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How many 4 digit numbers are there, if it is known that the [#permalink]
 28 Jan 2010, 15:45
Question Stats:
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65% (01:56) wrong based on 3 sessions
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]
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: Problem Solving [#permalink]
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. Cheers
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Re: Problem Solving [#permalink]
30 Jan 2010, 10:40
Because 0 is divisible by 3.
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How many 4 digit numbers are there, if it is known that the [#permalink]
31 May 2012, 00:04
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: How many 4 digit numbers are there [#permalink]
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. Regards, Shouvik.
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Re: How many 4 digit numbers are there, if it is known that the [#permalink]
31 May 2012, 01:32
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Re: Problem Solving [#permalink]
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]
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]
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]
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]
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]
05 Aug 2012, 10:19
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: Problem Solving
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05 Aug 2012, 10:19
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