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Re: M01-10 [#permalink]
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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??

This is how I did..same approach as you took...this might help you understand what went wrong

_ single digit -> 1 possibility
_ _ two digits -> (1 possibility * 10 possibilities) + (9 possibilities * 1 possibility) -> (9, as we cannot place a 0) = 19
_ _ _ three digits -> (1 possibility * 10 possibilities * 10 possibilities) + (9 possibilities * 1 possibility * 10 possibilities) + (9 possibilities * 10 possibilities * 1 possibility) = 100 + 90 + 90 = 280

add up, 1 + 19 + 280 -> 300
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Re: M01-10 [#permalink]
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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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M01-10 [#permalink]
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swatjazz wrote:
Bunuel wrote:
Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304


Multiple approaches are possible to solve this problem. Here are two:

Approach #1:

Let's consider the numbers from 0 to 999, which are written using three digits:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

There are a total of 1000 numbers, each of which uses 3 digits. Therefore, there are a total of \(3*1000=3000\) digits used in all these numbers. Since there is no reason for any digit to be favored over another, each of the 10 digits should be used an equal number of times. Hence, each digit (including 7) is used \(\frac{3000}{10}=300\) times.

Approach #2:

Within the range of 1-100, the digit 7 appears 10 times as the units digit (7, 17, 27, ..., 97) and 10 more times as the tens digit (70, 71, 72, ..., 79). Therefore, in the first 100 numbers, the digit 7 is written a total of 10 + 10 = 20 times.

Within each block of 100 numbers (e.g., 1-100, 101-200, ..., 901-1000), the digit 7 appears 20 times as the units or tens digit. Hence, it appears as the units or tens digit 20*10 = 200 times in the 10 blocks. Additionally, the digit 7 appears 100 times as the hundreds digit (700, 701, 702, ..., 799).

Thus, the digit 7 is written a total of 200 + 100 = 300 times in the range of 0-999.


Answer: D


Hi Bunuel,

In approach # 1, i didn't quite understand why did we use 3 digit numbers.
Can you please explain?
Apologies if it's a silly question.


Not a silly question at all.

Let me try to elaborate: this approach worked because when we write the numbers from 0 to 999 in the form XXX each digit take the values from 0 to 9 which provides that in the end all digits are used equal # of times.

For the range 100 to 999 it won't be so. We can solve for this range in the following way:
XX7 - 7 in the units place - first digit can take 9 values (from 1 to 9) and second digit can take 10 values (from 0 to 9) --> total numbers with 7 in the units place: 9*10=90;

X7X - 7 in the tens place - first digit can take 9 values (from 1 to 9) and third digit can take 10 values (from 0 to 9) --> total numbers with 7 in the tens place: 9*10=90;

7XX - 7 in the hundreds place - second digit can take 10 values (from 0 to 9) and third digit can take 10 values (from 0 to 9) --> total numbers with 7 in the hundreds place: 10*10=100.

TOTAL: 90+90+100=280.

Hope it helps.
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Re: M01-10 [#permalink]
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Bunuel wrote:
How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304


Hoozan

I will prefer to answer it using method-1

Method-1:

From 1 to 100: DIgit 7 is written 20 times
From 101 to 200: DIgit 7 is written 20 times
From 201 to 300: DIgit 7 is written 20 times
From 301 to 400: DIgit 7 is written 20 times
From 401 to 500: DIgit 7 is written 20 times
From 501 to 600: DIgit 7 is written 20 times
From 601 to 699: DIgit 7 is written 20 times
From 700 to 800: DIgit 7 is written 120 times ('7' comes at hundred's place in each number)
From 801 to 900: DIgit 7 is written 20 times
From 901 to 1000: DIgit 7 is written 20 times

Total Usages of digit 7 = 20*9+120 = 300

Method-2:

Case-1: Numbers using Digit 7 only once
Total Such cases = 3C1 = 3
'7' At Unit's place : 9 * 9 * 1 = 81
(Unit place can be filled only using digit 7
Ten's place can be filled in 9 places using any digit from 0 to 9 except 7
Hundred's place can be filled in 9 places using any digit from 0 to 9 except 7
Total Such numbers = 3*81 = 243

Case-2: Numbers using Digit 7 Twice
'7' At two place : Total Cases = 3C2 = 3 (choosing two places out of three digit places for digit '7')
Total Numbers = 9 Numbers
(Third place can be filled using any digit from 0 to 9 except '7')
ie. Total Such numbers = 3*9 = 27

Case-3: 7 is used at all the three places in a three digit number
Total Such numbers = 1 number

Total Usages of digit '7' = 243*1 + 27*2 + 1*3 = 300

Answer: Option D
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Re: M01-10 [#permalink]
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hessen923 wrote:
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

I make the answer C.


7 is not in 72 numbers (from 10 to 99) but out of 18 numbers with 7, one, 77, has two 7's, so 7 appears from 1 to 99: 18 + 1 + 1 =20 times.

Similarly you are missing that 177, 277, 377, ... contain two 7's and 777 contain three 7's.
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Re: M01-10 [#permalink]
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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??
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Re: M01-10 [#permalink]
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Bunuel wrote:
How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304


Here's one way to look at it.
Write all of the numbers as 3-digit numbers.
That is, 000, 001, 002, 003, .... 998, 999

ASIDE: Yes, I started at 000 and ended at 999, even though though the question asks us to consider the numbers from 1 to 1000. HOWEVER, notice that 000 and 1000 don't have any 7's, which means the outcome will be the same.

First, there are 1000 integers from 000 to 999
There are 3 digits in each integer.
So, there is a TOTAL of 3000 individual digits. (since 1000 x 3 = 3000)

Each of the 10 digits is equally represented, so the 7 will account for 1/10 of all digits.

1/10 of 3000 = 300

So, there are 300 0's, 300 1's, 300 2's, 300 3's, . . ., and 300 9's in the integers from 000 to 999

Answer: D

Cheers,
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Re: M01-10 [#permalink]
Why doesn't the following approach work?

Clearly, 7 appears once between 1 and 9. Then, there are 90 integers between 10 and 99. How many times does 7 NOT appear? 8 x 9, which is 72. 90-72 is 18. Therefore 7 appears 18 + 1 = 19 times between 1 and 99. There are 900 integers from 100 to 999. How many times does 7 NOT appear? 8 x 9 x 9 = 648. 900-648 = 252.
Therefore, 7 appears 252 times between 100 and 999.

252 + 19 = 271.

I make the answer C.
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M01-10 [#permalink]
Bunuel wrote:
Official Solution:

How many times will the digit 7 be written when listing the integers from 1 to 1000?

A. 110
B. 111
C. 271
D. 300
E. 304


Multiple approaches are possible to solve this problem. Here are two:

Approach #1:

Let's consider the numbers from 0 to 999, which are written using three digits:

1. 000

2. 001

3. 002

4. 003

...

1000. 999

There are a total of 1000 numbers, each of which uses 3 digits. Therefore, there are a total of \(3*1000=3000\) digits used in all these numbers. Since there is no reason for any digit to be favored over another, each of the 10 digits should be used an equal number of times. Hence, each digit (including 7) is used \(\frac{3000}{10}=300\) times.

Approach #2:

Within the range of 1-100, the digit 7 appears 10 times as the units digit (7, 17, 27, ..., 97) and 10 more times as the tens digit (70, 71, 72, ..., 79). Therefore, in the first 100 numbers, the digit 7 is written a total of 10 + 10 = 20 times.

Within each block of 100 numbers (e.g., 1-100, 101-200, ..., 901-1000), the digit 7 appears 20 times as the units or tens digit. Hence, it appears as the units or tens digit 20*10 = 200 times in the 10 blocks. Additionally, the digit 7 appears 100 times as the hundreds digit (700, 701, 702, ..., 799).

Thus, the digit 7 is written a total of 200 + 100 = 300 times in the range of 0-999.


Answer: D


Hi Bunuel,

In approach # 1, i didn't quite understand why did we use 3 digit numbers.
Can you please explain?
Apologies if it's a silly question.
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Re: M01-10 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M01-10 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M01-10 [#permalink]
GMATinsight I remember you guys solving a similar question by using Combinations. Could you possibly show how we could apply that out here?

What I did was:

For 7 in the Units place

0 0 0 7 = 0 in all places and 7 in the units place so 1
0 0 9 7 = 0 in last two places, 9 options for Tens place (except 0) and 7 for the units so 9
0 9 10 7 = 0 in the last place, 9 options for hundreds place (except 0), 10 options for tens place, and 7 for units place, so 90

Number of numbers in which 7 is written in the units place = 100

For 7 in the tens place and no 7 in the units place

0 0 7 0 = 0 in all places and 7 in the tens place so 1
0 9 7 9 = 0 in the last place, 9 options in the hundreds place (except 0), 7 in the tens place, and 9 options for the units place (Except 7) so 81

So the number of numbers which has 7 in the tens place and no 7 in units place is 90

Not sure If I am correct or how to process further
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Re: M01-10 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M01-10 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M01-10 [#permalink]
I have solved the ques as below,

1) 10 way*10way*1 way = 100 (when 7 is placed at unit's place) (10 - 0 to 9 digits)
2) 10 ways * 1 way* 10 ways = 100 (when 7 is placed at ten'2 place)
3) 1 way * 10 ways * 10 ways = 100 (when 7 is placed at hundred's place)

Total = 100+100+100 = 300 ways (answer)
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Re: M01-10 [#permalink]
Hello Bunuel,

Can you please share some questions of similar type to practice?
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