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# How many odd numbers are there from the integers 700 to 999

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Director
Joined: 15 Aug 2005
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How many odd numbers are there from the integers 700 to 999 [#permalink]

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05 Oct 2005, 02:49
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How many odd numbers are there from the integers 700 to 999 (inclusive)? If every digit of such an odd number must be non-zero and different.

A. 88
B. 89
C. 90
D. 91
E. 92

Any shortcuts for this that im missing???
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Cheers, Rahul.

Senior Manager
Joined: 30 Oct 2004
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05 Oct 2005, 05:34
I got D) 91

This may be a longer method... but I prefer using a table.

I hope I didnt miss any

I'm sure there is an easier way of doing this....
Attachments

odd.gif [ 3.83 KiB | Viewed 1536 times ]

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-Vikram

Director
Joined: 15 Aug 2005
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05 Oct 2005, 05:59
Vikramm, the answer is right. I did the same way! Im hoping theres so genius on this forum who comes up with a shortcut!!
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Cheers, Rahul.

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Joined: 24 Sep 2005
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05 Oct 2005, 07:54
my solution is shorter or not, it's for you to judge , but you may try this :

+let abc( 1<=a,b,c<=9) be the representative of such numbers. abc is odd, thus we have 300 odd numbers from 700-999
+when a=b=c , we have abc= 111c . There're 2 such numbers in all (777,999)
+when a=b<> c , we have 700<=abc=110b+c<=999==> b can only be 7,8,9. For each case, c can be 1,3,5,7,9 since abc is odd. Thus there're 3*5=15 cases of c overall. Except for 777 and 999, we have 13 abc left
+when a<>b=c , we have 700<=abc= 100a+11c<=999====> a can only be 7,8,9.For each case, c can be 1,3,5,7,9. Thus we have 15 cases of c in all. Except for 777 and 999, we have 13 abc left.
+when a=c<>b, we have 700<=abc= 101c +b <=999 ====> c can only be 7,9 . For each case, b can be 1--->9. Thus we have 18 cases of abc. Except for 777 and 999, we have 16 left.
+when b=0, we have 15 a0c.

Overall, the number of abc is (150-2-13-13-16-15)= 91.
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Joined: 28 Dec 2004
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05 Oct 2005, 08:01
look lets say our odd number is XYZ...

for X we have 3 choices...7, 8, 9

for Y we have (remember different numbers, so we cant use 7,8,9) 6 numbers..

for Z we have to have an odd digit, so 5 of them

so 3*6*5 +1 (since its says inclusive)=91.....

Last edited by FN on 05 Oct 2005, 08:26, edited 1 time in total.
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Joined: 24 Sep 2005
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05 Oct 2005, 08:07
wow, i'm impressed ....thanks for your great solution!!! ...ac, why didn't i think of combination
Manager
Joined: 28 Jun 2005
Posts: 65
Location: New York, NY

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05 Oct 2005, 08:18
fresinha12 wrote:
look lets say our odd number is XYZ...

for X we have 3 choices...7, 8, 9

for Y we have (remember different numbers, so we can use 7,8,9) 6 numbers..

for Z we have to have an odd digit, so 5 of them

so 3*6*5 +1 (since its says inclusive)=91.....

thanks for this explanation! i totally didn't consider using a combination. i'm having a little trouble understanding how you got 6 numbers for Y if you can use 7,8, or 9. do you mean cannot? if you can use 7,8, and 9 isn't the number of choices 9? (1,2,3,4,5,6,7,8,9) am i missing something?
Senior Manager
Joined: 30 Oct 2004
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05 Oct 2005, 08:20
fresinha12 wrote:
look lets say our odd number is XYZ...

for X we have 3 choices...7, 8, 9

for Y we have (remember different numbers, so we can use 7,8,9) 6 numbers..

for Z we have to have an odd digit, so 5 of them

so 3*6*5 +1 (since its says inclusive)=91.....

I knew there was an easy way!!! Thanks fresinha12!! Clear and concise
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-Vikram

Manager
Joined: 04 Oct 2005
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05 Oct 2005, 08:23
MsStephanie wrote:
fresinha12 wrote:
look lets say our odd number is XYZ...

for X we have 3 choices...7, 8, 9

for Y we have (remember different numbers, so we can use 7,8,9) 6 numbers..

for Z we have to have an odd digit, so 5 of them

so 3*6*5 +1 (since its says inclusive)=91.....

thanks for this explanation! i totally didn't consider using a combination. i'm having a little trouble understanding how you got 6 numbers for Y if you can use 7,8, or 9. do you mean cannot? if you can use 7,8, and 9 isn't the number of choices 9? (1,2,3,4,5,6,7,8,9) am i missing something?

You cannot use 7,8 and 9 for the X digit... since all digits must be distinct..
so 7yz up to 999...thus, you can only use numbers below 7, otherwise the three digits would not be distinct. 76Z would be fine
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05 Oct 2005, 08:28
just edited...I meant Cannot use 7,8,9....so (9-3)=6 numbers that are available for Y
Manager
Joined: 19 Sep 2005
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05 Oct 2005, 10:07
I'm doing something wrong cause I get an answer htat isn't there- 102
Take the possibilities for the 700s:
-For 70 there are 4 possibilites for the one's place (1,3,5,9)
-For the 71 there are 3 (3,5,9)
-for 72 there are 4 again (1,3,5,) and so on
-when you get to the 77 option- don't consider any of these possible

For the 800s:
80: 5 possibilities (1,3,5,7,9)
81: 4 possibilities (3,5,7,9)
and so on- exclude all the 88 options

for the 900's
same exact number as for the 700s

So for different possibilities considering the different number options in the 10's place:
0: 13 total
1: 10 total
2: 13 total
3: 10 total
4: 13 total
5: 10 total
6: 13 total
7: 7 total
8: 8 total
9: 7 total

That adds up to 102 right? I've got to be missing something silly but can't see it.
Manager
Joined: 19 Sep 2005
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05 Oct 2005, 10:15
ahhh- the importance of reading the question carefully:
no integer can be zero!
thanks
Senior Manager
Joined: 26 Jul 2005
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05 Oct 2005, 11:30
fresinha12 - nice answer and explanation!
Intern
Joined: 04 Oct 2005
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05 Oct 2005, 15:04
Why do you have to add the one since it says inclusive? It seems to me the 3*6*5=90 would be sufficient.
Senior Manager
Joined: 26 Jul 2005
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05 Oct 2005, 17:54
I think that's what inclusive means (including itself) so you add a 1.
Director
Joined: 15 Aug 2005
Posts: 796
Location: Singapore

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05 Oct 2005, 18:01
fresinha12 wrote:
look lets say our odd number is XYZ...

for X we have 3 choices...7, 8, 9

for Y we have (remember different numbers, so we cant use 7,8,9) 6 numbers..

for Z we have to have an odd digit, so 5 of them

so 3*6*5 +1 (since its says inclusive)=91.....

I knew I was not expecting too much in hoping for a genuis to give me a short cut!! Great expln! thanks man!!
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Cheers, Rahul.

Director
Joined: 23 Jun 2005
Posts: 841
GMAT 1: 740 Q48 V42

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05 Oct 2005, 18:24
That is a really nice explanation, Thanks fresinha!
VP
Joined: 22 Aug 2005
Posts: 1112
Location: CA

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05 Oct 2005, 18:53
D. 91. Here is the way I solved:
Attachments

comb.doc [26 KiB]

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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05 Oct 2005, 21:39
duttsit wrote:
D. 91. Here is the way I solved:

Nice one. I just changed something in your document. I think it's a typing mistake.
Attachments

comb_289.doc [26.5 KiB]

Manager
Joined: 03 Aug 2005
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06 Oct 2005, 08:49
I do not think fresinha12's reasoning is correct.

First I do not understand why to add one because the stem says that 700 and 699 are included since neither of those numbers are odd with no numbers repeated.

Second, we cannot exclude 7,8 and 9. for instance, 789 is a valid number.

Third: we do not have five choices for the last number, if the first number is 7, we cannot choose 7 for the last one.

If we take numbers from 700 to 799 and use this reasoning:

1*9*5 +1 =46 what is wrong.

In summary, I think it is a mere coincidence that we get the right result from that reasoning.
06 Oct 2005, 08:49

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