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Of the three-digit positive integers whose three digits are all differ

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Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post Updated on: 25 Apr 2016, 11:36
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Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help

Originally posted by Noxy416 on 25 Apr 2016, 11:28.
Last edited by ENGRTOMBA2018 on 25 Apr 2016, 11:36, edited 2 times in total.
Reformatted the question and the title.
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 26 Apr 2016, 01:10
5
56
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help


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Hope it helps.
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Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 25 Apr 2016, 11:42
26
12
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 25 Apr 2016, 12:22
Thanks Engr2012 for helping with this one. Will keep in mind the posting rules.

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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 26 Apr 2016, 02:10
4
1
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help


Hi,
since you want to know where you have gone wrong in your method..
hundreds digit - 3-- correct
units digit - 4 -- correct but ONLY in two cases where hundreds digit is 7 and 9-- MEANS you will have to add later for hundreds digit as 8, since 8 will have 5 ODDs as units digit..
I assume you mean tens digit as 8..NO, we are looking for Non-zero, so 10 - 2( ONE of hundreds and ONE of units) - 1( zero) = 7..

Now our answer is -- 3*4*7 = 84..
But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

Total = 84+7 = 91
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 26 Apr 2016, 11:36
1
Thanks Bunuel.
Thanks Chetan2u for explaining my mistake.
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 05 May 2016, 04:50
Hi chetan2u,

I did this question from 2 approach. By 1 I am not getting the correct ans.
Let me tell you both of my approaches

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

Approach 2

for 700-799= 1*7*4= 28
for 800-899= 1*7*5 = 35
for 900-999= 1*7*4= 28
28+35+28 = 91 which is correct answer

__________________________________________________-
In your earlier response

But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

I see that you to added 7 to 84.

But I am still not getting, whats arong in my approach 1. or what extra numbers do I need to add in 84.

Can you please assist.

Thanks
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 05 May 2016, 04:56
2
PrakharGMAT wrote:
Hi chetan2u,

I did this question from 2 approach. By 1 I am not getting the correct ans.
Let me tell you both of my approaches

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

Approach 2

for 700-799= 1*7*4= 28
for 800-899= 1*7*5 = 35
for 900-999= 1*7*4= 28
28+35+28 = 91 which is correct answer

__________________________________________________-
In your earlier response

But we have to add for 5th Odd in units place for 8 as hundred's digit = 1*7*1 = 7..

I see that you to added 7 to 84.

But I am still not getting, whats arong in my approach 1. or what extra numbers do I need to add in 84.

Can you please assist.

Thanks



Hi,

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options -Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

you can HALF ONLY when the numbers of ODD and EVEN are same,..
here it is not the same..
UNITS digit can be any of 10, 0 to 9, but we are excluding 0.. so there are 5 ODD- 1,3,5,7,9- and 4 EVEN - 2,4,6,8..
Again when 7 and 9 are first digit , the numbers would be different and when it is 8, it will be different
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 05 May 2016, 05:04
1
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PrakharGMAT wrote:
Hi chetan2u,

Approach 1

For thousand's digit we have 3 options- 7,8 and 9
For Unit's digit we have 8 options - Excluding 0 and the first digit
For Hundered's digit we have 7 options - Excluding 0 ,first and second digit
So= 3x8x7= 168 (Include both Odd and Even terms)
So, 168/2= 84 which is not the answer :(

Can you please assist.

Thanks


Hi PrakharGMAT,
If you want to follow approach I..
first find for 7 and 9..
hundreds digit - 7 or 9- 2 ways
units digit - 1,2,3,4,5,6,8,7/9 - so 8 --- 4 are even and 4 are ODD..
tens digit - remaining 7..
total ways = 2*8*7 = 112..
half of these will be ODD so 112/2 = 56

Now if hundreds digit is 8..
units digit = 9 but ONLY 5 will be ODD
tens digit remaining 7..
so total = 1*7*5=35

Overall = 56+35 = 91.. :) :)
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 24 May 2016, 20:02
1
Hi,
Please explain how are the possible number of options "7" for tens place.
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 01 Jun 2016, 04:27
1
1
Namita06 wrote:
Hi,
Please explain how are the possible number of options "7" for tens place.


I actually had the same question, and I think the reason is as follows:

For 7AB, A & B cannot be 7, but also, AB cannot be the same, right? What it means is:
A cannot be equal B, 0 or 7, thus 10-3 = 7
B cannot be even & equal 7 or A, but we've already accounted for A & B difference above, thus B has 4 options.

I hope it helps!
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 01 Jun 2016, 21:34
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3
To find odd integers greater than 700:

In 700s:
Units can be any of 4 (1,3,5,9)
Hundreds is obviously just 1 way (7)
So, tens can be any of the remaining 7.
So, total = 4*1*7 = 28

In 800s:
Units can be any of 5 (1,3,5,7,9)
Hundreds is obviously just 1 way (8)
So, tens can be any of the remaining 7.
So, total = 5*1*7 = 35

In 900s:
Units can be any of 4 (1,3,5,7)
Hundreds is obviously just 1 way (9)
So, tens can be any of the remaining 7.
So, total = 4*1*7 = 28

So, grand total = 28 + 35 + 28 = 91
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 05 Jun 2016, 12:44
Engr2012 wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.


Can you please explain how you are getting to the "Number of combinations for type" 1-3?
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 12 Jul 2016, 02:47
1
Bunuel wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help


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Hope it helps.



It took me hours to solve and understand the concepts behind all these sums. It more than helped me. Atleast now when i encounter such sums i can try solving it under 2 mins.
Thanks a tonne
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 26 Aug 2016, 03:05
1
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help




i tried to solve this way:


700-1000 there 300 numbers
300/2 odds so 150 numbers
710-720-730 .... 10 *3 =30 numbers so 150-30=120
than 701-702-703 ..... 10*3=30 numbers so 120-30=90
90 is close to 91.
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 26 Aug 2016, 14:31
3
We know the integers must be greater than 700 and they must be odd; we can divide them into 6 groups: 7EO, 7OO, 8EO, 8OO, 9EO and 9OO where E denotes an even digit and O an odd digit. Let’s determine the number of integers in each group.

7EO:
The hundreds digit is 7, so it’s only 1 choice. Since all the digits are nonzero, E can be 4 numbers: 2, 4, 6 or 8. Since the digits are all different, O can be 4 numbers: 1, 3, 5 or 9. Therefore, there are 1 x 4 x 4 = 16 such integers.

7OO:
Since all the digits are different, O can only be 4 numbers: 1, 3, 5 or 9. However, the second O can only be 3 numbers since it cannot be the same as the first O. Therefore, there are 1 x 4 x 3 = 12 such integers.

8EO:
The hundreds digit is 8, so there is only 1 option. Since all the digits are different and nonzero, E can only be 3 numbers: 2, 4, or 6. Since the last digit is O, an odd digit, O can any of the 5 odd digits: 1, 3, 5, 7 or 9. Therefore, there are 1 x 3 x 5 = 15 such integers.

8OO:
The first O can be any of the 5 odd digits: 1, 3, 5, 7 or 9. However, the second O can only be 4 numbers since it cannot be the same as the first O. Therefore, there are 1 x 5 x 4 = 20 such integers.

9EO:
This is analogous to 7EO, so there should be 16 such integers.

9OO:
This is analogous to 9EO, so there should be 12 such integers.

Therefore, there are 16 + 12 + 15 + 20 + 16 + 12 = 91 odd integers that are greater than 700 where all digits are different and nonzero.

Answer: B
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Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 30 Oct 2016, 07:40
1
According to the constraints we have 3 options for the thousands digit (7, 8, 9), 9 options for the hundreds digit (except 0), and 5 options for the units digit (it needs to be an odd number) in total with additional constraint that all digits should be different.
Lets look at two possible cases.

First (thousands digit is odd)
2 x 7 x 4 = 56
___ ____ ____
(7,9) (9-2) (5-1)

Thousands digit is even
1 x 7 x 5 = 35
___ ____ ____
(8) (9-2) (5)

Total 56+35=91
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 09 Apr 2017, 11:50
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



For the 700s:
_ _ _

I determined the possible options for each slot. The number of options for the first slot is 1 (7). For the second slot the number can be even or odd. Starting with the odd case, the number of options is 4 (1, 3, 5, 9). This means that the number of options available for the third slot is 3. 1x4x3=12.
Now, if the second slot is even, there are 4 options (0 cannot be counted). The number of options for the third slot is 4 (only one odd number is used up - by the first slot). 1x4x4=16.

12+16=28. This number of options is the same for the 900s. So, now we just need to find the number of options for the 800s:

The number of options for the first slot is 1 (8). For the second slot the number can be even or odd. Starting with the odd case, the number of options is 5 (1, 3, 5, 7, 9). This means that the number of options available for the third slot is 4. 1x5x4=20.
Now, if the second slot is even, there are 3 options (0 cannot be counted). The number of options for the third slot is 5 (no other odd numbers used up). 1x3x5=15.

20+15=35.

Total = 28+28+35=91.

Kudos if you agree! Please comment if you disagree with the method and/or have improvements.
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 21 May 2017, 18:44
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Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help

Start with the digit under most constraints
1. Units digit can have 4 values i.e., 1,3,5,9 and 1,3,5,7 when the hundreds digit is 7 and 9 resp and can have 5 values i.e, 1,3,5,7,9 when the hundreds digit is 8
2. Tens digit cannot have the value of unit and hundreds digit and cannot be 0. So can have 7 values
3. So when hundreds digit is 7 or 9, number of possibilities is 4*7 + 4*7 = 56 and when hundreds digit is 8, number of possibilities is 5*7=35
4. Total number of possibilities is 91
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Re: Of the three-digit positive integers whose three digits are all differ  [#permalink]

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New post 04 Nov 2017, 04:51
ENGRTOMBA2018 wrote:
Noxy416 wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?

(A) 84
(B) 91
(C) 100
(D) 105
(E) 243

What am i doing wrong in this question?

Number of options for the hundreds digit =3
Number of options for the Units digit =4
Number of options for the Units digit = 8

Total number of options- 3 x 8 x 4 = 96
Please help



Please follow posting guidelines (link in my signatures). Do make sure to mention the correct title and put your analyses either in the next post or under spoilers.


As for the question, you need to find the number of combinations possible such that:

1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9

Based on this, the numbers can be of the following 3 types:

Type 1: 7AB
Type 2: 8EF
Type 3: 9CD

For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).

Number of combinations for type 1 : 1*7*4 = 28

Number of combinations for type 2: 1*7*5 = 35

Number of combinations for type 3 : 1*7*4 = 28

Thus, total numbers possible = 28+35+28 = 91.

B is thus the correct answer.

Hope this helps.


Hi,

I am a little confused with the explanation. For number of combinations for type 2: why is it 1*7*5
Don't we have to make sure that all numbers are different, non zero and odd in type 2 as well?

What am I missing?
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Re: Of the three-digit positive integers whose three digits are all differ &nbs [#permalink] 04 Nov 2017, 04:51

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