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# m03 #12

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Senior Manager
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m03 #12 [#permalink]

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16 Apr 2009, 03:18
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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:(

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Re: m03 #12 [#permalink]

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08 Nov 2010, 07:05
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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.

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Re: m03 #12 [#permalink]

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11 Nov 2011, 13:08
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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
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Re: m03 #12 [#permalink]

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09 Nov 2010, 04:58
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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

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Re: m03 #12 [#permalink]

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16 Apr 2009, 09:35
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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.

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Senior Manager
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Re: m03 #12 [#permalink]

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09 Nov 2010, 03:57
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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.
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Re: m03 #12 [#permalink]

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12 Nov 2012, 08:26
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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

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Re: m03 #12 [#permalink]

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14 Nov 2012, 10:42
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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.

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Re: m03 #12 [#permalink]

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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!!
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Re: m03 #12 [#permalink]

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08 Nov 2010, 11:18
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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

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Re: m03 #12 [#permalink]

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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.
Rado wrote:
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.

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Re: m03 #12 [#permalink]

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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.

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Re: m03 #12 [#permalink]

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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

therefore answer should be C
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Re: m03 #12   [#permalink] 16 Nov 2011, 02:35
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# m03 #12

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