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Q1. How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated? a) 336 b) 784 c) 1120 d) 1804 e) 1936

Q2. How many 4-digit numbers can be formed, so that each contains exactly 3 distinct digits? a) 1944 b) 3240 c) 3850 d) 3888 e) 4216

Note: Found these questions. But not the OAs. Can anyone help? Thanks in advance.

Note: the above questions are beyond the GMAT scope.

How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated? A. 336 B. 784 C. 1120 D. 1804 E. 1936

I believe with "no two digits are repeated" the question means that all 4 digits are distinct.

Number is divisible by 4 if the last two digits form a number divisible by 4.

Therefore last two digits can be: 00; 04; 08; 12 16; ... 96.

Basically multiples of 4 in the range 0-96, inclusive. Multiples of 4 in a range 0-96, inclusive are \(\frac{last \ multiple \ in \ the \ range \ - \ first \ multiple \ in \ the \ range}{4}+1=25\) (for more on this issue: totally-basic-94862.html#p730075).

But 3 numbers out of these 25 are not formed with distinct digits: 00, 44, and 88. Hence the numbers we are looking for can have only 22 endings.

If there is 0 in the ending (04, 08, 20, 40, 60, 80 - total 6 such numbers), then the first and second digit can take 8 and 7 choices each = 56 choices total. As there are 6 numbers with 0 in ending, hence total 6*56=336.

If there is no 0 in the ending (total 22 - 6 with zero = 16 such numbers), then the first digit can take 7 choices (10 - 2 digits in the ending and zero, so total 3 digits = 7, as 4-digit number can not start with zero) and the second digit can take 7 choices too (10 digits - 3 digits we've already used) = 7*7=49 choices total. As there are 16 numbers without zero in ending, hence total 16*49=784.

TOTAL: \(336+784=1120\).

Answer: C.

How many 4-digit numbers can be formed, so that each contains exactly 3 distinct digits? A. 1944 B. 3240 C. 3850 D. 3888 E. 4216

As there should be 3 distinct digits in 4, the number will have 2 same digits and other 2 distinct - \(aabc\) (1123, 3477, ...)

\(C^3_{10}=120\) - # of ways to choose 3 distinct digits out of 10; \(C^1_3=3\) - # of ways to choose which digit will be represented twice; \(\frac{4!}{2!}=12\) - # of permutation of digits \(aabc\);

\(C^3_{10}*C^1_3*\frac{4!}{2!}=4320\).

Now, out of these 4320 numbers some will start with zero, but 4-digit number can not start with zero as in this case it becomes 3 digit number. So how many out of these 4320 start with zero? As no digit has any preferences so equal numbers will start with each digit: 1/10 will start with 1, 1/10 with 2, 1/10 with 3 and so on. Thus 1/10=0.1 of 4320 will start with 0 and 0.9 of 4320 will start with digit other than zero.

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21 Dec 2013, 09:57

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01 Jan 2014, 16:12

This is how I proceed for those two :

First one;

I used a simple methode:

- You choose the number of integers divided by 4 between 0 and 1000. This number is 250 (use the formula) - You multiply this number by 9 (because your number cannot start with a 0, so it is between 1000 and 9999). - You find 2250. But you are seeking for distinct numbers, so you need to divided you results by 2 (in order to be divded by 4 a number needs to have it last TWO digits divided by four such as 36 12...), which gives 1125.

You can now chose answer C! (approximation method)

for 2:

- For the first number you can chose 1 among 9 (without the zero) than you can chose 9 again (adding back the zero) than 8. - You need then to chose one 3 digit among the 3. - You have therefore 9*9*8*3. But you need to multiply by 4 (permutation). But watch out, you can end with a zero as a primary digit! So what do you do? You divide by 2 (to avoid those two cases when the first digit = 0) - Final number 3888.

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05 Mar 2014, 02:18

2

This post received KUDOS

Alt way

1. 4 digit different, multiple of 4, 4 digit numbers ?? calculate the even , four digit differ number, - - - - , 8 * 8 * 7 * 5, { (unit, 0,2,4,6,8) , thousand ( not 0 and one already come at unit)/2 = 1120 why /2 because every second no in even series is multiople of 4

2. - - - - 9 * 9 * 8 * 1 * 6 { start from left, at thousand, 9 digits (not included 0) , at hundred 9 (0 included) , at tens 8, at unit 1( same digit) why multiplied with 6 - same digits (1,2;13;14;23;24;34) OR we may apply ( 4!/2)/2 why again divided by 2 again because if for example in 1st and 2 2nd and 2nd and 1st are same will creat same no

Re: How many even 4-digit numbers can be formed, so that the [#permalink]

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31 Aug 2014, 10:33

Bunuel wrote:

jusjmkol740 wrote:

Q1. How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated? a) 336 b) 784 c) 1120 d) 1804 e) 1936

Q2. How many 4-digit numbers can be formed, so that each contains exactly 3 distinct digits? a) 1944 b) 3240 c) 3850 d) 3888 e) 4216

Note: Found these questions. But not the OAs. Can anyone help? Thanks in advance.

Note: the above questions are beyond the GMAT scope.

How many even 4-digit numbers can be formed, so that the numbers are divisible by 4 and no two digits are repeated? A. 336 B. 784 C. 1120 D. 1804 E. 1936

I believe with "no two digits are repeated" the question means that all 4 digits are distinct.

Number is divisible by 4 if the last two digits form a number divisible by 4.

Therefore last two digits can be: 00; 04; 08; 12 16; ... 96.

Basically multiples of 4 in the range 0-96, inclusive. Multiples of 4 in a range 0-96, inclusive are \(\frac{last \ multiple \ in \ the \ range \ - \ first \ multiple \ in \ the \ range}{4}+1=25\) (for more on this issue: totally-basic-94862.html#p730075).

But 3 numbers out of these 25 are not formed with distinct digits: 00, 44, and 88. Hence the numbers we are looking for can have only 22 endings.

If there is 0 in the ending (04, 08, 20, 40, 60, 80 - total 6 such numbers), then the first and second digit can take 8 and 7 choices each = 56 choices total. As there are 6 numbers with 0 in ending, hence total 6*56=336.

If there is no 0 in the ending (total 22 - 6 with zero = 16 such numbers), then the first digit can take 7 choices (10 - 2 digits in the ending and zero, so total 3 digits = 7, as 4-digit number can not start with zero) and the second digit can take 7 choices too (10 digits - 3 digits we've already used) = 7*7=49 choices total. As there are 16 numbers without zero in ending, hence total 16*49=784.

TOTAL: \(336+784=1120\).

Answer: C.

How many 4-digit numbers can be formed, so that each contains exactly 3 distinct digits? A. 1944 B. 3240 C. 3850 D. 3888 E. 4216

As there should be 3 distinct digits in 4, the number will have 2 same digits and other 2 distinct - \(aabc\) (1123, 3477, ...)

\(C^3_{10}=120\) - # of ways to choose 3 distinct digits out of 10; \(C^1_3=3\) - # of ways to choose which digit will be represented twice; \(\frac{4!}{2!}=12\) - # of permutation of digits \(aabc\);

\(C^3_{10}*C^1_3*\frac{4!}{2!}=4320\).

Now, out of these 4320 numbers some will start with zero, but 4-digit number can not start with zero as in this case it becomes 3 digit number. So how many out of these 4320 start with zero? As no digit has any preferences so equal numbers will start with each digit: 1/10 will start with 1, 1/10 with 2, 1/10 with 3 and so on. Thus 1/10=0.1 of 4320 will start with 0 and 0.9 of 4320 will start with digit other than zero.

TOTAL: \(4320*0.9=3888\).

Answer: D.

Hope it helps.

Hi Bunuel, Do you have another method for the second question? One where you calculate once for the options without "0" and once for the options with "0"? I would like to see how you do that because that's what I defaulted to, but can't get it to work...

gmatclubot

Re: How many even 4-digit numbers can be formed, so that the
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31 Aug 2014, 10:33

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