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# problem..exam in 10 days:(

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Manager
Joined: 14 Jun 2009
Posts: 94

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02 Aug 2009, 02:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?

i always have issues with solving questions like these..where i have to find numbers b/w a certain huge range...need pointers for such problems...!!!!
Senior Manager
Affiliations: ACA, CPA
Joined: 26 Apr 2009
Posts: 433
Location: Vagabond
Schools: BC
WE 1: Big4, Audit
WE 2: Banking
Re: problem..exam in 10 days:( [#permalink]

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02 Aug 2009, 02:44
Is the problem complete ? any ref to the place of 5 ie units/tens etc?

arvs212 wrote:
In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?

i always have issues with solving questions like these..where i have to find numbers b/w a certain huge range...need pointers for such problems...!!!!

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Senior Manager
Affiliations: ACA, CPA
Joined: 26 Apr 2009
Posts: 433
Location: Vagabond
Schools: BC
WE 1: Big4, Audit
WE 2: Banking
Re: problem..exam in 10 days:( [#permalink]

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02 Aug 2009, 02:52
1
KUDOS
This is the way i would solve it.

We are essentially looking at 101 to 999.

1st digit - can be (1-9 ; except 5) so --> 8
2nd digit - can be (0-9 ; except 5) so --> 9
3rd digit - has to be odd except 5 (1,3,7,9) so --> 4

So we can have a total of 8x9x4 possible numbers.
Surely took me more than 2 mins to solve it!

Am i right ?
_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Manager
Joined: 14 Apr 2008
Posts: 51
Re: problem..exam in 10 days:( [#permalink]

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02 Aug 2009, 03:15
no of digits b/w 100 n 1000 donot contain 5 = 9*10*1 =90
no of digits b/w 100 n 1000 are odd= 500-50=450
integers are odd and do not contain the digit "5" = 450-90= 360
Manager
Joined: 14 Jun 2009
Posts: 94
Re: problem..exam in 10 days:( [#permalink]

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02 Aug 2009, 04:38
sniper ur right...

way i did it...>>

say the number is ABC since 101-999

C..(has 4 possibilities taking only odds except 5)
B..(0-9, except 5, giving us 9 choices)
A..(1-9, except 5 giving us 8 options)

9*8*4

Last edited by arvs212 on 02 Aug 2009, 05:22, edited 1 time in total.
Senior Manager
Affiliations: ACA, CPA
Joined: 26 Apr 2009
Posts: 433
Location: Vagabond
Schools: BC
WE 1: Big4, Audit
WE 2: Banking
Re: problem..exam in 10 days:( [#permalink]

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02 Aug 2009, 05:06
After you get the numbers... the problem boils down to simple combinatorics

Like - if one has 2 shirts and 2 ties...how many different possible ways can one dress up --- 2x2

I hope you get the logic

arvs212 wrote:
sniper ur right...but can u explain it further as to why u multiplied the three...

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Re: problem..exam in 10 days:(   [#permalink] 02 Aug 2009, 05:06
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# problem..exam in 10 days:(

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