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m10 Q13

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m10 Q13 [#permalink] New post 20 Dec 2008, 15:42
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How many odd three-digit integers greater than 800 are there such that all their digits are different?

(A) 40
(B) 56
(C) 72
(D) 81
(E) 104

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

If the number begins with 8, there are 5*8 = 40 possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit).

If the number begins with 9, there are 4*8 = 32 possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit).

In all, there are 40 + 32 = 72 numbers that satisfy the constraints.
The correct answer is C.

Now, when I calculate all the possible choices I come up with 81, meaning answer choice D.

Here is my reasoning:

All odd numbers in 800s
- You have 9 choices for the second digit (0,1,2,3,4,5,6,7,9)
- You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices
= (5 x 4) + (4 x 5) = 40. (5 is for all the even and 4 is for all the odd)

All odd numbers in 900s
- You have 9 choices for the second digit (0,1,2,3,4,5,6,7,8)
- You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices
= (5 x 5) + (4 x 4) = 41. (5 is for all the even and 4 is for all the odd)

Total: 40 + 41 = 81


Where I am going wrong??

Thanks.
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Re: m10 Q13 [#permalink] New post 20 Dec 2008, 18:09
smarinov wrote:
Hi guys,

for some reason my calculations don't match the explanations for the correct answer.

Here is the question:

How many odd three-digit integers greater than 800 are there such that all their digits are different?

(C) 2008 GMAT Club - m10#13

* 40
* 56
* 72
* 81
* 104

If the number begins with 8, there are 5*8 = 40 possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit).

If the number begins with 9, there are 4*8 = 32 possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit).

In all, there are 40 + 32 = 72 numbers that satisfy the constraints.
The correct answer is C.

Now, when I calculate all the possible choices I come up with 81, meaning answer choice D.

Here is my reasoning:

All odd numbers in 800s
- You have 9 choices for the second digit (0,1,2,3,4,5,6,7,9)
- You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices
= (5 x 4) + (4 x 5) = 40. (5 is for all the even and 4 is for all the odd)

All odd numbers in 900s
- You have 9 choices for the second digit (0,1,2,3,4,5,6,7,8)
- You have 5 choices for the third digit (1,3,5,7,9) unless the second digit matches the third digit in that case you got 4 choices
= (5 x 5) + (4 x 4) = 41. (5 is for all the even and 4 is for all the odd)

Total: 40 + 41 = 81


Where I am going wrong??

Thanks.



You shouldn't count 9 in the units digit, it's already in hundreds
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Re: m10 Q13 [#permalink] New post 20 Dec 2008, 19:51
ah! there you go, thanks Dauren!
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Re: m10 Q13 [#permalink] New post 20 Jan 2009, 06:28
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smarinov wrote:
How many odd three-digit integers greater than 800 are there such that all their digits are different?

(C) 2008 GMAT Club - m10#13

* 40
* 56
* 72
* 81
* 104




C. Here is how I solved it, but other methods would be welcome.

For numbers between 800-899:
Hundreds: one (i.e. 8)
Tens: nine (i.e. 0-9, but not 8)
Units: five (i.e. 1,3,5,7,9)
1x9x5=45 numbers; there are five numbers we need to exclude from this list: 811, 833, 855, 877, 899.
So, between 800 and 899, there are 40 numbers that satisfy the constraints.

For numbers between 900-999:
Hundreds: one (i.e. 9)
Tens: nine (i.e. 0-8)
Units: four (i.e. 1,3,5,7)
1x9x4=36 numbers; there are four numbers we need to exclude from this list: 911, 933, 955, 977.
So, between 900 and 999, there are 32 numbers that satisfy the constraints.

Overall we have 72 numbers.
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Re: m10 Q13 [#permalink] New post 23 Feb 2010, 08:17
How much time did you guys need to solve this problem ?
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Re: m10 Q13 [#permalink] New post 23 Feb 2010, 10:31
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Instead of getting confused with the odd number stuff just calculate the total number of three-digit integers greater than 800 such that all their digits are different.

Here is how we do it.
_ _ _

the first place (hundreds place) can be filled by 8 or 9 ie 2 ways
the second place can be filled in 9 ways (since one integer is already take we only have to choose from 9 instead of 10)
the third place can be filled in 8 ways (since two integers are already take we have to choose from 8 digits)

hence total = 2 * 9 * 8 = 144.

Now there are equal number of odd and even numbers, so just 144/2 = 72 is the total number of odd numbers required.

This took me less than 30 seconds.
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Re: m10 Q13 [#permalink] New post 23 Feb 2010, 11:47
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Not a big fan of permutations! Anyway, answer is C

Multiply and add up your choices for each spot just like with a combination lock.

Hundreds Spot: 2 choices (8 & 9)
Tens Spot: 9 choices (0,1,2,3,4,5,6,7 & [8 or 9])
Units Spot: 4 choices (1,3,5,7,9)... yes. there are 5 odd choices but you'd subtract one so the hundreds spot doesn't double up

So 2 * 9 * 4 = 72
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Re: m10 Q13 [#permalink] New post 21 Sep 2010, 22:23
gaddam506 wrote:
Instead of getting confused with the odd number stuff just calculate the total number of three-digit integers greater than 800 such that all their digits are different.

Here is how we do it.
_ _ _

the first place (hundreds place) can be filled by 8 or 9 ie 2 ways
the second place can be filled in 9 ways (since one integer is already take we only have to choose from 9 instead of 10)
the third place can be filled in 8 ways (since two integers are already take we have to choose from 8 digits)

hence total = 2 * 9 * 8 = 144.

Now there are equal number of odd and even numbers, so just 144/2 = 72 is the total number of odd numbers required.

This took me less than 30 seconds.



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Re: m10 Q13 [#permalink] New post 28 Feb 2011, 20:02
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gaddam506 wrote:
Instead of getting confused with the odd number stuff just calculate the total number of three-digit integers greater than 800 such that all their digits are different.

Here is how we do it.
_ _ _

the first place (hundreds place) can be filled by 8 or 9 ie 2 ways
the second place can be filled in 9 ways (since one integer is already take we only have to choose from 9 instead of 10)
the third place can be filled in 8 ways (since two integers are already take we have to choose from 8 digits)

hence total = 2 * 9 * 8 = 144.

Now there are equal number of odd and even numbers, so just 144/2 = 72 is the total number of odd numbers required.

This took me less than 30 seconds.


Great explanation!!! +1
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Re: m10 Q13 [#permalink] New post 01 Mar 2011, 04:50
Case I - 100's digit is 8

801, 803, 805, 807, 809 etc.

So 100's place 1 way, unit's place - 5 ways

10's ways - 8 ways

= 40

Case II - 100's digit is 9

901, 903, 905, 907 etc.

So 100's place 1 way, unit's place - 4 ways

10's place - 8

= 32 ways

So total - 72 ways
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Re: m10 Q13 [#permalink] New post 01 Mar 2011, 06:11
1. Hundreds digit = 8.
Unit digit - 1,3,5,7,9 (5 ways)
Tens digit - 8 ways
Total = 1 * 8 * 5 = 40

2. Hundreds digit = 9.
Unit digit - 1,3,5,7 (4 ways)
Tens digit - 8 ways
Total = 1 * 8 * 4 = 32

Grand total = 40 + 32 = 72
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Re: m10 Q13 [#permalink] New post 05 Mar 2011, 11:16
The answer is C

possible first integers 2 (8 or 9)
possible second integers 9 (one of 0 to 9 is taken as first integer)
possible third integers 8 (two of 0 to 9 is already considered)

possible numbers = 2*8*9 = 144

as there are equal number of even and odd integers answer is 72
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Re: m10 Q13 [#permalink] New post 01 Mar 2012, 11:48
Ans- 72

Hundreds place- 2 options (8 or 9)
Tens place-9 options(0,1,2,3,4,5,6,7, 8 or 9)
Units place should be odd- 4 options (1,3,5,7)

Hence 2*9*4 = 72
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Re: m10 Q13 [#permalink] New post 01 Mar 2012, 15:45
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smarinov wrote:
How many odd three-digit integers greater than 800 are there such that all their digits are different?

(A) 40
(B) 56
(C) 72
(D) 81
(E) 104


This question was also posted in PS forum (how-many-odd-three-digit-integers-greater-than-800-are-there-94655.html) below is my solution from there:

In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9;
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*5*8 = 40.

In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.
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Re: m10 Q13 [#permalink] New post 01 Mar 2012, 17:56
This took me way too long to solve! But I eventually got it. Was difficult for me.
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Re: m10 Q13 [#permalink] New post 05 Mar 2013, 20:09
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For 3-digit integers greater than 800, two series are there 800 and 900...

For number starting with 8...first digit is 8...second digit may have 9 choices (0-9,except 8)
and third digit depends on second digit...
For second digit as even (0,2,4,6)...numbers = 4*5 = 20
for second digit as odd(1,3,5,7,9)...numbers = 5*4 = 20

For odd three-digit integers greater than 800 and less than 900, the numbers satisfies the criteria are = 20+ 20= 40

For number starting with 9...first digit is 9...second digit may have 9 choices (0-8)
and third digit depends on second digit...
For second digit as even (0,2,4,6,8)...numbers = 5*4 = 20 (since there are four odd numbers only, i.e. 1,3,5,7)
for second digit as odd(1,3,5,7)...numbers = 4*3 = 12 (since there are three odd numbers only)

For odd three-digit integers greater than 900 and less than 999, the numbers satisfies the criteria are = 20+ 12= 32


So, the total numbers satisfying the criteria are 40+32 = 72

Answer is C
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Re: m10 Q13 [#permalink] New post 07 Mar 2013, 22:41
gmat1220 wrote:
1. Hundreds digit = 8.
Unit digit - 1,3,5,7,9 (5 ways)
Tens digit - 8 ways
Total = 1 * 8 * 5 = 40

2. Hundreds digit = 9.
Unit digit - 1,3,5,7 (4 ways)
Tens digit - 8 ways
Total = 1 * 8 * 4 = 32

Grand total = 40 + 32 = 72


I did the same way. But it took me more than 3 minutes to come up with the idea. Good question!
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Re: m10 Q13 [#permalink] New post 25 Feb 2014, 05:03
I guess the best way is to count to certain extent and generalize the situation like

801,803,805,807,809
813,815,817,819
821,823,825,827,829
similarly
901,903,905,907 and 913,915,917 generalize and count
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Re: m10 Q13 [#permalink] New post 21 Apr 2014, 08:43
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Start with 8: 5*8=40
Start with 9: 4*8=32
-> total =72
Re: m10 Q13   [#permalink] 21 Apr 2014, 08:43
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