how many 3-dig. odd integers do not contain digit "5"?

[rennster]

This is driving me crazy.
Here's the question:
How many 3 digit integers are odd and do not contain the digit "5"?

These are my thoughts so far:
I know there are 449 positive odd three-digit integers and each digit has 4 odd positive integers.

 

[mmm4444bot]

Re: counting problem

EDIT: This post is not entirely correct. I hope I got it right on my subsequent attempt!

I know there are 449 positive odd three-digit integers

Hi rennster:

449 is not correct, but you're only off by one.

Let's forget about negative numbers for now to make the discussion easier; we can always double our final result later to include the negative counterparts.

There are different ways to count the number of 3-digit odd integers; here's the method that popped into my mind first.

We know that this set of integers is {101, 103, 105, ... 999}.

A formula for ALL of the odd integers is $\displaystyle 2n - 1$ as n goes from 1 to infinity.

We're interested in {101, 103, 105, ...}. We get these numbers by adding the odd integers to 100.

$\displaystyle 100 + 2n - 1$ simplifies to $\displaystyle 2n + 99$

Therefore, a formula for the numbers {101, 103, 105, ...} is

$\displaystyle 2n + 99$ for $\displaystyle n = {1, 2, 3, ...}$

We want our sequence of numbers to stop at 999. What value of n generates the number 999?

$\displaystyle 2n + 99 = 999$

$\displaystyle 2n = 900$

$\displaystyle n = 450$

Therefore, there are 450 three-digit odd integers because there is a one-to-one correspondence between the sets {1, 2, 3, ... 450} and {101, 103, 105, ... 999}.

Perhaps you arrived at 449 instead because you did the following.

$\displaystyle \frac{999-101}{2} = 449$

The error in logic with this approach is that subtraction does not account for the numbers at each end of an interval when counting. Subtraction only gives you the difference between the numbers at each end of an interval.

For example, how many integers are there from 1 through 10?

You might be tempted to calculate 10 minus 1 and report that there are nine integers from 1 through 10. That's wrong.

... each digit has 4 odd positive integers.

Huh? :shock:

Okay -- maybe you are trying to say something about the fact that after removing all of the numbers that end in the digit 5, the last digit of the remaining numbers cycle through the other four odd numbers less than 10 (i.e., the last digit follows the pattern 1, 3, 7, 9, 1, 3, 7, 9, 1, 3, 7, 9, ...).

If this is what you were thinking, then you're on the right track.

In the formula $\displaystyle 2n + 99$, the last digit of the resulting numbers cycles through 1, 3, 5, 7, 9 for every five increments of n.

n = 1 to 5 ? 101, 103, 105, 107, 109
n = 6 to 10 ? 111, 113, 115, 117, 119
n = 11 to 15 ? 121, 123, 125, 127, 129
n = 16 to 20 ? 131, 133, 135, 137 ,139
and so on ...

See the pattern?

In other words, there is exactly one number that ends in the digit 5 generated each time n increases through five values.

Therefore, the total number of times that n increases through five values while going from 1 to 450 is the same as the total number of integers that end in the digit 5 within the set {101, 103, 105, ... 999}.

It's very simple to calculate this (i.e., the number of 5-value increases that n goes through on its way from 1 to 450).

Subtracting it from the number of 3-digit odd integers results in the number of 3-digit odd integers that do not end in the digit 5.

Doubling this last calculation accounts for both the negatives and the positives, and that gives the answer to your original question.

So, what do you get :?:

Cheers,

~ Mark

 

[mmm4444bot]

I Misread Your Original Problem

Whoops!

Somebody just pointed out that I misread your original problem.

Somehow, I got the idea in my head that we're talking about 3-digit integers that END in 5.

Okay, let me scramble for a few minutes to see if I can salvage anything from my first post ...

(Arrrrgh)

 

[mmm4444bot]

Re: counting problem

Hi rennster:

Again, I apologize for my prevous error. (Now I'm a sitting duck for the regulars on this board. :wink

Well, hopefully you learned something in my initial attempt. I'll go ahead now and complete the entire task.

Let's break our set of 3-digit odd integers into nine subsets as follows.

{101, 103, 105, ... 199}
{201, 203, 205, ... 299}
{301, 303, 305, ... 399}
{401, 403, 405, ... 499}
{501, 503, 505, ... 599}
{601, 603, 605, ... 699}
{701, 703, 705, ... 799}
{801, 803, 805, ... 899}
{901, 903, 905, ... 999}

Now let's see how many numbers we need to throw out because they contain a digit 5.

Clearly, we need to throw out the entire subset {501, 503, 505, ... 599} because every element in this set contains the digit 5.

This subset contains 50 elements.

Using the approach in my initial post, we have the formula

$\displaystyle 2n + 99$ for $\displaystyle n = 1 to 50$

to generate the numbers in the subset {101, 103, 105, ... 199}

Every time n advances through 5 values, a number ending in the digit 5 is generated.

50/5 = 10, so there are 10 elements in the set {101, 103, 105, ... 199} that end in the digit 5.

There are four additional numbers that contain a 5 (the ones which contain a 5 as the middle digit: 151, 153, 157, and 159).

In other words, we have to throw out 14 numbers from the subset (101, 103, 105, ... 199} to purge it of all numbers that contain the digit 5.

Since there are eight subsets that follow the same pattern, we have to throw out 8 * 14, or 112 numbers total.

50 + 112 = 162

There are 162 odd 3-digit integers that contain the digit 5.

450 - 162 = 288

Double this amount to account for the negative odd 3-digit integers that do not contain the digit 5.

I now claim that the answer to your original question is 576.

(If this is wrong, then I'm a dead duck.)

:lol:

~ Mark

 

[Deleted member 4993]

Re: counting problem

Another way to look at the problem.

I missed the part where '5 was excluded. - corrected now

Three digit numbers are made of numbers 0 thru 10 distributed among three slots (repeatation allowed - and with certain restrictions)).

Considering only positive numbers

Then the first slot (100 th position) can be filled with 8 numbers (1 - 9 and 0 & 5 not allowed)

the 10th position can be filled 9 ways ( 0-9 - 5 not allowed)

the unit position can be filled 4 ways (1,3,7 & 9 - giving you odd numbers)

So the total number of positive three digit odd numbers (without 5) = 8 * 9 * 4 = 288

Now - your study guide is incorrect in giving you the answer as 288 - because the set of odd (or even numbers) include negative numbers.

 

[rennster]

Re: counting problem

according to the study guide the answer was 288. Thanks for the explanations. They really help as the guide only presents questions but don't explain the process.

 

[galactus]

Re: counting problem

We can not have a 0 or a 5 for the first digit so there are 8 choices for the first.

There are 9 choices for the second because, again, no 5.

There are 4 choices for the last because it must be an odd number. So, we have 1,3,7,9 for the last digit.

(8)(9)(4)=288