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

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28 Jan 2010, 15:45

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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?

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

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 ?

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

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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

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

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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...
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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?
_________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

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.

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

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17 Jul 2013, 00:58

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
_________________

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

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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)

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.

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02 Sep 2014, 20:50

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