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How many times will the digit 7 be written when listing the [#permalink]

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12 Jul 2010, 21:12

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What is the best approach to solve problems such as these two examples below. If someone can solve them and show me the steps they took, I would appreciate it greatly. The GMAT Club test explanations were a little difficult for me to digest.

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

"How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit?"

What is the best approach to solve problems such as these two examples below. If someone can solve them and show me the steps they took, I would appreciate it greatly. The GMAT Club test explanations were a little difficult for me to digest.

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

"How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit?"

Answer: 1339

Thanks Club!

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

Many approaches are possible. For example:

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: 300.

2. How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit?

There is one number in hundred with 2 in th tens digit and 1 in the units digit: 21, 121, 221, 321, ...

The difference between 324,700 and 458,600 is 458,600-324,700=133,900 - one number per each hundred gives 133,900/100=1,339 numbers.

Schools: University of Texas at Austin, Michigan State

Re: Digits Problem Difficulty in GMAT Club test 1 [#permalink]

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14 Jul 2010, 21:57

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When we write the numbers 1 to 1000 do we use 3 digits for each number?

What about: 0 1 2 3 4 5 6 7 8 9

and

10...99

The reason it works out is because we don't use zero later

so you could think of 1 as 001 and 10 as 010

However, if the question changed to how many 7's from 100 to 1000.

Then you have 900 numbers with 3 digits = 3*900 = 2700 total digits

2700 / 10 = 270

when we actually have 280 7's from 100 to 1000. right?

I came up with the same answer for the posted question 300, by counting the 7's in different groups

1 to 9= Only one 7 (7) 10 to 99= 19 (17,27,37,47,57,67,87,97 and 70, 71, 72, 73, 74, 75, 76, 78, 79 and 77 (2 sevens)) then I knew from 100 to 1000 these 20 sevens would repeat 9 more times plus all the 7's in the hundreds digits 100 sevens from 700 to 799

1 (7) +19 (tens and ones digits from 10 to 99) +180 (tens and ones digits from 100 to 1000) +100 (hundreds digits) =300

When we write the numbers 1 to 1000 do we use 3 digits for each number?

What about: 0 1 2 3 4 5 6 7 8 9

and

10...99

The reason it works out is because we don't use zero later

so you could think of 1 as 001 and 10 as 010

However, if the question changed to how many 7's from 100 to 1000.

Then you have 900 numbers with 3 digits = 3*900 = 2700 total digits

2700 / 10 = 270

when we actually have 280 7's from 100 to 1000. right?

I came up with the same answer for the posted question 300, by counting the 7's in different groups

1 to 9= Only one 7 (7) 10 to 99= 19 (17,27,37,47,57,67,87,97 and 70, 71, 72, 73, 74, 75, 76, 78, 79 and 77 (2 sevens)) then I knew from 100 to 1000 these 20 sevens would repeat 9 more times plus all the 7's in the hundreds digits 100 sevens from 700 to 799

1 (7) +19 (tens and ones digits from 10 to 99) +180 (tens and ones digits from 100 to 1000) +100 (hundreds digits) =300

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.

Re: Digits Problem Difficulty in GMAT Club test 1 [#permalink]

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14 Jan 2012, 22:44

Each hundred can have 1 such number with unit digit as 1 and Tenth digit as 2, like 21, 121, 321 so we need to find number of hundred between 2 numbers, Answer 1339.

Re: Digits Problem Difficulty in GMAT Club test 1 [#permalink]

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10 Aug 2013, 19:17

Bunuel wrote:

TallJTinChina wrote:

When we write the numbers 1 to 1000 do we use 3 digits for each number?

What about: 0 1 2 3 4 5 6 7 8 9

and

10...99

The reason it works out is because we don't use zero later

so you could think of 1 as 001 and 10 as 010

However, if the question changed to how many 7's from 100 to 1000.

Then you have 900 numbers with 3 digits = 3*900 = 2700 total digits

2700 / 10 = 270

when we actually have 280 7's from 100 to 1000. right?

I came up with the same answer for the posted question 300, by counting the 7's in different groups

1 to 9= Only one 7 (7) 10 to 99= 19 (17,27,37,47,57,67,87,97 and 70, 71, 72, 73, 74, 75, 76, 78, 79 and 77 (2 sevens)) then I knew from 100 to 1000 these 20 sevens would repeat 9 more times plus all the 7's in the hundreds digits 100 sevens from 700 to 799

1 (7) +19 (tens and ones digits from 10 to 99) +180 (tens and ones digits from 100 to 1000) +100 (hundreds digits) =300

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.

Are the numbers are not getting repeated in the above three ranges? For example , the no 777 will be part of all the three ranges above and is being counted thrice.

Re: Digits Problem Difficulty in GMAT Club test 1 [#permalink]

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11 Aug 2013, 15:10

Bunuel wrote:

tonebeeze wrote:

What is the best approach to solve problems such as these two examples below. If someone can solve them and show me the steps they took, I would appreciate it greatly. The GMAT Club test explanations were a little difficult for me to digest.

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

"How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit?"

Answer: 1339

Thanks Club!

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

Many approaches are possible. For example:

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: 300.

2. How many integers between 324,700 and 458,600 have a 2 in the tens digit and a 1 in the units digit?

There is one number in hundred with 2 in th tens digit and 1 in the units digit: 21, 121, 221, 321, ...

The difference between 324,700 and 458,600 is 458,600-324,700=133,900 - one number per each hundred gives 133,900/100=1,339 numbers.

i just have one question and its about this line "Now, why should ANY digit have preferences over another?" . Do you think from 1 to 1000, 1 and 0 haven't a little priority over others in aspects of their number ? i am telling because of especially the last number that is 1000 . it's not a 3 digit number but still here....... 7 was not here that's why couldn't make problem.....

By the way, one thing to tell and it's day by day i am learning lot from your uploaded files...............thanks to you again bunnel.....
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Re: How many times will the digit 7 be written when listing the [#permalink]

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15 Aug 2014, 16:50

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Re: How many times will the digit 7 be written when listing the [#permalink]

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07 Feb 2016, 01:07

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Re: How many times will the digit 7 be written when listing the [#permalink]

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19 Jul 2017, 19:20

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: How many times will the digit 7 be written when listing the [#permalink]

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28 Aug 2017, 04:30

keenys wrote:

Bunuel wrote:

TallJTinChina wrote:

When we write the numbers 1 to 1000 do we use 3 digits for each number?

What about: 0 1 2 3 4 5 6 7 8 9

and

10...99

The reason it works out is because we don't use zero later

so you could think of 1 as 001 and 10 as 010

However, if the question changed to how many 7's from 100 to 1000.

Then you have 900 numbers with 3 digits = 3*900 = 2700 total digits

2700 / 10 = 270

when we actually have 280 7's from 100 to 1000. right?

I came up with the same answer for the posted question 300, by counting the 7's in different groups

1 to 9= Only one 7 (7) 10 to 99= 19 (17,27,37,47,57,67,87,97 and 70, 71, 72, 73, 74, 75, 76, 78, 79 and 77 (2 sevens)) then I knew from 100 to 1000 these 20 sevens would repeat 9 more times plus all the 7's in the hundreds digits 100 sevens from 700 to 799

1 (7) +19 (tens and ones digits from 10 to 99) +180 (tens and ones digits from 100 to 1000) +100 (hundreds digits) =300

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.

Are the numbers are not getting repeated in the above three ranges? For example , the no 777 will be part of all the three ranges above and is being counted thrice.

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