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problem..exam in 10 days:(

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problem..exam in 10 days:( [#permalink]

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New post 02 Aug 2009, 03:01
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...!!!!

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Re: problem..exam in 10 days:( [#permalink]

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New post 02 Aug 2009, 03:44
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.

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Re: problem..exam in 10 days:( [#permalink]

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New post 02 Aug 2009, 03:52
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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 ?
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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.

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Re: problem..exam in 10 days:( [#permalink]

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New post 02 Aug 2009, 04:15
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

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Re: problem..exam in 10 days:( [#permalink]

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New post 02 Aug 2009, 05:38
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, 06:22, edited 1 time in total.

Kudos [?]: 32 [0], given: 9

Senior Manager
Senior Manager
User avatar
Affiliations: ACA, CPA
Joined: 26 Apr 2009
Posts: 433

Kudos [?]: 94 [0], given: 41

Location: Vagabond
Schools: BC
WE 1: Big4, Audit
WE 2: Banking
Re: problem..exam in 10 days:( [#permalink]

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New post 02 Aug 2009, 06:06
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...

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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.

Kudos [?]: 94 [0], given: 41

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