer, the answer should 6561.

And the algorithm used by "Senator" is a computer program written probably in C++ or php, therefore i dont think he would be able to explain you the program unless you have any prior knowledge of computer programming.

Let me know if you still have any doubts.

Great. Thanks. That's an awesome explanation.

Is there a list of all formula's or shortcut methods?

Thanks

[uptowngirl92]

How many integers from 1 to 10000, inclusive, do not conain digit 8?

Hey guys for this type of questions can we proceed by finding the no. of nos. WITH 8 and then substracting??
The nos. bet 0001 to 9999 containing 8 will be 3000 :roll: