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# M01-10

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Intern
Joined: 03 Sep 2017
Posts: 18
Location: Brazil
GMAT 1: 730 Q49 V41

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21 May 2018, 09:25
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

Many approaches are possible. For example:

Approach #1:

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) $$\frac{3000}{10}=300$$ times.

Hello Bunuel

Could you kindly explain why did you consider the interval 0 to 999, when the question was about the interval from 1 to 1000?
I understood the ''no favorites'' approach, and i think its a really clever way of solving it. But the interval change got me confused.

Thank you,
Math Expert
Joined: 02 Sep 2009
Posts: 59588

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21 May 2018, 21:28
vitorcbarbieri 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

Many approaches are possible. For example:

Approach #1:

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) $$\frac{3000}{10}=300$$ times.

Hello Bunuel

Could you kindly explain why did you consider the interval 0 to 999, when the question was about the interval from 1 to 1000?
I understood the ''no favorites'' approach, and i think its a really clever way of solving it. But the interval change got me confused.

Thank you,

Since neither 0 not 1000 has 7 in it, then excluding/including these values into the range won't change the answer.
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Intern
Joined: 22 Sep 2016
Posts: 46
Location: India
GMAT 1: 680 Q44 V39

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03 May 2019, 22:15
Bunuel chetan2u

The answer will be 300 for all the digits except 0 and 1, right? For 0, the answer will be 303 (Considering 3 zeroes in 1000) and for 1, the answer will be 301 (Considering one 1 in 1000).

Let me know if my understanding is correct?
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Best,
Spiritual Yoda
Math Expert
Joined: 02 Aug 2009
Posts: 8284

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03 May 2019, 23:38
1
SpiritualYoda wrote:
Bunuel chetan2u

The answer will be 300 for all the digits except 0 and 1, right? For 0, the answer will be 303 (Considering 3 zeroes in 1000) and for 1, the answer will be 301 (Considering one 1 in 1000).

Let me know if my understanding is correct?

Hi..
if we write all as 000 to 999, we will have 3000 digits as 3(number of digits in each number )*1000(total numbers)=3000..
In these 3000, all 10 digits will be 3000/10=300..

But we do not have 0s in single digit and two digits..
so 000,as we do not have 0, so 3 0s missing
00a.. from 1 to 9, they are not written as 001,002...009, so 2*9=18 x 0s missing
0ab.....from 10 to 99, they are not written as 010,011...099, so 1*90=90 x 0s missing
Total 3+18+90=111 number of 0s are less..

Thus, from 1 to 999, we have 300 of all 9 digits - 1 to 9, while we have 300-111=189 of 0s.
digit 0 is used 189+3=192 times
digit 1 is used 300+1=301 times
digits 2 to 9, 300 times each

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Joined: 14 Feb 2017
Posts: 1314
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
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29 May 2019, 16:25
Hey Bunuel, what is meant by "In 10 hundreds 7" in approach 3? I'm having a hard time understanding this.
Intern
Joined: 22 Dec 2018
Posts: 20
WE: Medicine and Health (Health Care)

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11 Sep 2019, 11:09
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

Many approaches are possible. For example:

Approach #1:

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) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the tens digit or the units digit. Case 1: 7 is the tens digit. There are 10 ways to place 7 as the tens digit of a two-digit number. Case 2: There are 9 ways to place the units digit. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

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 $$200+100=300$$.

Hi Bunuel,

In approach # 1, i didn't quite understand why did we use 3 digit numbers.
Apologies if it's a silly question.
Math Expert
Joined: 02 Sep 2009
Posts: 59588

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17 Sep 2019, 05:54
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

Many approaches are possible. For example:

Approach #1:

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) $$\frac{3000}{10}=300$$ times.

Approach #2:

There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.

One-digit numbers: 7 is the only one-digit number.

Two-digit numbers: 7 could be the tens digit or the units digit. Case 1: 7 is the tens digit. There are 10 ways to place 7 as the tens digit of a two-digit number. Case 2: There are 9 ways to place the units digit. Thus, for two-digit numbers we have: $$10+9=19$$ numbers that contain a 7.

Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.

Approach #3:

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 $$200+100=300$$.

Hi Bunuel,

In approach # 1, i didn't quite understand why did we use 3 digit numbers.
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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Joined: 11 Sep 2019
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09 Nov 2019, 05:48
Bunuel I solved with this approach. Please correct me if I am wrong.

we can not use thousands digit as 7, because of 7XXX > 1000
So we have only 3 places i.e Units, Tens, Hundreds place, where we can put 7 as a digit.
Also we have total 10 digits (0-9) to choose from.
Now, in the no 0XXX,

1) If 7 is at units place i.e 0XX7
Tens digit can be any one of the 10 available digits (0-9)
Hence Tens place can be placed in 10 different ways.
Similarly, Hundreds place in 10 different ways
So, the no. 0XX7 can be formed using 10*10 = 100 different ways

(2) Exactly as mentioned in (1), the no. 0X7X can be formed using 100 different ways.

(3) And hence the no. 07XX in 100 different ways.

So total No. of times the 7 was appeared between 0 to 1000 is (1)+(2)+(3) i.e. 100+100+100=300
M01-10   [#permalink] 09 Nov 2019, 05:48

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# M01-10

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