What is the 120th digit when writing consecutive positive even integer

Question

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

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Official Answer

B

gmatophobia · Accepted Answer

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

WizakoVishwajit · Answer

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
 2468 101214161820222426283032343638404244 … … … …

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.
2468 10 12 14 16 18 20222426283032343638404244 … … … …
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 =   =   = 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 … …98 100 102 104 106 …

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

Ikemen · Answer

Note that each 100 numbers contains (100-2)/2 +1 =50 even numbers, for numbers between 2 to 100, we have 5 single digits, and 45 double digits thats (5)+(2)(45)=95 digits. 120-95 = 25 digits remianing, we know that for each number between 102 and 200 there are three digits, so 25/3 mod 1, therefore we are back to the first digit of three digits numbers. Hence B.

VibhuAnurag · Answer

4 single digit numbers --> 2, 4, 6, 8
From 10 to 98 --> (98 - 10)/2 + 1 = (88/2) + 1 = 44 + 1 = 45 numbers = 45 * 2 = 90 digits

Now, we need another (120 - 4 - 90 = 26 digits) to be covered by 3 digit numbers.

26/3 = 8 numbers and then the 2nd digit of the 9th number --> 100, 102, ..., 116 is the 9th number
--> The 120th digit is 1.

The answer is B.

What is the 120th digit when writing consecutive positive even integer

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What is the 120th digit when writing consecutive positive even integer

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