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21 Mar 2024, 21:36
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:15) correct
41%
(02:27)
wrong
based on 242
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Time
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Not Attempted Yet
What is the 120th digit when writing consecutive positive even integers starting from 2 onwards?
A) 0
B) 1
C) 2
D) 3
E) 4
A) 0
B) 1
C) 2
D) 3
E) 4
Source: http://www.GMATinsight.com
(Inspired from Official GMAT FOCUS Question)---
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New Addition: Video on Everything about Data Insights Section
(Inspired from Official GMAT FOCUS Question)---
GMATinsight (4.9/5 star rating on Google)
Providing Focused GMAT Prep (Online and Offline) for GMAT Focus along with 100% successful Admissions counselling
http://www.GMATinsight.com | http://www.AdmitSure.com
http://www.Youtube.com/GMATinsight (LIKE and SUBSCRIBE the channel for 1100+ topic-wise sorted Videos)
New Addition: Video on Everything about Data Insights Section
22 Mar 2024, 23:23
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If we go on writing even positive integers one after the other in ascending order we will get a number that looks like -
2468101214161820222426283032343638404244 … … … …
We need to identify 120th digit of this number.
If we observe, numbers from 2 to 8 (2, 4, 6 and 8) they are single digit even numbers. While numbers from 10 to 98 are two-digit even numbers. Means, each two-digit even number will account for two digits in the given number and further digits will be given by 3-digit even numbers and so on.
Hence, in the given number
Digits occupied by 1-digit even numbers =4
2468101214161820222426283032343638404244 … … … …
Out of 120 digits to be counted, 4 are by single digit numbers. Remaining 116 digits will be counted using 2-digit even and / or 3-digit numbers.
As each 2-digit even number accounts for 2 digits and we can see that from 10 to 98 there are 45 two-digit even numbers.
2468101214161820222426283032343638404244 … … … …
10, 12, 14, 16, 18 … …. This is an Arithmetic progression
Here, First Term (a) = 10, Last term = 98 Common difference (d) = 2
Number of terms = [ (Last Term - First Term)/2) + 1] = [ (98-10)/2 +1] = 44 +1 = 45
These 45 2-digit numbers account for 90 digits.
Out of 116 remaining digits 90 are occupied by two-digit even numbers. Hence, Digits remaining to be counted = 116 - 90 = 26
These 26 digits will be counted now by counting digits of 3-digit even numbers.
2468101214161820222426283032343638404244 … …98100102104106 …
As each 3-digit even number will occupy 3 digits and we need to count 26th digit starting from 100
We divide 26 by 3.
We get, 26 = 3 x 8 +2
This means first 8 3-digit even numbers will account for 24 digits and middle digit of 9th 3-digit number (second digit) will be the 120th digit (answer required) of the number. Three-digit numbers in order are
100, 102, 104, 106, 108, 110, 112, 114, 116
We can see that 9th 3-digit even number is 116 and its second digit = 1.
Hence, 120th digit of the given number = 1
2468101214161820222426283032343638404244 … … … …
We need to identify 120th digit of this number.
If we observe, numbers from 2 to 8 (2, 4, 6 and 8) they are single digit even numbers. While numbers from 10 to 98 are two-digit even numbers. Means, each two-digit even number will account for two digits in the given number and further digits will be given by 3-digit even numbers and so on.
Hence, in the given number
Digits occupied by 1-digit even numbers =4
2468101214161820222426283032343638404244 … … … …
Out of 120 digits to be counted, 4 are by single digit numbers. Remaining 116 digits will be counted using 2-digit even and / or 3-digit numbers.
As each 2-digit even number accounts for 2 digits and we can see that from 10 to 98 there are 45 two-digit even numbers.
2468101214161820222426283032343638404244 … … … …
10, 12, 14, 16, 18 … …. This is an Arithmetic progression
Here, First Term (a) = 10, Last term = 98 Common difference (d) = 2
Number of terms = [ (Last Term - First Term)/2) + 1] = [ (98-10)/2 +1] = 44 +1 = 45
These 45 2-digit numbers account for 90 digits.
Out of 116 remaining digits 90 are occupied by two-digit even numbers. Hence, Digits remaining to be counted = 116 - 90 = 26
These 26 digits will be counted now by counting digits of 3-digit even numbers.
2468101214161820222426283032343638404244 … …98100102104106 …
As each 3-digit even number will occupy 3 digits and we need to count 26th digit starting from 100
We divide 26 by 3.
We get, 26 = 3 x 8 +2
This means first 8 3-digit even numbers will account for 24 digits and middle digit of 9th 3-digit number (second digit) will be the 120th digit (answer required) of the number. Three-digit numbers in order are
100, 102, 104, 106, 108, 110, 112, 114, 116
We can see that 9th 3-digit even number is 116 and its second digit = 1.
Hence, 120th digit of the given number = 1
23 Mar 2024, 00:53
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GMATinsight
Single Digit Even Numbers = 2, 4, 6, 8 ⇒ 4 Nos.
Number of digits = 1 * 4 = 4
2-Digit Even Numbers
Between 10 to 19, we have the following even numbers ⇒ 10, 12, 14, 16, and 18
Number of digits = 2 * 5 = 10
Between 10 and 99, we have 9 such sets.
Number of digits = 10 * 9 = 90
3-Digit Even Numbers
Between 100 to 109, we have the following even numbers ⇒ 100, 102, 104, 106, and 108
Number of digits = 3 * 5 = 15
Number of digits so far = 4 + 90 + 15 = 109
Remaining = 120 - 109 = 11
\(110 \quad 112 \quad 114 \quad 116 \)
1 in 116 is the 120th digit
Option B
