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How many even 3 digit integers greater than 700 with distinct non zero

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How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 02 Sep 2004, 15:21
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How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 20 Jun 2012, 01:59
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twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77


Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for XYZ, where Z is an even number.

If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;

If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;

Total: 56+21=77 numbers.

Answer: E.

Hope it's clear.
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 03 Sep 2004, 07:47
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Exactly, the numbers need to be distinct. My answer:
The below are no. of ways of filling units digit, tenth digit and hundredth digit, in that order.

701-799 = 4.7.1 = 28
800-899 = 3.7.1 = 21
900-999 = 4.7.1 = 28
Total = 77

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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 02 Sep 2004, 19:29
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1. Number of 3 digit even numbers >than 700 and <1000 = 149. We can use Arthmetic progression to find 149. How? 702 is the first even number and 998 is the last even number. The difference is 2 between the numbers. So, we have 702 + (n-1)2 = 998 or n=149.

2. Number of 3 digit even numbers that need to eliminated = 41, How? 702, 704, 706, 708, 710 => 5 numbers x 3 (for 70x, 80x and 90x series)= 15. Also, 720, 730...790 - 8 numbers x 3 = 24. Also 800 and 900 - 2 numbers. So, 15 + 24 + 2 = 41.

3. So net result is 149-41 = 108.

4. C is the answer.
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 02 Sep 2004, 23:53
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How many even 3 digit integers greater than 700 with distinct non zero digits are there ?
A 729
B 243
C 108
D 88
E 77

I agree with ur explanation but using another method i m getting D...plz tell me wat is going wrong

the no.s are even, distinct n each place has a non-zero no.....

If the 1st hundreds digit is 7.......units can be (2,4,6,8)...i.e 4 possibilities....n tens can be any of the remaining 8 digits....(7 is in hundredes n one even no. occupies tens place)....so total options 8*4=32

for 8 in hundreds........8*3 options.....coz 8 in hundreds is already an even no.....so units can be any of the 3 remaining even....so that gives 24

for 9 in hundreds...again 8*4=32


so 32+24+32=88.........

although it seems long and may not seem to be a GMAT question which takes less time........it took no more than 30 sec to solve....

why this difference in answers....plz clarify!!!!
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 03 Sep 2004, 00:23
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I have C as well.

the only hundred digit that are possible are 7, 8 and 9 (3)
the only ten digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9 (9)
the only unit digit that are possible are 2, 4, 6, 8 (4)

what I did is: 3 * 9 * 4 = 108

With a lot of practise these questions are better to understand.

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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 03 Sep 2004, 02:56
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OA is E

Please note that these 3 digits have to be distinct so 771 for instance is not valid

It took me 10 minutes after 2 mistakes :

even integers in the range : 149
0 as last digit (including 00) = 29
0 as tens digit = 12 (702 704 706 708, ...)
Distinct digits :
7>9 : hundreds digit
1>9 : tens digit
2- 4-6-8 :units digit

=31

Result = 149-29-12-31=77
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 20 Jun 2012, 01:40
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Hi,

My soltuion is as follows:
You have to find the even (3 digit) numbers greater than 700, with non-zero digits as well as distict digits. So, available digits would be (1 to 9)
Even numbers starting with 7 = 1*7*4 = 28 (Hundredth digit is 7 - so, only 1 choice, unit digit can be (2, 4, 6, 8). Now tens digit will not be 7 & a digit chosen at units place - 7 possibilities)
Even numbers starting with 8 = 1*7*3 = 21
Even numbers starting with 9 = 1*7*4 = 28
Total numbers = 77

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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 24 Jul 2013, 13:39
venksune wrote:
1. Number of 3 digit even numbers >than 700 and <1000 = 149. We can use Arthmetic progression to find 149. How? 702 is the first even number and 998 is the last even number. The difference is 2 between the numbers. So, we have 702 + (n-1)2 = 998 or n=149.

2. Number of 3 digit even numbers that need to eliminated = 41, How? 702, 704, 706, 708, 710 => 5 numbers x 3 (for 70x, 80x and 90x series)= 15. Also, 720, 730...790 - 8 numbers x 3 = 24. Also 800 and 900 - 2 numbers. So, 15 + 24 + 2 = 41.

3. So net result is 149-41 = 108.

4. C is the answer.


I am clear on the explanation provided by Bunuel for this question. However I am trying to understand the original answer provided using Arithmetic Progression. Where does the ( n-1)*2 come from ? Please explain and clarify.

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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 24 Jul 2013, 22:46
1
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hb wrote:
venksune wrote:
1. Number of 3 digit even numbers >than 700 and <1000 = 149. We can use Arthmetic progression to find 149. How? 702 is the first even number and 998 is the last even number. The difference is 2 between the numbers. So, we have 702 + (n-1)2 = 998 or n=149.

2. Number of 3 digit even numbers that need to eliminated = 41, How? 702, 704, 706, 708, 710 => 5 numbers x 3 (for 70x, 80x and 90x series)= 15. Also, 720, 730...790 - 8 numbers x 3 = 24. Also 800 and 900 - 2 numbers. So, 15 + 24 + 2 = 41.

3. So net result is 149-41 = 108.

4. C is the answer.


I am clear on the explanation provided by Bunuel for this question. However I am trying to understand the original answer provided using Arithmetic Progression. Where does the ( n-1)*2 come from ? Please explain and clarify.

Image Posted from GMAT ToolKit



First note that the explanation you are referring to has incorrect answer. Also, this method of solving is extremely treacherous - there are far too many limitations on the acceptable numbers to use this method. You must follow Bunuel's method for this question.

As for (n - 1)*2, we are given this particular sequence: 702, 704, 706, ..... 996, 998
We need to find how many numbers there are here. One way of doing that is using arithmetic progression concept.
nth term = First term + (n - 1)*Common difference
(last term) 998 = 702 + (n - 1)*2
Here, n will be the total number of terms.

For more on AP and the related formulas, check: http://www.veritasprep.com/blog/2012/03 ... gressions/
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 15 Feb 2014, 04:07
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twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77

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File comment: The correct answer is E (see the attached file)
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 16 Feb 2014, 13:02
How I was able to guess the right answer:

Find no. of even 3-digit #'s greater than 700:

(1000-702)/2+1 = 150

Subtract numbers with 0 in tens place (Excluding multiples of 100):

I.e. any even number ending in 02-08. 4*3 = 12

So 150-12 = 138

Subtract numbers with 0 in units place (again, excluding multiples of 100)

I.E. 710, 720, 730, etc... 9*3 = 27

So 138-27 = 111

Subtract multiples of 100

I.e. 800,900,100

So 111-3= 108

Subtract even numbers with idential digits in the tens and units place

I.e. 722, 744, 766, 788. 4*3 = 12

So 108-12 = 96

Subtract even numbers with identical digits in hundreds and tens place

I.e. 772, 774, 882, 884, etc...4*3 = 12

96-12=88

Subtract numbers with 3 identical digits

The only one in the set is 888.

88-1= 87. Once I got to this point I stopped, because the only remaining answer choice that could possibly be right is E (77). To finish the calculation, you subtract the no. of even numbers with identical digits in the hundreds and ones place (i.e. 808, 818, 828, etc...) and you get 77.

Maybe not the ideal way to solve the problem but it's how I worked through it while never having tried a problem like this before. Took me about 4 minutes.
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 03 Mar 2014, 01:45
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Alt solution

lets consider only 2,4,6 as unit digit as of now
--- (ways to fill unit 3, tens 7, hundred 3) +
now consider only 8 as unit digit
--- (ways to fill unit 1, tens 7, hundred 2) = 63+14 = 77
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 18 Apr 2014, 13:08
3
1
XYZ

No of ways to enter hundreds digit = 3 ( 7, 8, 9 ; Because the number has to be more than 700)

No of ways to enter units digit = 2, 4, 6, 8 (We cannot enter 0 as all digits are non zero, and the numbers are to be even so 1, 3, 5, 7, 9 are out)

No of ways to enter tenth digit = 10 - 3 (All digits are to be distinct. Hence we cannot have 0 and the two digits used in units and hundreds place).

Hence total number of digits = \(3 * 7 * 4\) = 84.

Wait if we have 8 on the units and hundreds place then we are repeating digits. Hence we must remove them as per our initial conditions that all digits are distinct.

No of such numbers

No of ways to enter hundreds digit = 1 (As were considering 8 in units and hundreds place)

No of ways to enter tenth digit = 7 (All digits are to be distinct. Hence we cannot have 0 and the two digits used in units and hundreds place).

No of ways to enter tenth digit = 1 (As were considering 8 in units and hundreds place)

No of such numbers = \(1 * 7 * 1 = 7\)


Answer = \(84 - 7 = 77\)
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 04 May 2014, 19:18
Bunuel wrote:
twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77


Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for XYZ, where Z is an even number.

If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;

If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;

Total: 56+21=77 numbers.

Answer: E.

Hope it's clear.



Hi Bunuel,

If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used).

Here what about 2 as second digit. For eg: 722. In this case second digit can't take 2. How are these numbers (such as 744,766,788,922,944) removed from 2*4*7 = 56 numbers?

If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used).

Here i didn't get the last part "(9 minus two digits we already used)" . Which 2 digits are we talking about? Here also how are we not considering 822,844 etc using the above approach?
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 05 May 2014, 00:05
thoufique wrote:
Bunuel wrote:
twixt wrote:
How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

A. 729
B. 243
C. 108
D. 88
E. 77


Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for XYZ, where Z is an even number.

If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;

If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;

Total: 56+21=77 numbers.

Answer: E.

Hope it's clear.



Hi Bunuel,

Here i didn't get the last part "(9 minus two digits we already used)" . Which 2 digits are we talking about? Here also how are we not considering 822,844 etc using the above approach?


We are told that X, Y and Z must be distinct non zero digits, thus XYZ cannot be 822 or 844.

As for your other question: 9 minus two digits we already used means that for Y we cannot use a digit we used for X and a digit we used for Z, so we can use only 9-2=7 values of it.

Hope it's clear.
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 29 May 2014, 18:21
But how about this. Question says "Distinct non-zero digits"
hence "Non-distinct zero digits" shall not be eliminated. (801,800 are all valid.)???
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 29 May 2014, 20:23
guruseen wrote:
But how about this. Question says "Distinct non-zero digits"
hence "Non-distinct zero digits" shall not be eliminated. (801,800 are all valid.)???


The intent of the question is that the digits are distinct and they are non-zero.
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Re: How many even 3 digit integers greater than 700 with distinct non zero  [#permalink]

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New post 27 May 2015, 11:42
Hi,

How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

Now, we have to find 3 digit[Distinct non-zero] even integers (Last digit 2,4,6 or 8) greater than 700 (i.e., first digit has to be 7,8 or 9).

So,
1.) No. of integers greater than 700 & less than 799, satisfying above mentioned conditions are:
==> 7 _ _
==> 7 _ (2,4,6,8)
==> 1 * 4 * 7 = 28 (4 digits--2,4,6, or 8 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

2.) No. of integers greater than 800 & less than 899, satisfying above mentioned conditions are:
==> 8 _ _
==> 8 _ (2,4,6)
==> 1 * 3 * 7 = 21 (3 digits--2,4 or 6 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

3.) No. of integers greater than 900 & less than 999, satisfying above mentioned conditions are:
==> 9 _ _
==> 9 _ (2,4,6,8)
==> 1 * 4 * 7 = 28 (4 digits--2,4,6, or 8 can be used to fill up the units digit and the tens place can be filled by remaining 7 digits as digits are distinct.).

So, adding them all we get total integers as : 28 + 21 + 21 = 77.
So, (E).

I hope that the explanation is clear.
Re: How many even 3 digit integers greater than 700 with distinct non zero &nbs [#permalink] 27 May 2015, 11:42

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