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M01-10

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Re: M01-10 [#permalink]

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New post 30 Jun 2016, 10:00
Using basic counting principles:

_ _ _

Total Numbers to be considered = 000 to 999

Case1: _ _ 7
Digits 0 - 9 can take first place in 10 ways. Similarly 0 - 9 digits can fill up the second place in 10 ways. Total ways = m * n = 10 * 10 = 100

Case2: _ 7 _
Total 100 ways

Case3: 7 _ _
Total 100 ways

Case1+2+3 =
[Reveal] Spoiler:
300 Ways

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Re: M01-10 [#permalink]

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New post 07 Jul 2016, 09:27
Bunuel
hello can we do it this way ?

since there are 10 digits and repetitions can be allowed .
keeping 7 in hundreds place >> 7 __ __ { 1*10*10}
keeping 7 in tens place >> __ 7[u]__ { 10*1*10}
keeping 7 in units place >> __ __ [u]7
{10*10*1}
adding above 3 we get 300

also 3c1 * 10*10

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Re M01-10 [#permalink]

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New post 14 Jul 2016, 12:44
I think this is a high-quality question and I agree with explanation. Third approach was easiest for me to understand.

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Re M01-10 [#permalink]

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New post 31 Jul 2016, 03:33
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Consider numbers from 1 to 1000 written as follows:


1. 0001

2. 0002

3. 0003

...

1000. 1000

We still have 1000 numbers. However, we used 4 digits per number, hence used total of 4∗1000=4000 digits.
4000/10=400 times.

What's wrong with the above inference?
Thanks.

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M01-10 [#permalink]

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New post 13 Aug 2016, 09:11
is this approach correct
there are 3 places x,x,x
consider 7 can be placed in each of the position
case 1 7 at hundredths place then rest 2 position can be filled by 10 numbers (0-9) there fore total in 100 ways (1X10X10) (possibilities are in and clause)
case2 7 at tenths place then rest 2 position can be filled by 10 numbers (0-9) there fore total in 100 ways (10X1X10) (possibilities are in and clause)
case 3 7 at ones place then rest 2 position can be filled by 10 numbers (0-9) there fore total in 100 ways (10X10X1) (possibilities are in and clause)
as these possibilities are in or clause so adding them up gives 300 ways

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Re: M01-10 [#permalink]

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New post 13 Aug 2016, 22:00
Haihai, I believe that's because when you have four digits the numbers could exceed 1000 such as 7,674 and so on. I believe what you did would be the approach to finding out how many times 7 is used between 1 and 10,000.

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Re: M01-10 [#permalink]

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New post 27 Dec 2016, 09:28
If you got 271 (like me), you were most likely tricked :).

This question: how many times does 7 appear?

"Typical" question of this type: how many digits with a 7 appear?

The typical question doesn't allow you to count 777 three times (or 77 twice); this question does.

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Re: M01-10 [#permalink]

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New post 27 Dec 2016, 09:46
Saurav Arora wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Consider numbers from 1 to 1000 written as follows:


1. 0001

2. 0002

3. 0003

...

1000. 1000

We still have 1000 numbers. However, we used 4 digits per number, hence used total of 4∗1000=4000 digits.
4000/10=400 times.

What's wrong with the above inference?
Thanks.


It doesn't work because you have 4 digits (and aren't accounting for 1001 to 9999). The concept applies as follows:

1 digit (0 to 9): 10 numbers, 1 digit per number = 10 digits --> 10/(10 digits) = each digit used 1 time
2 digits (00 to 99): 10*10=100 numbers, 2 digits per number = 200 digits --> 200/(10 digits) = each digit used 20 times
3 digits (000 to 999): 10*10*10=1000 numbers, 3 digits per number = 3000 digits --> 3000/(10 digits) = each digit used 300 times
4 digits (0000 to 9999): 10*10*10*10=10000 numbers, 4 digits per number = 40000 digits --> 40000/(10 digits) = each digit used 4000 times
..
n digits: 10^n numbers, n digits per number = n*10^n digits --> (n*10^n)/10 = each digit used n*10^(n-1) times

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Re: M01-10 [#permalink]

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New post 14 Feb 2017, 08:22
POSSIBILE COMBO NUMBER OF 7's use X as every single digit number except 7 -->X=0,1,2,3,4,5,6,8,9 --> 9 POSSIBLE NUMBERS
7XX 81
XX7 81
X7X 81
X77 9x2
77X 9x2
7X7 9x2
777 3
_____
300

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Re: M01-10 [#permalink]

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New post 01 May 2017, 19:57
Hi,

Are we not counting 77 twice in Approach 3?

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Re: M01-10 [#permalink]

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New post 06 May 2017, 14:05
(7)=1
9 (7)=9
(7) 10=10
9 10 (7)=90
9 (7) 10=90
(7) 10 10 = 100
total=300
ans:D

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Re: M01-10 [#permalink]

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New post 03 Jun 2017, 08:22
For a 3 digit number, 7 can be at the unit place, tens place or hundred place.
7 at hundred place, that means total combination will be:- 1*10*10 (first digit can only be seven but second and third can be any from 0-9)
7 at tens place, that means total combination will be:- 9*1*10 (first digit can be any number from 1-9, second will only be 7 and third will be any number from 0-9)
7 at unit place, that means total combination will be:- 9*10*1 (first digit can be any number from 1-9, second will any number from 0-9 and third will be 7)
So total 3 digit number with 7 at some place = 100+90+90=280 (You can calculate this for any digit)

For a 2 digit number, 7 can be at unit place, or tens place
Going with the same logic as described above:-
7 at tens place total combination will be:- 1*10 (first digit can only be seven but second can be any from 0-9)
7 at unit place total combination will be:- 9*1 (first digit can be any number from 1-9 and second digit is 7)
So for 2 digit number total combination will be 10+9=19 (You can calculate this for any digit)

For single digit number, there is only one possibility
So total is 280+19+1=300.
D is the answer.

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Re: M01-10 [#permalink]

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New post 14 Jul 2017, 06:45
On counting three digit numbers,total is 280 but this also includes 777 thrice,which means we should subtract 2 777's leading to answer of 298.

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Re: M01-10 [#permalink]

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New post 14 Jul 2017, 06:48
Kumargouravnayak wrote:
On counting three digit numbers,total is 280 but this also includes 777 thrice,which means we should subtract 2 777's leading to answer of 298.


The correct answer is D - 300. Please read the whole thread before posting. Thank you.
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Re: M01-10 [#permalink]

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New post 04 Sep 2017, 11:40
WaterFlowsUp wrote:
possibilities are

X X 7 = 9*9*1
X 7 X = 9*1*9
7 X X = 1*9*9
7 7 X = 1*1*9
7 X 7 = 1*9*1
X 7 7 = 9*1*1
777 = 1
if we add all of them = 81+81+81+9+9+9+1=271
where did i go wrong??


in cases 77X, 7X7, X77 and 777 the digit appears 1*1*9 * 2, 1*9*1*2, 9*1*1*2, 1*3. You do this the answer is exactly 300.

Note- Question is about number of 7, not the numbers which have seven. eg. 777 has 3 sevens.

Hope it helps.
:-) :-) :-)

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Re M01-10 [#permalink]

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New post 02 Oct 2017, 21:05
I think this is a high-quality question and I don't agree with the explanation.

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Re: M01-10 [#permalink]

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New post 02 Oct 2017, 21:12
amar.igtr wrote:
I think this is a high-quality question and I don't agree with the explanation.


Hi amar.igtr,

Thank you for posting. One request though: please be specific, why don't you agree with the solution? So, far you've posted two similar replies and none of them were correct. Thank you.
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Re: M01-10   [#permalink] 02 Oct 2017, 21:12

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