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How many 4 digit numbers are there, if it is known that the

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How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 28 Jan 2010, 15:45
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D
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How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

A. 20
B. 150
C. 225
D. 300
E. 320
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Re: Problem Solving  [#permalink]

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New post 28 Jan 2010, 16:19
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How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?
A. 20
B. 150
C. 225
D. 300
E. 320

4 options for the first digit: 2, 4, 6, 8;
5 options for the second digit: 1, 3, 5, 7, 9;
4 options for the third digit: 2, 3, 5, 7;
4 options for the fourth digit: 0, 3, 6, 9.

Four digit # possible without the restriction (about the digit 2): 4*5*4*4=320

Numbers with two 2-s, 2X2X 1*5*1*4=20.

Thus there are 320-20=300 such numbers.

Answer: D.
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Re: How many 4 digit numbers are there  [#permalink]

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New post 31 May 2012, 01:23
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Joy111 wrote:
How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

A)20
B)150
C)225
D)300
E)320


Hi,

The first digit can be 2, 4, 6, 8.
The second digit can be 1, 3, 5, 7, 9
The third digit can be 2, 3, 5, 7
The fourth digit can be 0, 3, 6, 9

Case 1: Using 2 as the 1st digit only.

The first digit can be 2, 4, 6, 8. No. of selections = 4
The second digit can be 1, 3, 5, 7, 9. No. of selections = 5
The third digit can be 3, 5, 7. No. of selections = 3
The fourth digit can be 0, 3, 6, 9. No. of selections = 4

Total number of numbers = 4x5x3x4 = 240

Case 2: Using 2 as the 3rd digit only.

The first digit can be 4, 6, 8. No. of selections = 3
The second digit can be 1, 3, 5, 7, 9. No. of selections = 5
The third digit can be 2. No. of selections = 1
The fourth digit can be 0, 3, 6, 9. No. of selections = 4

Total number of numbers = 3x5x1x4 = 60

Hence, total number of numbers = 240 + 60 = 300

Answer is D.

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Re: Problem Solving  [#permalink]

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New post 30 Jan 2010, 08:07
Bunuel wrote:
sudai wrote:
Hi All,

Need your help in solving the below problem!

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

The answer is 300, but why....

Thank you very much!!


4 options for the first digit: 2, 4, 6, 8;
5 options for the second digit: 1, 3, 5, 7, 9;
4 options for the third digit: 2, 3, 5, 7;
4 options for the fourth digit: 0, 3, 6, 9.

Four digit # possible without the restriction (about the digit 2): 4*5*4*4=320

Numbers with two 2-s, 2X2X 1*5*1*4=20.

Thus there are 320-20=300 such numbers.


Why do we have to consider 0 for the last digit? Shouldn't it be only 3,6, and 9 ?

Please explain.

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Re: Problem Solving  [#permalink]

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New post 30 Jan 2010, 10:40
1
Because 0 is divisible by 3.
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Re: Problem Solving  [#permalink]

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New post 31 May 2012, 04:03
cipher wrote:

Why do we have to consider 0 for the last digit? Shouldn't it be only 3,6, and 9 ?

Please explain.

Cheers


I missed the '0' :(

hope to remember it in future
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 25 Jul 2012, 10:56
"4 options for the first digit: 2, 4, 6, 8;"

Aren't there 5 options ? 0 is even as far as know and so meets also the condition ?
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 25 Jul 2012, 12:24
Trying to save some time...

For the second digit, 5 possibilities: 1, 3, 5, 7, 9
For the last (fourth) digit, 4 possibilities: 0, 3, 6, 9

Therefore, the total number of possibilities should be a multiple of 20 = 5 * 4.
A is out, being too small, there are more possibilities for the first and the third digit.

So, I have to chose between D and E.

First digit, 4 possibilities: 2, 4, 6, 8
Third digit also 4 possibilities: 2, 3, 5, 7

It would give 4 * 4 = 16, and 16 * 20 = 320, but because we have to allow only one digit of 2, the final number should be less than 320.
Only 300 is left.

Well, I guess being lazy to carry out all the computations it isn't always safe, but sometimes it feels so good to take even a little shortcut...
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 26 Jul 2012, 04:40
Alexmsi wrote:
"4 options for the first digit: 2, 4, 6, 8;"

Aren't there 5 options ? 0 is even as far as know and so meets also the condition ?


0 is even indeed.
But if you take 0 for your first digit, it means you have a 3digits number and we are looking for a 4 digits number.
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Re: Problem Solving  [#permalink]

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New post 26 Jul 2012, 08:22
Bunuel wrote:
How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?
A. 20
B. 150
C. 225
D. 300
E. 320

4 options for the first digit: 2, 4, 6, 8;
5 options for the second digit: 1, 3, 5, 7, 9;
4 options for the third digit: 2, 3, 5, 7;
4 options for the fourth digit: 0, 3, 6, 9.

Four digit # possible without the restriction (about the digit 2): 4*5*4*4=320

Numbers with two 2-s, 2X2X 1*5*1*4=20.

Thus there are 320-20=300 such numbers.

Answer: D.


Hello Bunuel, can you please help me understand how u calculated for the restriction?
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Re: Problem Solving  [#permalink]

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New post 05 Aug 2012, 10:19
1
harshavmrg wrote:
Bunuel wrote:
How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?
A. 20
B. 150
C. 225
D. 300
E. 320

4 options for the first digit: 2, 4, 6, 8;
5 options for the second digit: 1, 3, 5, 7, 9;
4 options for the third digit: 2, 3, 5, 7;
4 options for the fourth digit: 0, 3, 6, 9.

Four digit # possible without the restriction (about the digit 2): 4*5*4*4=320

Numbers with two 2-s, 2X2X 1*5*1*4=20.

Thus there are 320-20=300 such numbers.

Answer: D.


Hello Bunuel, can you please help me understand how u calculated for the restriction?



Restriction where the 4 digit number has two twos.
The two 2s can occur in a digit as shown by Bunuel- 2X2X
Therefore number of such numbers 1*5*1*4=20
Hope its clear now.
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 17 Jul 2013, 00:58
1
sudai wrote:
How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

A. 20
B. 150
C. 225
D. 300
E. 320


1. first digit is even - 2,4,6,8
2. second digit is odd - 1,3,5,7,9
3. third is prime - 2,3,5,7
4. fourth is divisible by 3- 0,3,6,9

In addition to the above 4 conditions we have the following:
digit 2 can be used only once

Assume two in (1) and (3) are both not used
the number of possibilities is 3*5*3*4=180
In addition to this either the two in (1) or the two in (3) is used
Number of possibilities if the two in (1) is used - 1*5*3*4=60
Number of possibilities if the two in (3) is used - 3*5*1*4=60
total number of possibilities= 180+60+60=300
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 17 Jul 2013, 08:45
Funny how this is a 700 level question...not much to it really.

Scenario #1: Start with a 2 as the first number. This leaves 5,3 and 4 choices respectively for the rest. Total = 60 #'s
Scenario #2: Start with a 2 as the third number. This leaves 3,5 and 4 choices respectively for the rest. Total = 60 #'s
Scenario #3: Remove 2 as an option for the 1st and 3rd numbers. This leaves 3,5,3 and 4 choices respectively for the rest. Total = 180#'s

Add them up....60+60+180 = 300 Total numbers can be formed. (D)
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 17 Jul 2013, 10:04
1
Since this has been solved a few times (all correctly), I'd like to spend a minute looking at the answer choices. This type of question is very hard to backsolve, but it's relatively easy to see where the trap answers lie.

How many 4 digit numbers are there, if it is known that the first digit is even, the second is odd, the third is prime, the fourth (units digit) is divisible by 3, and the digit 2 can be used only once?

A. 20
B. 150
C. 225
D. 300
E. 320

The correct answer is D, but if you neglect the restriction of 2 being used twice, the total number of options is E: 320. If you go the other way and over-emphasize this restriction, you end up with answer choice A: 20. C is the choice if you forget that zero is also a multiple of 3 (classic GMAT trap). You'd then have 3/4 as many choices and would get to C. B is a little harder to get to without making multiple mistakes, but it may be a fairly tempting number if you're guessing blindly.

It's obviously crucial to determine which answer choice is the correct one, but there is value in analyzing the other choices and seeing where the GMAT thinks your brain may go. Remember this exam is nothing if not foreseeable and preparable.

Hope this helps!
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 30 Sep 2014, 07:42
Did

0,2,4,6,8
1,3,5,7,9
2,3,5,7 (removed 2 as condition says)
0,3,6,9

5*5*3*4=300

P.S. GMATprep considers 0 as even integer, so if use Bunuel's approach, the correct answer is 380
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How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 05 Dec 2016, 06:01
Hi,

I dont quite get why we dont use combination here. meaning there are 10 numbers for the first digit, but we can select only 4, hence 10C4. Hence, the total number of ways shouldnt be:

10C4 x 10C5 x 10C4 x 10C4

?

Can anyone explain to me?
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Re: How many 4 digit numbers are there, if it is known that the  [#permalink]

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New post 06 Feb 2018, 12:03
Hi All,

This question has a lot of details to it, but the math involved isn't too bad (however, you will have to account for a series of 4-digit numbers that are NOT allowed, since they contain more than one 2 in their digits).

We're given a series of facts about the 4 digit number:
The first digit is EVEN: 2, 4, 6 or 8 (but not 0, since a 4-digit number can't start with 0)
The second digit is ODD: 1, 3, 5, 7, 9
The third digit is PRIME: 2, 3, 5, 7
The fourth digit is divisible by 3: 3, 6, 9, 0

If there were NO other restrictions, then the total number of 4-digit numbers would be:

(4)(5)(4)(4) = 320 options

However, there IS a restriction - the digit "2" can be used no more than once. Thus, any number that includes MORE than one 2 has to be removed... Thankfully, there aren't that many numbers that fit that description. If the first and third digits are both 2s, then the there are...

(1)(5)(1)(4) = 20 numbers with two 2s in the digits. Those 20 options have to be removed, which leaves us with...

320 - 20 = 300 4-digit numbers.

Final Answer:

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