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How many odd three-digit integers greater than 800 are there such that

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

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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

M25-20

Originally posted by vr4indian on 09 Oct 2008, 10:30.
Last edited by Bunuel on 26 Sep 2019, 00:10, edited 2 times in total.
Renamed the topic and edited the question.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 23 May 2010, 01:23
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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. 60
C. 72
D. 81
E. 104

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.

Hope it's clear.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 23 May 2010, 05:00
16
I would post another way with little short method.
First of all let's find out all numbers greater then 800 and having all different digits.
For hundreth place you have 2 choices, for tenth digit you have (10-1)=9 choices as one digit out of 10 have already used in hundreth place, for unit's digit you have (10-2) = 8 choices
Total three digit numbers with all different digits greater then 800 = 2*8*9 = 144
As 144 is an even number, half of these will be odd and half of this will be even.
So answer will be 72
Ans: C (72)

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

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New post 23 May 2010, 03:44
1
Bunuel wrote:
dimitri92 wrote:
Ok so this is a gmatclub test but I don't get the explanation. Can someone help?


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

40
60
72
81
104


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.

Hope it's clear.


wow amazing approach ... now this is crystal clear !!
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 24 May 2010, 17:29
Let's xyz is our integer.

x e {8,9}
y e {0...9}
z e {1,3,5,7,9}

So, we have 1[x]*5[z]*(10-2)[y] = 40 integers that begin with 8 and 1[x]*4[z]*(10-2)[y] = 32 integers that begin with 9.
N = 40 + 32 = 72
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 23 Jun 2010, 19:53
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i'm still not getting the logic..i find myself convincing myself of different things for examplle

for 800s
first digit can be 8 only
2nd digit can be 0,1,2,3,4,5,6,7,9
3rd digit can be 1,3,5,7,9
1x9x5

for 900s
1st 9
2nd 0,1,2,3,4,5,6,7,8
3rd 1,3,5,7
1x9x4

how is the answer not 45 + 36 ?
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 24 Jun 2010, 05:42
1
kobra92 wrote:
i'm still not getting the logic..i find myself convincing myself of different things for examplle

for 800s
first digit can be 8 only
2nd digit can be 0,1,2,3,4,5,6,7,9
3rd digit can be 1,3,5,7,9
1x9x5

for 900s
1st 9
2nd 0,1,2,3,4,5,6,7,8
3rd 1,3,5,7
1x9x4

how is the answer not 45 + 36 ?


This approach is not correct. If you say that (in the range 800-900) for the second digit you have 9 choices, then for the third digit you'll have sometimes 5 choices (in case you choose even for the second) and sometimes 4 choices (in case you choose odd for the second), so you can not write 1*9*5.

Correct way would be to count the choices for the third digit first - 5 choices (1, 3, 5, 7, 9) and only then to count the choices for the second digit - 8 choices (10-first digit-third digit=8) --> 1*5*8=40.

Similarly for 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.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 23 Aug 2010, 13:43
Hey Bunuel, in the solution below, I did not understand the highlighted part.

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

In the range 800 - 900:
1 choice for the first digit: 8;
[highlight]5 choices for the third digit: 1, 3, 5, 7, 9;[/highlight]
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;
[highlight]4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);[/highlight]
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 23 Aug 2010, 13:56
seekmba wrote:
Hey Bunuel, in the solution below, I did not understand the highlighted part.

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

In the range 800 - 900:
1 choice for the first digit: 8;
[highlight]5 choices for the third digit: 1, 3, 5, 7, 9;[/highlight]
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;
[highlight]4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit);[/highlight]
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.

1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.


Odd numbers should end with odd digit.

Hence third digit should be 1, 3, 5, 7 or 9 - 5 choices, but for range 900-999 we can not use 9 as we used it as first digit so only 4 options are available.

Hope it's clear.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 23 Aug 2010, 20:40
Bunuel wrote:
dimitri92 wrote:
Ok so this is a gmatclub test but I don't get the explanation. Can someone help?


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

40
60
72
81
104


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.

Hope it's clear.


Bunuel,
Why did you split up the three digit number into two different subsets and then proceed to solve the problem. Is there some subtle logic behind this.

Please do share your thoughts on this.

This was my approach to solve this problem

x y z is the three digit number. Unit digit z has 4 choices (1, 3, 5 and 7), y has 7 choices (0, 1, 3, 5, 2, 4, 6 leaving out 7 with the assumption that it is the number chosen for the unit digit.) and x has 2 choices (8 and 9).

Of course multiplying these choices does not lead to any of the answer choice.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 24 Aug 2010, 06:07
ezhilkumarank wrote:
Bunuel,
Why did you split up the three digit number into two different subsets and then proceed to solve the problem. Is there some subtle logic behind this.

Please do share your thoughts on this.

This was my approach to solve this problem

x y z is the three digit number. Unit digit z has 4 choices (1, 3, 5 and 7), y has 7 choices (0, 1, 3, 5, 2, 4, 6 leaving out 7 with the assumption that it is the number chosen for the unit digit.) and x has 2 choices (8 and 9).

Of course multiplying these choices does not lead to any of the answer choice.


There are 5 choices for units digit when hundreds digit is 8 and 4 when hundreds digit is 9. So you should count numbers in these 2 ranges separately.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 30 Aug 2010, 04:49
Hey Bunuel, in the solution below, I did not understand

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

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. --> this should be 9 digits right why havent u considered 0

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. -->-> this should be 9 digits right why havent u considered 0


1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 30 Aug 2010, 05:51
Divyababu wrote:
Hey Bunuel, in the solution below, I did not understand

How do you get only 5 choices in the range 800 - 900 and 4 choices in the range 900 - 999? Please explain.

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. --> this should be 9 digits right why havent u considered 0

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. -->-> this should be 9 digits right why havent u considered 0


1*4*8 = 32.

Total: 40+32 = 72.

Answer: C.

Hope it's clear.


Odd numbers should end with odd digit.

Hence third digit should be 1, 3, 5, 7 or 9 - 5 choices, but for range 900-999 we can not use 9 as we used it as first digit so only 4 options are available.

Hope it's clear.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 27 Nov 2011, 21:36
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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 ?
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 28 Nov 2011, 04:25
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2
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.
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 09 Dec 2011, 11:14
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 ?
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 10 Dec 2011, 23:18
devinawilliam83 wrote:
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 such that  [#permalink]

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New post 05 Apr 2012, 11:49
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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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post 05 Apr 2012, 12:17
nikhil007 wrote:
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?


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.

Hope it's clear.

In case of any question please post it here: how-many-odd-three-digit-integers-greater-than-800-are-there-94655.html
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Re: How many odd three-digit integers greater than 800 are there such that  [#permalink]

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New post Updated on: 01 Aug 2012, 07:39
Here is my solution, using reasoning based on symmetry:

The total number of three digit numbers, greater than 800 with all three digits distinct, is 2 * 9 * 8 = 144.
We have two choices for the first digit (8 or 9), 9 choices for the second digit (it must be different from the first digit) and 8 choices for the third digit, as it should be different from the two previous digits).

Now, here is where, I think, symmetry can help:

Among those starting with 8, there are less even numbers, as the last digit cannot be 8 (so, only 4 choices), while odd choices for the last digit are 5.
For the numbers starting with 9, the situation is reversed, as there are only 4 choices for the third digit for odd numbers and 5 choices for the even numbers.
If we put all the numbers together, at the end, we have a balanced outcome, there must be the same number of each type.
It means that among the above 144 numbers, there are as many even as odd numbers, so 72 of each type.

Therefore, Answer C.

Did you meet questions where some type of reasoning based on symmetry can be used?
Please, POST!
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Originally posted by EvaJager on 01 Aug 2012, 04:07.
Last edited by EvaJager on 01 Aug 2012, 07:39, edited 1 time in total.
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Re: How many odd three-digit integers greater than 800 are there such that   [#permalink] 01 Aug 2012, 04:07

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