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# The integer n is formed by writing the integers 1 through 3

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Director
Joined: 29 Nov 2012
Posts: 682
The integer n is formed by writing the integers 1 through 3  [#permalink]

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Updated on: 19 Jun 2013, 07:29
2
7
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Difficulty:

95% (hard)

Question Stats:

47% (02:24) correct 53% (02:23) wrong based on 125 sessions

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The integer n is formed by writing the integers 1 through 333 in order as shown below.

n=123456789101112.......331332333

counting from the left, what is the 333rd digit of n?

A. 0
B. 1
C. 3
D. 7
E. 8

Originally posted by fozzzy on 18 Jun 2013, 23:25.
Last edited by fozzzy on 19 Jun 2013, 07:29, edited 1 time in total.
Intern
Joined: 12 Jun 2013
Posts: 1
Re: The integer n is formed by writing the integers 1 through 3  [#permalink]

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19 Jun 2013, 07:36
6
3
Here's how I went about solving this problem:

Single digit integers: 1-9 => 9

Double digit integers: 10-99 => 90 * 2 => 180

99 is the (180 +9 =) 189th digit

333 - 189 = 144 digits are left to go

Now you are left with only 3 digit integers from here on so you can divide the 144 by 3 to see how many 3 digit numbers there are (this starts at 100 and moves on:

144 / 3 = 48

Then:

99 + 48 = 147

Thus the 333 digit is the last digit in the number, 7

##### General Discussion
Director
Joined: 14 Dec 2012
Posts: 679
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34
GPA: 3.6
Re: The integer n is formbed by writing the integers 1 through 3  [#permalink]

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19 Jun 2013, 01:01
1
1
fozzzy wrote:
The integer n is formed by writing the integers 1 through 333 in order as shown below.

$$n=123456789101112.......331332333$$

counting from the left, what is the 333rd digit of n?

a) 0
b) 1
c) 3
d) 7
e) 8

Any alternative solution

1-9(9 digits)
now 10-19,20-29,.....,90-99....each set have 10 numbers with 2 digit....and there are 9 such sets=>therefore total digit in each set will be 10*2=20
and total digit in all 9 sets will be=20*9=180
uptil now we reached up to 180+9=189th digit.
we have been asked 333rd digit therefore remaining digits=333-189=144
now 100-109,110-119,120-129,130-139,140-149==>we have five sets having 10 numbers and each number has 3 digits==>therefore total no. of digit from 100-149 will be equal to 5*3*10=150
so 123456..........99100101....149==>1to 99 we came to 189 th digit therefore 149 will reach up to 339th digit
123456......100101....147148149=>9 is the 339th digit so by counting back 7 will be the 333rd digit.

i know thw explanation seems to be but when you will think its not so big.
press kudos if it helped.
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Director
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Posts: 682
Re: The integer n is formbed by writing the integers 1 through 3  [#permalink]

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19 Jun 2013, 01:09
I had the same solution in mind was wondering if there is any other way to solve this question / any faster way.
Director
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Posts: 679
Location: India
Concentration: General Management, Operations
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Re: The integer n is formbed by writing the integers 1 through 3  [#permalink]

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19 Jun 2013, 01:35
1
fozzzy wrote:
I had the same solution in mind was wondering if there is any other way to solve this question / any faster way.

the other way you can approach this is:
123456789....
make a pair of 3 by making each number a 3 digit number
000
001
002
003
004
005
006
007
008
009
010.... and so on
....by doing this we can make that every other 30th digit will be a 9
and for doing this we are adding 1*3+2*9+1*90=112 zeroes...therefore 333rd digit now will be 333+111=444 th digit
now 30*15=450....therefore 450th digit will be 9
and 15 th series will be of 141,142....149
so ........147148149.....in this 9 is the 450th digit therefore 444 th will be 7
hope it helps.
_________________
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Director
Joined: 29 Nov 2012
Posts: 682
Re: The integer n is formbed by writing the integers 1 through 3  [#permalink]

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19 Jun 2013, 07:18
shaileshmishra wrote:
fozzzy wrote:
I had the same solution in mind was wondering if there is any other way to solve this question / any faster way.

the other way you can approach this is:
123456789....
make a pair of 3 by making each number a 3 digit number
000
001
002
003
004
005
006
007
008
009
010.... and so on
....by doing this we can make that every other 30th digit will be a 9
and for doing this we are adding 1*3+2*9+1*90=112 zeroes...therefore 333rd digit now will be 333+111=444 th digit
now 30*15=450....therefore 450th digit will be 9
and 15 th series will be of 141,142....149
so ........147148149.....in this 9 is the 450th digit therefore 444 th will be 7
hope it helps.

Great approach! +1 kudos
Manager
Joined: 14 Sep 2014
Posts: 86
WE: Engineering (Consulting)
Re: The integer n is formed by writing the integers 1 through 3  [#permalink]

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26 Nov 2014, 21:09
1-9 = 9 nos. with 1 digit
10 - 99 = 90 nos. with 2 digits so total 180 digits
that means from 1 - 99 there are 189 digits
now remaining 333 -180 = 144 digits
all numbers [100-999] will have 3 digits so 1
44/3 = 48th number after 99
which is 147 will be 331 332 and 333rd digit
so 7 is ans.
Senior Manager
Joined: 03 Apr 2013
Posts: 263
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
Re: The integer n is formed by writing the integers 1 through 3  [#permalink]

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25 Jun 2016, 07:02
fozzzy wrote:
The integer n is formed by writing the integers 1 through 333 in order as shown below.

n=123456789101112.......331332333

counting from the left, what is the 333rd digit of n?

A. 0
B. 1
C. 3
D. 7
E. 8

Fast way to solve this..

We need the 333rd digit in this series..

Let's count a few things first..

1. Single Digit Numbers

Numbers = 9 , Digits per Number = 1

Total Digits = 9

2. Double Digit Numbers

Numbers = 90(calculated as 99-10+1), Digits per Number = 2

Total Digits = $$90*2=180$$

So how many digits have we come so far?

180+9=189

How much more do we need to go?

333-189=144 digits

Now, as we know that we will have all 3-Digit Numbers with us, in other words, 3 digits will make up 1 number.

$$144/3 = 48$$

which means we are looking at the 48th 3-Digit Number, moreover, as the division didn't leave a remainder, we are looking at the last digit of this 48th 3-Digit Number.

48th 3-digit number = 147, last digit = 7, and Voila!
VP
Joined: 07 Dec 2014
Posts: 1230
The integer n is formed by writing the integers 1 through 3  [#permalink]

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25 Jun 2016, 17:29
numbers 1-100 contain 10+9*20=190 digits
numbers 1-200 contain 190+10*30=490 digits
(190+490)/2=340 digits
100+200/2=150 numbers
the 340th digit is the digit 1 in number 150
working backwards 7 digits from digit 340,
→9→4→1→8→4→1→7,
the 333rd digit is the digit 7 in number 147
Intern
Joined: 04 Feb 2018
Posts: 46
The integer n is formed by writing the integers 1 through 3  [#permalink]

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14 Oct 2019, 14:20
Number of single digits = 1-9 = 9
Number of 2 digits 10-99 = 99-10+1=90x2 = 180 digits
Total digits so far = 9+189=180
Balance from 333 = 333-180 = 144
The 333rd digit will be found among the 3-digit integers and based on the analysis so far it is obvious that the 3 digit integers have a total of 144 digits.
Number of 3 digit integers = (x-100+1) x 3=144
where x represents the 3 digit integer containing the 333rd digit and the 333rd digit is equal to the units digit of x.
x-99 = 144/3=48
x = 48+99=147
The units digit of 147 is 7 implying that the 333rd digit must be 7.
The integer n is formed by writing the integers 1 through 3   [#permalink] 14 Oct 2019, 14:20
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