Working with positive integers less than 5,000...

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rem45acp

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I am completely stuck on this problem. I got to letter c and I'm not even sure if these are correct!

a) How many positive integers are there less than 5,000? (My answer 4,999, because zero is not positive or negative)
b) How many OF THESE have four digits? (My answer is 4,999-999 = 4,000)
c) How many OF THESE have no repeated digits?

My reasoning here is there can be 4 choices for the first digit (1 through 4), 9 choices for the second digit (the chosen first digit cannot be repeated but zero can now be used), then 8 choices for the third digit and 7 for the fourth digit, or 4 x 9 x 8 x 7 = 2016 possible numbers.

d) How many OF THESE are divisible by 5? (Stuck!)
e) How many of those divisible by 5 are even? (Stuck!)
f) How many of those divisible by 5 are odd (Stuck!)

Any help would greatly be appreciated!!
 
rem45acp said:
I am completely stuck on this problem. I got to letter c and I'm not even sure if these are correct!

a) How many positive integers are there less than 5,000? (My answer 4,999, because zero is not positive or negative)
b) How many OF THESE have four digits? (My answer is 4,999-999 = 4,000)
c) How many OF THESE have no repeated digits?

My reasoning here is there can be 4 choices for the first digit (1 through 4), 9 choices for the second digit (the chosen first digit cannot be repeated but zero can now be used), then 8 choices for the third digit and 7 for the fourth digit, or 4 x 9 x 8 x 7 = 2016 possible numbers.

d) How many OF THESE are divisible by 5? (Stuck!)
e) How many of those divisible by 5 are even? (Stuck!)
f) How many of those divisible by 5 are odd (Stuck!)

Any help would greatly be appreciated!!
Click to expand...

Numbers divisible by '5' has '5' or '0' at their unit digit.

Now think a little bit about the other two.....
 
I know that it must end in 5 or 0, but aren't there gaps in the 2016 possible numbers? They're not consecutive? Perhaps is it 2016/5 and round it down?

EDIT: What do you mean by unit digit?
 
rem45acp said:
b) How many OF THESE have four digits? (My answer is 4,999-999 = 4,000)
c) How many \(\displaystyle > > > \)OF THESE \(\displaystyle < < < \)have no repeated digits?

My reasoning here is there can be 4 choices for the first digit (1 through 4), . . .
Click to expand...

rem45acp,

please answer this:

You see that phrase I highlighted. I can take "OF THESE" to mean
it is referring back to the 4,999 integers, not the 4,000 integers having
exactly four digits.

Which is it? The second use of "OF THESE" is ambiguous to which
set of integers is being referred to.
 
OF THESE means in this case use the answer to the previous question, and "of those" means use the answer from d) to answer e) and f)
 
Denis said:
You also need the ones that have 1, 2 and 3 digits.
Click to expand...

I don't think that I do, since the next question uses the answer to the previous one. Hence, how many in the set of those 2016 numbers are divisible by 5. Since there are obviously many numbers missing from that set, I don't know how to figure this out without listing them all out.
 
rem45acp said:
I don't think that I do, since the next question uses the answer to the previous one. Hence, how many in the set of those 2016 numbers are divisible by 5. Since there are obviously many numbers missing from that set, I don't know how to figure this out without listing them all out.
Click to expand...

How many of those have 5 at unit digit?

The thousand digit has 4 choices

The hundred digit has 8 choices (cannot have 5 and the digit in thousand position)

The ten digit has 7 choices

The unit digit has 1 choice (5)

So the number of numbers = 4 * 8 * 7 = 224
 
Hello, rem45acp!

a) How many positive integers are there less than 5,000?
(My answer 4,999, because zero is not positive or negative) ..Correct!

b) How many OF THESE have four digits?
(My answer is 4,999-999 = 4,000) . Right!

c) How many OF THESE have no repeated digits?
My reasoning here is there can be 4 choices for the first digit (1 through 4),
9 choices for the second digit (the chosen first digit cannot be repeated but zero can now be used),
then 8 choices for the third digit and, 7 for the fourth digit,
or 4 x 9 x 8 x 7 = 2016 possible numbers. . Excellent!
Click to expand...


d) How many OF THESE are divisible by 5?
Click to expand...
To be divisible by 5, the number must end in 0 or 5.

Numbers ending in 0: .\(\displaystyle \_\;\_\;\_\;0\)
. . 1st digit: 4 choices n {1, 2, 3, 4}
. . 2nd digit: 8 choices . (Not the 1st digit and not 0)
. . 3rd digit: 7 choices n (Not 1st digit, not 2nd digit, and not 0)
Hence:.\(\displaystyle 4\cdot8\cdot7 \,=\,224\) numbers ending in 0.

Numbers ending in 5: .\(\displaystyle \_\;\_\;\_\;5\)
. . 1st digit: 4 choices n {1, 2, 3, 4}
. . 2nd digit: 8 choices . (Not the 1st digit and not 5)
. . 3rd digit: 7 choices n (Not 1st digit, not 2nd digit, and not 5)
Hence:.\(\displaystyle 4\cdot8\cdot7 \,=\,224\) numbers ending in 5.

Therefore:.\(\displaystyle 224 + 224 \,=\,448\) numbers divisible by 5.


e) How many of those divisible by 5 are even?
f) How many of those divisible by 5 are odd?
Click to expand...
We've already answered these questions.

. . \(\displaystyle \begin{array}{cc}\text{The numbers ending in 0 are even:} & 224 \\ \text{The numbers ending in 5 are odd:} & 224 \end{array}\)
 
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