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

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.
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Re: How many odd three-digit integers greater than 800 are there [#permalink]

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09 Oct 2008, 09:54

IanStewart wrote:

vr4indian wrote:

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

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.

Ian why are we computing them separately ? can we not do them together ?

Thanks
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Re: How many odd three-digit integers greater than 800 are there [#permalink]

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09 Oct 2008, 10:53

vr4indian wrote:

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

40 56 72 81 104

Assume: X = {x | 800 < x < 1,000 and all x's digits are different}

Set of number = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Possible digits for x = 1, 3, 5, 7, and 9

We have to separate into two cases.

Case 1: 800 < x < 899 ---------- because the first digit is 8 or an even number, there are 5 odd number available for the last digit. Possible numbers = 1 x 8 x 5 = 40 numbers

Case 2: 900 < x < 1,000 ---------- because the first digit is 9 or an odd number, there are only 4 odd numbers available for the last digit. Possible numbers = 1 x 8 x 4 = 32 numbers

Re: How many odd three-digit integers greater than 800 are there [#permalink]

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23 Jan 2010, 00:19

IanStewart wrote:

vr4indian wrote:

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

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

Re: How many odd three-digit integers greater than 800 are there [#permalink]

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27 Nov 2011, 20:36

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Can we do it like this ? _ _ _ 2 Possibilities for 100th digit 8 and 9, 9 Possibilities for 10th Digit, 8 Possibilities for Unit digit. that gives us 2 * 9 * 8 = 144, Since we need only odd numbers 144/2 = 72. And thats the right answer. Can anyone tel me if i am missing something or this method is alright ?

Can we do it like this ? _ _ _ 2 Possibilities for 100th digit 8 and 9, 9 Possibilities for 10th Digit, 8 Possibilities for Unit digit. that gives us 2 * 9 * 8 = 144, Since we need only odd numbers 144/2 = 72. And thats the right answer. Can anyone tel me if i am missing something or this method is alright ?

Yes, you are missing something and there is a reason why Ian did what he did. Take numbers in the range 500 - 600. How many numbers are there such that all 3 digits are different? 1 (hundred's digit) * 9 (ten's digit) * 8 (unit's digit) = 72 How many of them are odd? 5 is odd so number of possibilities for unit's digit is 4. [highlight]Number of odd numbers = 1 * 8 * 4 = 32[/highlight] Number of even numbers = 72 - 32 = 40

Now take numbers in the range 800 - 900. Total numbers where all digits are different = 72 (as before) [highlight]Number of odd numbers = 1 * 8 * 5 = 40[/highlight] (now there are 5 possibilities for the unit's digit) Number of even numbers = 72 - 40 = 32

So the number of even and odd numbers are not half of the total numbers in each range. You see why above. The number of odd numbers depends on the range (whether the hundred's digit is odd or even).

The reason you still got your answer is that the range is 800 - 1000 which has the 800 - 900 and the 900 - 1000 range in it. It averaged out to be half because in the first range 40 numbers are odd and in the second range 32 are odd so they averaged out to 36 i.e. half of 72.

If instead the question says how many 3 digit numbers above 700 are odd with all digits different, the answer will be 104, not 108. Your method will not work in this case. So be careful.
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Re: How many odd three-digit integers greater than 800 are there [#permalink]

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28 Nov 2011, 04:04

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Thanks Karishma, Actually i missed out the "ODD Numbers" part in the question earlier and my answer was 144, but none of the answers were matching. Then I saw 72 which is equal to 144/2 so i thought its okay to do this way. Now I understand it better

Re: How many odd three-digit integers greater than 800 are there [#permalink]

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28 Nov 2011, 07:36

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separate odd and even integers (0,1,2,4,6,8) and (1,3,5,7,9) if the 100's place is 8 then the number of ways 1*4(all even except 8)*5(all odd) + 1*5(all odd)*4(four odds left to be chosen)=20+20 =40 if the 100's place is 9 1*5(all even)*4(all odd except 9) + 1*4(all odd)*3(three odd left)=20+12=32

Re: How many odd three-digit integers greater than 800 are there [#permalink]

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09 Dec 2011, 10:14

VeritasPrepKarishma wrote:

arjunbt wrote:

Can we do it like this ? _ _ _ 2 Possibilities for 100th digit 8 and 9, 9 Possibilities for 10th Digit, 8 Possibilities for Unit digit. that gives us 2 * 9 * 8 = 144, Since we need only odd numbers 144/2 = 72. And thats the right answer. Can anyone tel me if i am missing something or this method is alright ?

Yes, you are missing something and there is a reason why Ian did what he did. Take numbers in the range 500 - 600. How many numbers are there such that all 3 digits are different? 1 (hundred's digit) * 9 (ten's digit) * 8 (unit's digit) = 72 How many of them are odd? 5 is odd so number of possibilities for unit's digit is 4. [highlight]Number of odd numbers = 1 * 8 * 4 = 32[/highlight] Number of even numbers = 72 - 32 = 40

Now take numbers in the range 800 - 900. Total numbers where all digits are different = 72 (as before) [highlight]Number of odd numbers = 1 * 8 * 5 = 40[/highlight] (now there are 5 possibilities for the unit's digit) Number of even numbers = 72 - 40 = 32

So the number of even and odd numbers are not half of the total numbers in each range. You see why above. The number of odd numbers depends on the range (whether the hundred's digit is odd or even).

The reason you still got your answer is that the range is 800 - 1000 which has the 800 - 900 and the 900 - 1000 range in it. It averaged out to be half because in the first range 40 numbers are odd and in the second range 32 are odd so they averaged out to 36 i.e. half of 72.

If instead the question says how many 3 digit numbers above 700 are odd with all digits different, the answer will be 104, not 108. Your method will not work in this case. So be careful.

I am definitely taking more than the prescribed 2 minutes to solve these questions. Am i missing something ?

Re: How many odd three-digit integers greater than 800 are there [#permalink]

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09 Dec 2011, 10:33

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WE have following slots: ABC(for 800) and XYZ (for 900) A is reserved for 8 and C has to be odd. X is reserved for 9 and Z has to be odd, But Z cannot be 9. Therefore, For A and X we have 1 choice. For C we have 5 choices. For Z we have 4 Choices. For B and Y we have 8 Choices. Therfore we have: (1C1*8C1*5C1)+(1C1*8C1*4C1) 40+32=72

I am definitely taking more than the prescribed 2 minutes to solve these questions. Am i missing something ?

There is nothing called the 'prescribed two minutes'. For the lower level question, you will take less than two mins and for the higher level ones, you will take more than that. Two minutes is just the average. Also, once you work on quite a few such questions, you will be able to do them in 2 minutes or less.
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Re: How many odd three-digit integers greater than 800 are there [#permalink]

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12 Dec 2011, 22:41

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OA is C 72 800 to 900: choice for the unit digit place must be odd(1,3,5,7,9) is 5 choices for the tens digit (exclude 8 and odd digit used at unit place) 8 choices

5*8 = 40 choices

From 900 to 1000: choice for the unit digit place must be odd(1,3,5,7,) is 4 choices for the tens digit (exclude 9 and odd digit used at unit place)8

Re: How many odd three-digit integers greater than 800 are there [#permalink]

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05 Apr 2012, 10:49

IanStewart wrote:

vr4indian wrote:

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

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.

Understood the concept here, but i just cannot understand one thing, regarding tens digit why do we have 8 choices? as far my understanding goes for 800 its 8(hundred digit) 0,2,4,6( 4 digits available for tens) as we cannot have any odd digit here else it will coincide with unit digit. and 1,3,5,7,9(five available for unit digit) can someone make me understand here?
_________________

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

40 56 72 81 104

From 800 to 900: 1 choice for the first digit (must be 8) 5 choices for the third digit (must be odd) 8 choices remaining for the second digit (must be different from the others)

1*5*8 = 40 possibilities.

From 900 to 1000: 1 choice for the first digit (must be 9) 4 choices for the third digit (must be odd and must not be 9) 8 choices remaining for the second digit (must be different from the others)

1*4*8 = 32 possibilities.

Thus 40+32 = 72 possibilities in total.

Understood the concept here, but i just cannot understand one thing, regarding tens digit why do we have 8 choices? as far my understanding goes for 800 its 8(hundred digit) 0,2,4,6( 4 digits available for tens) as we cannot have any odd digit here else it will coincide with unit digit. and 1,3,5,7,9(five available for unit digit) can someone make me understand here?

We can have odd digit for tens digit, though it must be different from that we used for units digit.

Next, since all 3 digit must be distinct then if you use one for hundreds and one for units then there are 8 choices left for tens digit.

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