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

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How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 30 Sep 2016, 06:28
3
6
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (01:07) correct 34% (01:20) wrong based on 340 sessions

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Re: How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 30 Sep 2016, 07:03
2
1
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
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Re: How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 30 Sep 2016, 09:44
1
Bunuel wrote:
How many 4-digit positive integers are there in which all 4 digits are even?

A. 625
B. 600
C. 500
D. 400
E. 256


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
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Re: How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 30 Sep 2016, 09:50
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.
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How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 30 Sep 2016, 13:46
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
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Re: How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 30 Sep 2016, 14:16
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
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Re: How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 01 Mar 2017, 02:09
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
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Re: How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 06 Mar 2018, 23:23
Bunuel wrote:
How many 4-digit positive integers are there in which all 4 digits are even?

A. 625
B. 600
C. 500
D. 400
E. 256


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
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Re: How many 4-digit positive integers are there in which all 4 digits are  [#permalink]

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New post 08 Jul 2018, 02:25
sajib2126 wrote:
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.



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?
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Re: How many 4-digit positive integers are there in which all 4 digits are &nbs [#permalink] 08 Jul 2018, 02:25
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