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# m03 #12

Author Message
Senior Manager
Joined: 18 Feb 2008
Posts: 497
Location: Kolkata
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Kudos [?]: 117 [0], given: 66

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16 Apr 2009, 03:18
2
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In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?

(A) 180
(B) 196
(C) 286
(D) 288
(E) 324

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Can somebody proceed as below and get the ans.?
100 to 1000 we have 901 integers out of which 451 are odd.From there deduct 500-599 series i.e 100 more nos and get to 351...after that i got stuck:(
Intern
Joined: 14 Oct 2010
Posts: 41
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Kudos [?]: 19 [17] , given: 11

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08 Nov 2010, 07:05
17
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D.

The easiest way to calculate this equation is by thinking how many integers are viable in hundreds, tens and units places, which results 8 * 9 * 4 = 288.

I hope this helps.
Director
Joined: 21 Dec 2009
Posts: 585
Concentration: Entrepreneurship, Finance
Followers: 18

Kudos [?]: 704 [5] , given: 20

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11 Nov 2011, 13:08
5
KUDOS
Range of Nos: 100 - 1000, but the largest odd digit is 999 (3 digit figure)
We need a 3-digit HTU that satisfies the condition therein.

H: cannot be 0, 5: 8 digits
T: can be any number except 5: 9 digits
U: can only take 1, 3, 7, and 9: 4 digits

Total of odd numbers not containing digit "5" = 8 x 9 x 4 = 288
_________________

KUDOS me if you feel my contribution has helped you.

Manager
Joined: 24 Aug 2010
Posts: 189
Location: Finland
Schools: Admitted: IESE($$),HEC, RSM,Esade WE 1: 3.5 years international Followers: 6 Kudos [?]: 94 [4] , given: 18 Re: m03 #12 [#permalink] ### Show Tags 09 Nov 2010, 04:58 4 This post received KUDOS Yah I understand how you got it and I still feel the Slot Method, at least for me, is the most appropriate technique for such type of questions. I think my answer is correct if repetition of digits is not allowed. In this problem, however, the repetition of digits is allowed. Units place in 4 ways. Hundreds place in 8 ways (0 and 5 are out). Tens place in 9 ways (5 is out) Answer: 4*8*9 = 288 Posted from my mobile device Manager Joined: 08 Feb 2009 Posts: 145 Schools: Anderson Followers: 3 Kudos [?]: 53 [3] , given: 3 Re: m03 #12 [#permalink] ### Show Tags 16 Apr 2009, 09:35 3 This post received KUDOS There is 450 Odd integers in total between 100 to 1000, inclusive. Among all the three digit ODD numbers, how many numbers have 5 in them: UNITS place = 10x9 = 90 (105,115,125.....195......995) TENS place = 4x9 = 36 (excluding the numbers counted above - 151,153,157,159......551,553,557,559.....959) HUNDREDS place = 50-10-4=36 (From 500 to 599, there are 50 ODD numbers. Out of which, 10 are counted in UNITS place(505,515...595), 4 are counted in TENS place (551,553,557,559)) 450-90-36-36 = 288. Good question overall. Senior Manager Joined: 19 Oct 2010 Posts: 262 Location: India GMAT 1: 560 Q36 V31 GPA: 3 Followers: 7 Kudos [?]: 80 [1] , given: 27 Re: m03 #12 [#permalink] ### Show Tags 09 Nov 2010, 03:57 1 This post received KUDOS This is how I worked it out. It's probably a bit long-winded, but it didn't take much time. I had an answer in under a minute. Worked out as follows: 1). First consider the number of odd numbers without the digit 5 from 100-110. This is 101, 103, 107 and 109, i.e. four numbers. 2). Next apply this rule for all the numbers between 100 and 200. Since there are 9 series of numbers between 100 and 200 (excluding the 150 - 160 series), we have, 4 x 9 = 36. 3). Now extend the rule to each series of 100 numbers between 100 and 1000 (remember to eliminate 500 - 600). This gives us 4 x 9 x 8 = 288. _________________ petrifiedbutstanding Intern Joined: 04 Sep 2009 Posts: 43 Followers: 1 Kudos [?]: 67 [1] , given: 9 Re: m03 #12 [#permalink] ### Show Tags 12 Nov 2012, 08:26 1 This post received KUDOS One easier way to do it, is the following: Just for a quick glance, and to see if we can take out some of the answer choices: Integers from 100 to 1000 inclusive: we have (1000 – 100 + 1= 901 integers). Odd Integers: [(999-101)/2]+1 = 450  we need to take out all the “5” digit… We should consider the following permutation: A x B x C Slot A: we have 8 possibilities (we cannot consider 0 and 5) Slot B: we have 9 possibilities (we cannot consider number 5) Slot C: we have 4 possibilities (we should only consider odd number, with exception of number 5) So, final product is: 8 x 9 x 4 = 288 IMO: D If you liked my post, consider giving me a Kudos! _________________ Aeros "Why are you trying so hard to fit in when you were born to stand out?" "Do or do not. There is no 'try'..." Intern Joined: 30 Aug 2012 Posts: 8 Concentration: General Management, Finance Followers: 0 Kudos [?]: 7 [1] , given: 3 Re: m03 #12 [#permalink] ### Show Tags 14 Nov 2012, 10:42 1 This post received KUDOS Just think this question in terms of forming a 3 digit odd number that does not include 5 since 100 and 1000 are even numbers 1: Since the number is odd, units digit should be odd = so available options = 4 (1,3,7,9) as 5 should be omitted. 2: Middle digit can be anything other than 5, so options = 9 3: hundreds digit cannot be 0 or 5 so options = 8 Total = 8*9*4 = 288 Hope this helps. Current Student Joined: 14 Apr 2008 Posts: 453 Schools: F2010 - HBS (R1 - denied w/o interview ), INSEAD (R1 - admitted), Wharton (R1 - waitlisted & ding), Ivey (R2 - admitted w/ 60% tuition) WE 1: 3.5yrs as a Strategy Consultant - Big 4 Followers: 14 Kudos [?]: 36 [0], given: 16 Re: m03 #12 [#permalink] ### Show Tags 11 May 2009, 20:26 goldeneagle94 wrote: There is 450 Odd integers in total between 100 to 1000, inclusive. Among all the three digit ODD numbers, how many numbers have 5 in them: UNITS place = 10x9 = 90 (105,115,125.....195......995) TENS place = 4x9 = 36 (excluding the numbers counted above - 151,153,157,159......551,553,557,559.....959) HUNDREDS place = 50-10-4=36 (From 500 to 599, there are 50 ODD numbers. Out of which, 10 are counted in UNITS place(505,515...595), 4 are counted in TENS place (551,553,557,559)) 450-90-36-36 = 288. Good question overall. Thanks for the explanation ... need to train my brain to segregate this into counting for units, tens and hundreds!! _________________ INSEAD Sept 2010 Interview Invite Nov 5, 2009 Admit & Matriculating Wharton Sept 2010 Interview Invite Oct 30, 2009 Waitlisted & Ding Harvard Sept 2010 Ding without Interview Ivey May 2010 Interview Invite Nov 23, 2009 Admit +$$

Manager
Joined: 24 Aug 2010
Posts: 189
Location: Finland
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WE 1: 3.5 years international
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Kudos [?]: 94 [0], given: 18

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08 Nov 2010, 11:18
suyashjhawar wrote:
In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?

(A) 180
(B) 196
(C) 286
(D) 288
(E) 324

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Can somebody proceed as below and get the ans.?
100 to 1000 we have 901 integers out of which 451 are odd.From there deduct 500-599 series i.e 100 more nos and get to 351...after that i got stuck:(

Ok. I think there are two ways of understanding this question.
1st: We have to find 3-digits numbers, which are odd and do not contain a 5 at the unit's place, and lie between 100 and 1000.
Everybody seems to be getting D as the correct answer but I am getting a completely different answer. Kindly point out the error in my explanation given below:
Basically we have to find the total number of 3-digit numbers, which are odd, and lie between 100 and 1000.
We have restrictions on the unit's place and the hundred's place. So I will fill those places first.
Unit's place or the third place can be filled in 4 different ways (1,3,7, or 9 can fill the unit's place).
Hundred's place or the first place can be filled in 8 different ways ( since 0 cannot occupy the first place).
Tens' place or the second place can be filled in 8 different ways (No restriction).
Hence, total number of ways = 4*8*8 = 256 ( not even an answer choice).

Now comes the 2nd part which I think is the correct interpretations to the question being asked.
2nd: Find 3-digit odd numbers, which do not contain 5 at all, and lie between 100 and 1000.
Units place can be filled in 4 ways. (1,3,7, or 9).
Hundred's place can be filled in 7 ways. (No 0 and no 5)
Tens place can be filled in 7 ways. (No 5)
Total number of ways = 4*7*7 = 196 - B
Intern
Joined: 25 Aug 2010
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09 Nov 2010, 08:29
goldeneagle94 wrote:
There is 450 Odd integers in total between 100 to 1000, inclusive.

Among all the three digit ODD numbers, how many numbers have 5 in them:

UNITS place = 10x9 = 90 (105,115,125.....195......995)
TENS place = 4x9 = 36 (excluding the numbers counted above - 151,153,157,159......551,553,557,559.....959)
HUNDREDS place = 50-10-4=36 (From 500 to 599, there are 50 ODD numbers. Out of which, 10 are counted in UNITS place(505,515...595), 4 are counted in TENS place (551,553,557,559))

450-90-36-36 = 288.

Good question overall.
thank you for your detailed explanation.
D.

The easiest way to calculate this equation is by thinking how many integers are viable in hundreds, tens and units places, which results 8 * 9 * 4 = 288.

I hope this helps.

oh, i found yours is a quicker way:)

PS: i guess the question is tricky by asking 'not containing digit 5'. originally i planned to calculate all 5s.
Manager
Joined: 21 Nov 2010
Posts: 128
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Kudos [?]: 5 [0], given: 12

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13 Nov 2011, 17:02
suyashjhawar wrote:
In a set of numbers from 100 to 1000 inclusive, how many integers are odd and do not contain the digit "5"?

(A) 180
(B) 196
(C) 286
(D) 288
(E) 324

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Can somebody proceed as below and get the ans.?
100 to 1000 we have 901 integers out of which 451 are odd.From there deduct 500-599 series i.e 100 more nos and get to 351...after that i got stuck:(

I definitely like the slot method on this problem, but to answer your question, 100-999 (no need for 1000 since it's even) you get 450 odd integers (999-100 + 1). Remember when deducting from the 500's, you only deduct the odds as you already deducted the evens, so that's 50. Then you deduct the 5s in the tens and units columns. Hope it makes more sense now.
Manager
Joined: 10 Jan 2011
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GMAT Date: 07-16-2012
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Kudos [?]: 63 [0], given: 25

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16 Nov 2011, 02:35
Unit digit with restriction= only four possible odd number allowed (1,3,7,9)
Tenth digit with restriction = 9 possiblilities (1,2,3,4,6,7,8,9,0)
hunderdth digit with restriction = 8 possibilities (1,2,3,4,6,7,8,9)

To number of odd number integers with digit 5 = 4*9*8 = 288

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-------Analyze why option A in SC wrong-------

Re: m03 #12   [#permalink] 16 Nov 2011, 02:35
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# m03 #12

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