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M25#20

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M25#20 [#permalink] New post 14 Mar 2009, 09:06
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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

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

Hope that helps!!
[Reveal] Spoiler: OA
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M25 #20 [#permalink] New post 17 Mar 2009, 10:01
Hello,

I know this answer has been recently discussed but that explanation confuses me even more.
____________________________________________________________________________________
How many odd three-digit integers greater than 800 are there such that all their digits are different?

72= Answer

I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you. :thanks
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Re: M25 #20 [#permalink] New post 18 Mar 2009, 21:00
dczuchta wrote:
Hello,

I know this answer has been recently discussed but that explanation confuses me even more.
____________________________________________________________________________________
How many odd three-digit integers greater than 800 are there such that all their digits are different?

72= Answer

I understand the explanation given, the answer divided into two parts. What I don't understand is how you come up with 72 when you don't break the answer in two. The way I see it: There are 2 options for the first spot (8 and 9), 4 options for the last spot (1,3,5,7, without the 9 because of option in first spot), and 8 options for middle spot (because you start with ten options and minus 2 for the numbers taken in the other two spots)= 64. Why is it 2*9*4 instead? Why are there 9 options for the middle spot? Can someone please explain that? Thank you. :thanks



First place = 8 or 9 = 2
Second place = 0 - 7 and either 8 or 9 = 8
Third place = 1, 3, 5, 7 and 9 but there could be odd integer in first or second place. so;

1. if 8 in first place and 4 even integers in second place = 1x4x5 = 20
2. if 8 in first place and one of 5 odds in second place and one of remaining 4 odds in third place = 1x5x4 = 20
3. if 9 in first place and one of 5 evens in second place and one of 4 odds in third place = 1x5x4 = 20
4. if 9 in first place and one of 4 odds in second place and one of remaining 3 odds in third place = 1x4x3 = 12

total = 72.
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Re: M25#20 [#permalink] New post 23 May 2009, 17:34
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I used the following method during my test:

If 8 is first digit,
there are 5 ways to fill third digit (1,3,5,7 or 9) and 8 ways to fill 2nd digit: 5x8 = 40

If 9 is first digit,
there are 4 ways to fill 3rd digit (1,3,5 or 7) and 8 ways to fill 2nd digit: 4x8 = 32

40+32 = 72
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Re: M25#20 [#permalink] New post 28 Jun 2009, 13:19
Hello,
Can you please help me understand as how there are 8 digits in the middle -
If first can be 8
then middle is 0,1,2,3,4,5,6,7,9 - which totals to 9. Can you please help
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Re: M25#20 [#permalink] New post 29 Jun 2010, 06:27
If first is 8 and last can be either of 3,5,7,9 (odd numbers only)
so in middle it will have only 8 numbers left to be filled out of 0-9

From your choices above (then middle is 0,1,2,3,4,5,6,7,9 - which totals to 9), you have taken all the 4 odd numbers (3,5,7,9) at same time. Which cannot be possible.
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Re: M25#20 [#permalink] New post 29 Jun 2010, 08:02
Constraints :-
1. Three digit number
2. Starting with 8 or 9
3. No repetition
4. Odd

When 8 is first digit:-
1. Second number even (0,2,4,6), third number (1,3,5,7,9)- total cases (4*5 =20)
2. Second number odd (1,3,5,7,9), third number (odd numbers -1=4) - total cases (5*4=20)

When 9 is first digit:-
3. Second number even (0,2,4,6,8), third number odd (1,3,5,7)- total cases (5*4 = 20)
4. Second number is odd (1,3,5,7), third number odd (odd numbers -1 =3) - total cases (4*3 = 12)

Total number of possible arrangements = 20+20+20+12 = 72

This is lengthy procedure and is risky because you may miss one case & answer will be completely different ,
IMO direct application of P&C theory will be better approach under time constraints .
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Re: M25#20 [#permalink] New post 29 Jun 2010, 08:55
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8 _ _
You need to pick 2 numbers from set { 0,1,2,3,4,5,6,7,9}. Hence 9C2 = 36. But u can order the 2 numbers around, so 36*2 = 72

9 _ _
You need to pick 2 numbers from set { 0,1,2,3,4,5,6,7,8}. Hence 9C2 = 36. But u can order the 2 numbers around, so 36*2 = 72

But you only need odd numbers so (72 + 72)/2 = 72.
Answer is C.
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Re: M25#20 [#permalink] New post 30 Jun 2010, 06:20
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I hope my approach is correct anyway;
Let N = total no of digits with different individual digits
N= 2x9x8 = 144
no of even = no of odds = 72. (D)
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Re: M25#20 [#permalink] New post 03 Jul 2010, 15:52
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I just used slot method: 2 choices for the hundreds digit (8 or 9), 9 choices for the tens digit (0,1,2,3,4,5,6,7, 8 or 9), and 4 choices for the tens digit (1 or 9, 3 or 9, 5 or 9 , 7 or 9) = 2 x 9 x 4 = 72.
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Re: M25#20 [#permalink] New post 02 Jul 2011, 06:40
Between 800 and 899 -- (1)*(8)*(5) = 40 (units place -- 1,3,5,7,9)
Between 900 and 999 -- (1)*(8)*(4) = 32 (units place -- 1,3,5,7)

72
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Re: M25#20 [#permalink] New post 03 Jul 2011, 05:11
first digit can be selected from (8,9) in 2 ways
second distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 9 ways since one digit is already selected
Third distinct digit can be selected from (0,1,2,3,4,5,6,7,8,9) in 8 ways since two digits are already selected .
So total number of ways digits can be slected is 2*8*9
But we need to consider odd number of digits so (2*8*9)/2=72
SO...ANSWER is C
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Re: M25#20 [#permalink] New post 26 Mar 2012, 07:20
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Expert's post
k8llykh 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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

Hope that helps!!


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.

Total for this range: 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.

Total for this range: 1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.
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Re: M25#20 [#permalink] New post 05 Jul 2012, 04:21
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Slot method.

If hundreds digit is 8, there are 5 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*5=40
If hundreds digit is 9, there are 4 options to make last digit (ones digit) odd. That leaves 8 options for tens digits. Total is 1*8*4=32
Total= 72

Answer is C.

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Re: M25#20 [#permalink] New post 05 Jul 2012, 09:52
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I just culculated (2*9*8)/2 = 72

(2*9*8) to get all nummers with all their digits different from 800 to 999
and than divided it by 2 to get only the odd numbers
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Re: M25#20 [#permalink] New post 08 Jul 2012, 08:58
k8llykh 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

[Reveal] Spoiler: OA
C

Source: GMAT Club Tests - hardest GMAT questions

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

Hope that helps!!


good problem...every time i get such problems as these, i try to solve in conservative(old manual) method.....i should try to find the right method which is also the faster method....ANSWER is 72 and no doubt abt it.
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Re: M25#20 [#permalink] New post 23 Jul 2012, 11:05
Bunuel wrote:
k8llykh 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

Hi everyone, I spent 3 minutes writing out all the possibilities and then getting to the right answer, 72. I then read the suggested solution which uses the combination theory, which is great and much quicker.

For anyone wondering how it works, here's a more detailed explanation of the "explain answer" part:

For digits beginning with 8 _ _ , you get 9*5 possilibities. Since you can't repeat 8, your second digit must be 0,1,2,3,4,5,6,7 or 9. Similarly, for the 3rd digit, you can only choose from 1,3,5,7 and 9. Therefore, the possibilities for 3-digit integers beginning with 8 _ _ are 9*5. However, you need to deduct repeated digits in the second and third digit places, i.e. 11,33,55,77, and 99. Therefore, it is (9*5)-5= 45-5= 40

Similarly, with the same logic for digits beginning with 9 _ _ , you get 9*4 possibilities, i.e. 0,1,2,3,4,5,6,7,8 for the 2nd digit and 1,3,5,7 for the 3rd digit. You can't use 9 in either the 2nd or the 3rd digit. Finally, take out the repeats, which are 11,33,55,77. 99 has already been deducted from your earlier calculations. Therefore for digits beginning with 9 _ _, you have (9*4) - 4 = 36 - 4 = 32 possibilites.

Add the 2 together, and you get 40+32=72. The answer!

Hope that helps!!


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.

Total for this range: 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.

Total for this range: 1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.


Hi Geek,

I wonder how u r solving all questions with proper explanations... HAts off for tat...

How u r solving by these strategy?
Does any math strategy is there for this types of problem and could you pls help me to read those things....

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.

Total for this range: 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.

Total for this range: 1*4*8 = 32.


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Re: M25#20 [#permalink] New post 31 Aug 2012, 19:49
I am kicking myself for getting this problem wrong on account of a really bad silly mistake. I missed reading the "odd" in the question stem and ended up with getting an answer not in the given choices.

The good thing is that I knew how to approach it using the combinatrics method but on account of poor reading of the question stem I ended up choosing a wrong answer.

Bunuel's way of selecting the 3rd digit after selecting the first (8/9) was particularly helpful and quick.

Missed this question only because of not reading with full concentration. Real bad mistake. :(
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Re: M25#20 [#permalink] New post 05 Jul 2013, 23:30
Hi everyone, most fastest way to solve this question is

8_ _, different odd no. will be 8 _ _.
first digit is 8, for odd number third digit will be (1,3,5,7,9) i.e. 5 possibility
2nd digit we can fill from (0-9), since we are looking for different digit so choice for 2nd digit will 10(0-9) - 1st digit-2nd digit with 8 possibility
8 _(8) _(5) = 8*5 = 40


similarly for 9 _ _

for 3rd digit choices are {1,3,5,7}
for 2nd digit choices are 10 - 1st digit-3rd digit = 8

so 8*4=32

so total = 72 different odd digits

Another simple wat to solve is
b/w 800-900 we have 50 odd no.
now we will substract the odd no. like {11,33,55,77,99}= 5
and 881,83,85,87,89 =5
i.e. 50-5-5 =40

b/w 900-999 we have 50 odd no
substract {11,33,55,77,99}=5
and 991,93,95,97,99 = 5
and like 919,929 .....989=8
50-5-5-8 = 32
so different odd number will be 40+32 = 72
Re: M25#20   [#permalink] 05 Jul 2013, 23:30
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