How many of the 343 smallest positive integers written in base 7 use 4 or 5 (or both) as a digit?

RektTheNoob
Sep 15, 2018

#1**+2 **

How many of the 343 smallest positive integers written in base 7 use 4 or 5 (or both) as a digit?

Ill add the word positive. Smallest positive integers.

343 base10

7^3=343

so 343 base10 = 1000 base 7

There are a total of 343 numbers.

Lets think about the base 7 number. And first I am only going take a look at the ones less than 1000 base 7

The 7^2 digit can be 0 to 6 that is 7 digits

The 7^1 digit can be 0 to 6

The units digit can be 0 to 6 so long as the number is less than 1000 base7

How many have numbers less than 1000 base 7 have NO 4 or 5.

Each digit can be 0,1,2,3 or 6

5*5*5 = 125

[I have actually included 000 and not included 1000 but one in one out makes no difference]

So that is 125 that do not have a 4 or a 5

343- 125 = 218

So I think **218 **do have a either a 4 or a 5 or both.

Melody
Sep 15, 2018

#2**+2 **

I don't think you got this one quite right Melody.

Consider the 1 digit numbers. The digit can't be 0, 4, 5. That leaves 4 possible

2 digit numbers have 4C1 x (5C1) = 20 possibilities

3 digit numbers have 4C1 x (5C1)^{2} = 100 possibilities

There is 1 4 digit number, 1000, and it is counted

This gives 4 + 20 + 100 + 1 = 125 numbers

Oh, duh. Ok, These are the numbers that don't match what you want.

We have to subtract this from the 343 and we get

343 - 125 = 218

Just like Melody got. Sorry for the confusion.

Rom
Sep 15, 2018

#4**+2 **

Hi Rom, I think you have interpreted the question incorrectly.

The question says we **must use** a 4 or a 5 or both.

If it is only one digit there are only 2 possibilities.

It must be 4 or 5. So I think that you have answered a different question.

I tried doing it a different way but I still got my original answer. :/

Melody
Sep 15, 2018

#5**+1 **

L:ets try again.

1000 does not have a 4 or a 5 so it is no good.

I want to know how many 3 digit base 7 numbers have at least one 4 or at least one 5

4@@ 6 *3 =18

4@# 6*5 *3 =90

44@ 6 *3 = 18

444 1

total = 127

same for 5s = 127

then

454 3

455 3

45* 5*3!=30

total = 36

one or the other or both = 2*127-36 = 218 ** So 218 have either a 4 or a 5 or both.**

There are 343 numbers altogether so that just leaves 125 who do not have either one.

Melody
Sep 15, 2018