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M01 #10 [#permalink] New post 29 Sep 2008, 23:43
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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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

I'm confused with the solution given. Seems like there's something wrong with their formula.

[Edit] Also discussed here gmat-club-ps-69165.html
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Re: mo1 #10 [#permalink] New post 30 Sep 2008, 01:30
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300 for me.

For single digit number, 7 will be written only once.

For two digit number, 7 can appear in 10s as well as unit place. Hence, number of times, 7 will appear = 9 * 1 (for 7 at unit place) + 1 * 10 (for 7 at 10s place) = 19

Similarly, for three digit number, number of times 7 will appear
= 1 * 10 * 10 (for 7 at 100s place) + 9 * 1 * 10 (for 7 at 10s place) + 9*10 * 1(for 7 at unit place) = 280

Hence, total number of times 7 will appear = 1 + 19 + 280 = 300.
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Re: mo1 #10 [#permalink] New post 30 Sep 2008, 06:50
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scthakur wrote:
300 for me.

For single digit number, 7 will be written only once.

For two digit number, 7 can appear in 10s as well as unit place. Hence, number of times, 7 will appear = 9 * 1 (for 7 at unit place) + 1 * 10 (for 7 at 10s place) = 19

Similarly, for three digit number, number of times 7 will appear
= 1 * 10 * 10 (for 7 at 100s place) + 9 * 1 * 10 (for 7 at 10s place) + 9*10 * 1(for 7 at unit place) = 280

Hence, total number of times 7 will appear = 1 + 19 + 280 = 300.


I don't get the part about the 3 digit numbers
If 1 * 10 * 10 can be anywhere from 700 to 799
9 * 1 * 10 can be from 170 to 970
and 9*10 * 1 can be from 107 to 997
Then wouldn't there be an overlap between these numbers?like wouldn't numbers like 777 be counted three times in this condition? I'm really confused
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Re: mo1 #10 [#permalink] New post 30 Sep 2008, 06:58
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1: 7,17,---97 : total 11 (in 77 we get 7 two times)
2: 70,71,79 = 11 - 2 = 9 (77 is repeated here so remove it once)

so from 1,100 there is 20 times 7

so 1 to 1000 there will be 20 * 10 = 200 time 7

3: 700 -799 : 100 times 7 (now we do not need to count 770 to 779), we already count it

so total no. of 7 will be 200 + 100.
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Re: mo1 #10 [#permalink] New post 30 Sep 2008, 09:31
redbeanaddict wrote:
scthakur wrote:
300 for me.

For single digit number, 7 will be written only once.

For two digit number, 7 can appear in 10s as well as unit place. Hence, number of times, 7 will appear = 9 * 1 (for 7 at unit place) + 1 * 10 (for 7 at 10s place) = 19

Similarly, for three digit number, number of times 7 will appear
= 1 * 10 * 10 (for 7 at 100s place) + 9 * 1 * 10 (for 7 at 10s place) + 9*10 * 1(for 7 at unit place) = 280

Hence, total number of times 7 will appear = 1 + 19 + 280 = 300.


I don't get the part about the 3 digit numbers
If 1 * 10 * 10 can be anywhere from 700 to 799
9 * 1 * 10 can be from 170 to 970
and 9*10 * 1 can be from 107 to 997
Then wouldn't there be an overlap between these numbers?like wouldn't numbers like 777 be counted three times in this condition? I'm really confused



No. There won't be. We are concerned with digits and not the numbers.

For example, number 777 will appear in the count where 100s digit is 7, in the count where 10th digit is 7 as well as in the count where unit digit is 7, thus making the total count of 7 as 3.
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Re: mo1 #10 [#permalink] New post 01 Oct 2008, 18:15
scthakur wrote:
redbeanaddict wrote:
scthakur wrote:
300 for me.

For single digit number, 7 will be written only once.

For two digit number, 7 can appear in 10s as well as unit place. Hence, number of times, 7 will appear = 9 * 1 (for 7 at unit place) + 1 * 10 (for 7 at 10s place) = 19

Similarly, for three digit number, number of times 7 will appear
= 1 * 10 * 10 (for 7 at 100s place) + 9 * 1 * 10 (for 7 at 10s place) + 9*10 * 1(for 7 at unit place) = 280

Hence, total number of times 7 will appear = 1 + 19 + 280 = 300.


I don't get the part about the 3 digit numbers
If 1 * 10 * 10 can be anywhere from 700 to 799
9 * 1 * 10 can be from 170 to 970
and 9*10 * 1 can be from 107 to 997
Then wouldn't there be an overlap between these numbers?like wouldn't numbers like 777 be counted three times in this condition? I'm really confused



No. There won't be. We are concerned with digits and not the numbers.

For example, number 777 will appear in the count where 100s digit is 7, in the count where 10th digit is 7 as well as in the count where unit digit is 7, thus making the total count of 7 as 3.

wow. I didnt notice that. very very helpful. Anyway, if we're counting the numbers, what could have been the answer?
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Re: mo1 #10 [#permalink] New post 01 Oct 2008, 18:51
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scthakur wrote:
300 for me.

For single digit number, 7 will be written only once.

For two digit number, 7 can appear in 10s as well as unit place. Hence, number of times, 7 will appear = 9 * 1 (for 7 at unit place) + 1 * 10 (for 7 at 10s place) = 19

Similarly, for three digit number, number of times 7 will appear
= 1 * 10 * 10 (for 7 at 100s place) + 9 * 1 * 10 (for 7 at 10s place) + 9*10 * 1(for 7 at unit place) = 280

Hence, total number of times 7 will appear = 1 + 19 + 280 = 300.


300 is also my answer

All digits are written same, note that 1 = 001

So we can calculate the total of all digits appears then divide by 10

Each number appears = 10 x 10 x 10, in each number there are 3 digits so 10 x 10 x 10 x 3 = 3000

/10 = 300
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Re: mo1 #10 [#permalink] New post 02 Oct 2008, 05:20
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redbeanaddict wrote:
aim2010 wrote:
redbeanaddict wrote:
wow. I didnt notice that. very very helpful. Anyway, if we're counting the numbers, what could have been the answer?


281

how did you get 281?


my apologies - it should be 271.

using a new approach i learned in this forum, imagine each number to be XXX where X can be 0 to 9:
numbers with 7 at units place only: 9 * 9 = 81
numbers with 7 at tens place only: 9 * 9 = 81
numbers with 7 at hundreds place only: 9 * 9 = 81
numbers with 7 at units and tens place only: 9
numbers with 7 at hundreds and tens place only: 9
numbers with 7 at units and hundreds place only: 9
numbers with 7 at all three places: 1
81 *3 + 9*3 + 1 = 271
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Re: m01 #10 [#permalink] New post 15 Dec 2008, 00:14
It's a tricky question indeed. Between 1-1000 we can find 7 in digits (7), tens (17,27,37, and so on), and hundreds digit (107, 177, and so on). Here how we break it down logically:

(1) digits, it's obvious that we can find only 1 digit with 7 in it which is 7.
(2) tens. we can find it in 17,27,...,97 => 9 ; and we can find it in 70,71,...79 =>10. Mind you, we find "7" twice in 77. I count it once when 7 was in the tens and another one when 7 was in the digit. Or mathematically, there are 10 possible combination of numbers when 7 is in tenth (hence 1*10); and there are 9 possible combination of numbers when 7 is in digits (0 is excluded, only 1-9)
(3) hundreds. Now this is the tricky part. let's say that there are 3 combinations on where we can put 7 which is 7xy or x7y or xy7. Apply a basic combination solution:
#1: when 7 is in hundred, there are 1*10*10 (meaning 10 possible number in ten (0-9) and 10 possible number in digit (0-9))
#2: when 7 is in ten, there are 9*1*10 (meaning 9 possible number in hundred (1-9) and 10 possible number in digit (0-9))
#3: when 7 is in digit, there are 9*10*1 (meaning 9 possible number in hundred (1-9) and 10 possible number in digit (0-9))

From (1)+(2)+(3) we got 1+ 19 + 1*10*10 + 9*1*10 + 9*10*1 = 300.
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m01. Q.10 [#permalink] New post 21 Feb 2009, 22:57
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Number 7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999.

There are 3 types of numbers we have to consider: 1 digit, 2 digits and 3 digits.

Number 7 is the only one-digit number.

For any two-digit number the 7 could be the first digit. In that case, there are 10 ways to place the second digit, i.e. 0-9. In case, 7 is the unit's digit there are 9 ways to place the first digit. Since, 0 cannot take the tens digit. Thus, 10+9 = 19 numbers that will have 7.

Similarly, for 3 digit numbers 1*10*10 = 100 and 9*1*10 = 90 and 9*10*1 = 90. Add all the 3 digit combinations and you get 280.

Total = 280 + 19 + 1 = 300.
------------------------------------------
How do we get 280 for the 3rd digit. I understand that we get 100, but the 90 + 90?? I dont know how we care counting this into the calculation.
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Re: m01. Q.10 [#permalink] New post 25 Feb 2009, 21:53
Forrester300 wrote:
How many times will the digit 7 be written when listing the integers from 1 to 1000?

(C) 2008 GMAT Club - m01#10

* 110
* 111
* 271
* 300
* 304

Number 7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999.

There are 3 types of numbers we have to consider: 1 digit, 2 digits and 3 digits.

Number 7 is the only one-digit number.

For any two-digit number the 7 could be the first digit. In that case, there are 10 ways to place the second digit, i.e. 0-9. In case, 7 is the unit's digit there are 9 ways to place the first digit. Since, 0 cannot take the tens digit. Thus, 10+9 = 19 numbers that will have 7.

Similarly, for 3 digit numbers 1*10*10 = 100 and 9*1*10 = 90 and 9*10*1 = 90. Add all the 3 digit combinations and you get 280.[/color]

Total = 280 + 19 + 1 = 300.
------------------------------------------
How do we get 280 for the 3rd digit. I understand that we get 100, but the 90 + 90?? I dont know how we care counting this into the calculation.


Why 280?
1. 100 is for 7 in 100s place i.e. 700, 701 .........799. so there are 100 sevens (7s). remember only 7 in 100s place.
2. 90 is 7 in 10s position, for ex: 70, 71...79, 170, 171.....179 .......970, 971........979. there are 10x10 = 100 sevens in tens digit but 10 sevens are already counted in 770, 771..........779. so it becomes 90.
3. Another 90 is 7 in unit position, for ex: 7, 17..97, 107, 117........197....... 907, 917............997. there are 10x10 = 100 sevens in tens digit but 10 sevens are already counted in 707, 717.........797. so it becomes 90.

total = 100+90+90 = 280



Alternatively:
1. 100 is for 7 in 100s place i.e. 700, 701 .........799. so there are 100 sevens (7s). remember only 7 in 100s place.
2. 90 is 7 in 10s position, for ex: 70, 71...79, 170, 171.....179 .......970, 971........979. there are 10x10 = 100
3. Another 90 is 7 in unit position, for ex: 7, 17..97, 107, 117........197....... 907, 917............997. there are 10x10 = 100

total = 100+100+100 = 300


Alternatively:

100's place = 1x10x10 = 100
10's place = 10x1x10 = 100
1's place = 10x10x1 = 100

so total = 300
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Re: mo1 #10 [#permalink] New post 18 May 2009, 16:31
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lylya4 wrote:

300 is also my answer

All digits are written same, note that 1 = 001

So we can calculate the total of all digits appears then divide by 10

Each number appears = 10 x 10 x 10, in each number there are 3 digits so 10 x 10 x 10 x 3 = 3000

/10 = 300


That smells like a smart way to solve it, but I don't understand it... would you elaborate please?
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Re: M01 #10 [#permalink] New post 18 May 2009, 16:42
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For me, the easiest way to solve this problem is by simply counting the number of sevens per digit and then adding them.

Units digit: (sevens between 0-100)*(sevens between 100-1000)=10*10=100
Teens digit: (sevens between 0-100)*(sevens between 100-1000)=10*10=100
Hundred digit:(sevens between 700-799)=100

So, 100+100+100=300

I find this way of solving the problem so much easier than the other ways that I am afraid it would be wrong.
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Re: M01 #10 [#permalink] New post 06 Nov 2009, 23:07
1. 1-100  9 + 11 = 20

2. 1 – 1000 9 + 11 = 20  20 * 9 = 180 (not include 700 – 800)
3. 700 – 800  100 + 20 = 120

180 + 120 = 300

300 for me
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Re: M01 #10 [#permalink] New post 01 Nov 2010, 07:38
redbeanaddict 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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

I'm confused with the solution given. Seems like there's something wrong with their formula.

[Edit] Also discussed here gmat-club-ps-69165.html

Its better to use the number of ways technique to fill in the spaces
for 3 digit numbers with a single 7
---
For example if 7 is in the Hundreds place then then calculate the number of ways in whuch the remaining spaces can be filled.Its a more structured way though laborious but then the question itself is cumbersome.
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Re: M01 #10 [#permalink] New post 02 Dec 2010, 03:52
The correct answer is C.

Instead of looking at 1 -> 1000, I break it down into 3 parts: one-digit numbers (1 -> 9), two-digit numbers (10 -> 99), three-digit numbers (100 -> 999). (1000 doesn't include any digit 7).

From 1 -> 9 : 1 integer (which is 7).

From 10 -> 99:
Let ab be any two-digit number (a is the tens digit and b is the units digit).
a : 1 -> 9 : 9 options (1 option = 7 and 8 options ≠ 7)
b : 0 -> 9 : 10 options (1 option = 7 and 9 options ≠ 7)

If a = 7, b ≠ 7, there are 1 x 9 = 9 integers (for example: 70, 71, 72...but not including 77)
If a = 7, b = 7, there is 1 x 1 = 1 integer (which is 77)
If a ≠ 7, b = 7, there are 8 x 1 = 8 integers (for example: 17, 27, 37..but not including 77)

So from 10 -> 99, there are 9 + 1 + 8 = 18 integers

From 100 -> 999:
Let abc be any three-digit number (a is the hundreds digit, b is the tens digit, and c is the units digit).
a : 1 -> 9 : 9 options (1 option = 7 and 8 options ≠ 7)
b : 0 -> 9 : 10 options (1 option = 7 and 9 options ≠ 7)
c : 0 -> 9 : 10 options (1 option = 7 and 9 options ≠ 7)

If a = 7, b ≠ 7, c ≠ 7, there are 1 x 9 x 9 = 81 integers
If a ≠ 7, b = 7, c ≠ 7, there are 8 x 1 x 9 = 72 integers
If a ≠ 7, b ≠ 7, c = 7, there are 8 x 9 x 1 = 72 integers

If a ≠ 7, b = 7, c = 7, there are 8 x 1 x 1 = 8 integers
If a = 7, b ≠ 7, c = 7, there are 1 x 9 x 1 = 9 integers
If a = 7, b = 7, c ≠ 7, there are 1 x 1 x 9 = 9 integers

If a = 7, b = 7, c = 7, there is 1 x 1 x 1 = 1 integer

So from 100 -> 999, there are 81 + 72 + 72 + 8 + 9 + 9 + 1 = 252 integers

In conclusion, the digit 7 is written 1 + 18 + 252 = 271 times when listing the integers from 1 to 1000.
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Re: M01 #10 [#permalink] New post 02 Dec 2010, 05:31
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The correct answer is C.

Instead of looking at 1 -> 1000, I break it down into 3 parts: one-digit numbers (1 -> 9), two-digit numbers (10 -> 99), three-digit numbers (100 -> 999). (1000 doesn't include any digit 7).

From 1 -> 9 : 1 integer (which is 7).

From 10 -> 99:
Let ab be any two-digit number (a is the tens digit and b is the units digit).
a : 1 -> 9 : 9 options (1 option = 7 and 8 options ≠ 7)
b : 0 -> 9 : 10 options (1 option = 7 and 9 options ≠ 7)

If a = 7, b ≠ 7, there are 1 x 9 = 9 integers (for example: 70, 71, 72...but not including 77)
If a = 7, b = 7, there is 1 x 1 = 1 integer (which is 77)
If a ≠ 7, b = 7, there are 8 x 1 = 8 integers (for example: 17, 27, 37..but not including 77)

So from 10 -> 99, there are 9 + 1 + 8 = 18 integers

From 100 -> 999:
Let abc be any three-digit number (a is the hundreds digit, b is the tens digit, and c is the units digit).
a : 1 -> 9 : 9 options (1 option = 7 and 8 options ≠ 7)
b : 0 -> 9 : 10 options (1 option = 7 and 9 options ≠ 7)
c : 0 -> 9 : 10 options (1 option = 7 and 9 options ≠ 7)

If a = 7, b ≠ 7, c ≠ 7, there are 1 x 9 x 9 = 81 integers
If a ≠ 7, b = 7, c ≠ 7, there are 8 x 1 x 9 = 72 integers
If a ≠ 7, b ≠ 7, c = 7, there are 8 x 9 x 1 = 72 integers

If a ≠ 7, b = 7, c = 7, there are 8 x 1 x 1 = 8 integers
If a = 7, b ≠ 7, c = 7, there are 1 x 9 x 1 = 9 integers
If a = 7, b = 7, c ≠ 7, there are 1 x 1 x 9 = 9 integers

If a = 7, b = 7, c = 7, there is 1 x 1 x 1 = 1 integer

So from 100 -> 999, there are 81 + 72 + 72 + 8 + 9 + 9 + 1 = 252 integers

In conclusion, the digit 7 is written 1 + 18 + 252 = 271 times when listing the integers from 1 to 1000.


OA for this question is D (300).

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

Consider numbers from 0 to 999 written as follows:
1. 000
2. 001
3. 002
4. 003
...
...
...
1000. 999

We have 1000 numbers. We used 3 digits per number, hence used total of 3*1000=3000 digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) 3000/10=300 times.

Answer: D (300).

Another approach:

In the range 0-100:
7 as units digit - 10 times (7, 17, 27, ..., 97);
7 as tens digit - 10 time (71, 72, 73, ..., 79);
So in first one hundred numbers 7 is written 10+10=20 times.

In 10 hundreds 7 as units or tens digit will be written 10*20=200 times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total 200+100=300.

Answer: D (300).

Hope it helps.
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Re: M01 #10 [#permalink] New post 03 Dec 2010, 18:51
Quote:
We have 1000 numbers. We used 3 digits per number, hence used total of 3*1000=3000 digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) 3000/10=300 times.

Answer: D (300).

Another approach:

In the range 0-100:
7 as units digit - 10 times (7, 17, 27, ..., 97);
7 as tens digit - 10 time (71, 72, 73, ..., 79);
So in first one hundred numbers 7 is written 10+10=20 times.

In 10 hundreds 7 as units or tens digit will be written 10*20=200 times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total 200+100=300.

Answer: D (300).

Hope it helps.


It can't be 300...I even wrote 1000 numbers (from 1 to 1000) and counted numbers which contain digit 7, it must be 271 numbers....( i used excel)
Quote:
7 as units digit - 10 times (7, 17, 27, ..., 97)
-> YES, it must be 10. So you included 77
Quote:
7 as tens digit - 10 times (71, 72, 73, ..., 79)
-> NO, it must be 9. You forgot to include 70 and you counted 77 twice here.
So in first one hundred numbers 7 is written 10+9=19 times.

Quote:
In 10 hundreds 7 as units or tens digit will be written 10*20=200 times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).
So you actually counted numbers from 700->799 twice. I agree there are 100 numbers from 700->799. If you count that, then in 8 hundreds 7 as units or tens digit will be written 8*19=152. For the hundreds digit, there are 10 numbers from 0->9, but if you include 0 and 7, then you count and numbers from 10->99 twice (if the hundreds digit is 0) and numbers from 700->799 twice (if the hundreds digit is 7).
Total: 10 + 9 + 100 + 152 = 271.
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Re: M01 #10 [#permalink] New post 04 Dec 2010, 01:30
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Quote:
We have 1000 numbers. We used 3 digits per number, hence used total of 3*1000=3000 digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) 3000/10=300 times.

Answer: D (300).

Another approach:

In the range 0-100:
7 as units digit - 10 times (7, 17, 27, ..., 97);
7 as tens digit - 10 time (71, 72, 73, ..., 79);
So in first one hundred numbers 7 is written 10+10=20 times.

In 10 hundreds 7 as units or tens digit will be written 10*20=200 times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total 200+100=300.

Answer: D (300).

Hope it helps.


It can't be 300...I even wrote 1000 numbers (from 1 to 1000) and counted numbers which contain digit 7, it must be 271 numbers....( i used excel)
Quote:
7 as units digit - 10 times (7, 17, 27, ..., 97)
-> YES, it must be 10. So you included 77
Quote:
7 as tens digit - 10 times (71, 72, 73, ..., 79)
-> NO, it must be 9. You forgot to include 70 and you counted 77 twice here.
So in first one hundred numbers 7 is written 10+9=19 times.

Quote:
In 10 hundreds 7 as units or tens digit will be written 10*20=200 times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).
So you actually counted numbers from 700->799 twice. I agree there are 100 numbers from 700->799. If you count that, then in 8 hundreds 7 as units or tens digit will be written 8*19=152. For the hundreds digit, there are 10 numbers from 0->9, but if you include 0 and 7, then you count and numbers from 10->99 twice (if the hundreds digit is 0) and numbers from 700->799 twice (if the hundreds digit is 7).
Total: 10 + 9 + 100 + 152 = 271.


Again: OA (Official Answer) for this question is D (300) and it's correct.

In the range 0-100:
7 as units digit - 10 times (7, 17, 27, ..., 97);
Here in 77, we counted only one 7 which is in units digit 77;

7 as tens digit - 10 times (71, 72, 73, ..., 79);
Here I haven't written 70, but haven't forgotten to count it: 70, 71, 72, 73, 74, 75, 76, 77, 78, and 79 - 10 numbers. We do not double count 77 here becous we count 7 which is now in tens digit 77;

So in first one hundred numbers 7 is written 10+10=20 times.

In 10 hundreds 7 as units or tens digit will be written 10*20=200 times.

Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799). Note that there is no double counting here for such number as 777 or 707, because for example we include 777 in 100 for it's 7 in hundreds digit.

Total 200+100=300.

Answer: D (300).

Hope it helps.
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Re: M01 #10 [#permalink] New post 06 Dec 2010, 01:51
Bunuel wrote:
phuongle wrote:
Quote:
We have 1000 numbers. We used 3 digits per number, hence used total of 3*1000=3000 digits. Now, why should ANY digit have preferences over another? We used each of 10 digits equal # of times, thus we used each digit (including 7) 3000/10=300 times.

Answer: D (300).

Another approach:

In the range 0-100:
7 as units digit - 10 times (7, 17, 27, ..., 97);
7 as tens digit - 10 time (71, 72, 73, ..., 79);
So in first one hundred numbers 7 is written 10+10=20 times.

In 10 hundreds 7 as units or tens digit will be written 10*20=200 times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).

Total 200+100=300.

Answer: D (300).

Hope it helps.


It can't be 300...I even wrote 1000 numbers (from 1 to 1000) and counted numbers which contain digit 7, it must be 271 numbers....( i used excel)
Quote:
7 as units digit - 10 times (7, 17, 27, ..., 97)
-> YES, it must be 10. So you included 77
Quote:
7 as tens digit - 10 times (71, 72, 73, ..., 79)
-> NO, it must be 9. You forgot to include 70 and you counted 77 twice here.
So in first one hundred numbers 7 is written 10+9=19 times.

Quote:
In 10 hundreds 7 as units or tens digit will be written 10*20=200 times. Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799).
So you actually counted numbers from 700->799 twice. I agree there are 100 numbers from 700->799. If you count that, then in 8 hundreds 7 as units or tens digit will be written 8*19=152. For the hundreds digit, there are 10 numbers from 0->9, but if you include 0 and 7, then you count and numbers from 10->99 twice (if the hundreds digit is 0) and numbers from 700->799 twice (if the hundreds digit is 7).
Total: 10 + 9 + 100 + 152 = 271.


Again: OA (Official Answer) for this question is D (300) and it's correct.

In the range 0-100:
7 as units digit - 10 times (7, 17, 27, ..., 97);
Here in 77, we counted only one 7 which is in units digit 77;

7 as tens digit - 10 times (71, 72, 73, ..., 79);
Here I haven't written 70, but haven't forgotten to count it: 70, 71, 72, 73, 74, 75, 76, 77, 78, and 79 - 10 numbers. We do not double count 77 here becous we count 7 which is now in tens digit 77;

So in first one hundred numbers 7 is written 10+10=20 times.

In 10 hundreds 7 as units or tens digit will be written 10*20=200 times.

Plus 100 times when 7 is written as hundreds digit (700, 701, 702, ..., 799). Note that there is no double counting here for such number as 777 or 707, because for example we include 777 in 100 for it's 7 in hundreds digit.

Total 200+100=300.

Answer: D (300).

Hope it helps.


I misunderstood the question. I tried to find the answer to the question "how many integers from 1 to 1000 contain digit 7?"
So you're right. It's 300.
Thanks.
Re: M01 #10   [#permalink] 06 Dec 2010, 01:51
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