How many 4-digit positive integers are there in which all 4 digits are

Positive integers - 2,4,6,8,0

Let the integers of a four digit positive number be ABCD

A can take four values (2,4,6,8)
B can take five values (0,2,4,6,8)
C can take five values (0,2,4,6,8)
D can take five values (0,2,4,6,8)

The total is 5*5*5*4

The answer according to me is 500

Take the task of creating the 4-digit numbers and break it into stages.

Stage 1: Select an even digit for the thousands digit.
Since 0 cannot be the first digit of a 4-digit number, this digit can be 2, 4, 6 or 8
So, we can complete stage 1 in 4 ways

Stage 2: Select an even digit for the hundreds digit.
This digit can be 0, 2, 4, 6, or 8
We can complete stage 2 in 5 ways

Stage 3: Select an even digit for the tens digit.
We can complete stage 3 in 5 ways

Stage 4: Select an even digit for the units digit.
We can complete stage 3 in 5 ways

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus a 4-digit) in (4)(5)(5)(5) ways (= 500 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEO

The 4 - digit integer should be comprised of the following digits:
2, 4 ,6, 8, 0

For it to be a 4 digits integer, the first digit cannot be zero.
Hence the following scenarios are possible

1st digit: (2,4,6,8)
2nd digit: (0,2,4,6,8)
3rd digit: (0,2,4,6,8)
4th digit: (0,2,4,6,8)

Total numbers possible = 4*5*5*5 = 500
Correct Option: C

E. 256

In the first place ( thousand place), i can choose 4 numbers : 2, 4, 6, 8 ( recall 0 is neither positive nor negative although O is even integer , don't confuse even with positive integer )
2nd place ( hundred place) can also be filled up by 4 number : 2, 4, 6, 8 and same logic applies to 3rd ( 10th ) and 4th( unit) . so number of ways is 4*4*4*4= 256 ways .

Important concepts : if a task has two steps , step 1 can be done in n ways and step 2 can be done in m ways , then number of easy the task can be done is n*m ways . Same concept applies in above question.

We need to find: # of 4-digit positive integers, all 4-digits are even

Digits can be : 0,2,4,6,8 (0 is an even number)
4 digit integer = "_ _ _ _"

first place (Thousands place ) can be filled with 4 even integers : 2,4,6,8 (Why not 0?? --> then the number will be a 3 digit number)
second place (Hundreds Place) can be filled with 5 even integers : 0,2,4,6,8 (there is no restriction on repetition of digits)
third place (Tens Place) can be filled with 5 even integers : 0,2,4,6,8
fourth place (Unit Place) can be filled with 5 even integers : 0,2,4,6,8

Hence total # : 4*5*5*5
=500

Hence C

4 options for digit 1 (2,4,6,8), 5 options for digits 2,3,&4 (2,4,6,8,0)

4*5*5*5=500
C

Even digits: 0,2,4,6,8
The thousands place can be filled in 4 ways
The hundreds place can be filled in 5 ways
The tens place can be filled in 5 ways
The ones place can be filled in 5 ways
Therefore total number of numbers = 4*5*5*5 = 500. Option C

Even Digits: 0,2,4,6,8 (Five digits total)

A 4-digit positive integer starts out the thousandths place. In the thousandths place, the digit cannot be 0, because this makes the number a three digit, rather than four digit.
Therefore, in the thousandths place, there is 4 possible options.
in the hundreds place, we have 5 possible options; the same goes for the tens and singles.
This information translates into this: 4x5x5x5 = 500

(C)

Hope this helps

Can anyone refer to this excellent point highlited?
zero is neither a negative or a positive number!
How come the answer considers 0 as positive?

MoriyaHa because if you think about it, the numbers start from 2,000 anyway (which will be a positive number). 0 is an even number and not a positive number. if you say 0 is a positive even integer, then technically you would include 0000 is as part of the 4 digit integers.

Solution:

The even digits are 0, 2, 4, 6, and 8. However, since the first digit (or the thousands place digit) can’t be 0, there are 4 choices for the thousands digit, while the other three places (hundreds, tens, and units) have 5 choices each. Therefore, the total number of 4-digit numbers with all even digits is 4 x 5 x 5 x 5 = 500.

Answer: C
_________________

If you have trouble finding a formula, you can analyze blocks of 1000 numbers (and quickly recognize a pattern):

0001 - 0999: There are no numbers in this section since they are all 3-digit integers, not 4-digit integers
1000 - 1999: There are no numbers in this section since the thousands digit (1) is always odd; therefore, it is not possible that all digits are even
2000 - 2999: Let's dig a bit deeper here:
The thousands digit is always 2: so 1 possibility
The hundreds, tens, and units digit can be 0, 2, 4, 6, 8: so each digit has 5 possibilities
1 * 5 * 5 * 5 = 125
3000 - 3999: same as 1000 - 1999
4000 - 4999: same as 2000 - 2999
...

Notice that you will have 125 different numbers that fit to our criteria in a section of 1000 numbers that begins with an even digit und 0 numbers in a section of 1000 numbers that begins with an odd digit:
2000 - 2999; 4000 - 4999; 6000 - 6999; 8000 - 8999
4 sections * 125 different numbers: 500 different numbers

PROPERTIES OF ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. Zero is neither positive nor negative (the only one of this kind)

4. Zero is divisible by EVERY integer except 0 itself (x/0 = 0, so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (x*0 = 0, so 0 is a multiple of any number, x)

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (x^0 = 1)

9. 0^0 case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), 0^n = 0.

11. If the exponent n is negative (n < 0), 0^n is undefined, because 0^n = 1/0^(-n) = 1/0. You CANNOT take 0 to the negative power.

12. 0! = 1! = 1.
_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________