HOW MANY TIMES DOES THE DIGIT 8 OCCUR BETWEEN 1 AND 100?

Digit Times

10 numbers end in 8 (8, 18, … 98)
10 more begin with 8 (80, 81, … 89)
Total number of 8’s is 20.

There are 9 1-digit numbers, and 90 2-digit numbers, for a total of 189 digits used to write all the numbers from 1 to 99. Each non-zero digit appears 20 times (for 9×20=180). The logic is identical to that above.
Zero appears only at the end of 10, 20 … 90 (for 9 times).

180 + 9 = 189, the total number of digits.

It’s helpful to count up all the cases for all the digits, to make sure you account for everything. That’s a good way to check your results.

19

19 times

Source(s): my head

It occurs once in every group of ten numbers plus in the eighties it appears an additional 10 times as the leading digit, so 10+10=20 times….

*really….

Answer 6

20
begin with 8 10
ending with 8 10

Answer 7

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20 times 8 , {8,18,28,38,48,58,68,78,80,81,82,83,84,85,86,87,89,98} 18 times +{88} 2 times; same goes for 2,3,4,5,6,7,9 s occurrences
11 times 0 , {10,20,30,40,50,60,70,80,90} 9 times +{100} 2 times;
21 times 1 , {1,10,[11],12,13,14,15,16,17,18,19,21,31,41,51,61,71,81,91,100}

Well you have 8,18,28,38,48,58,68,78,80,81,82,83,84,85,86,87,88,89,98 so it total there are19 if I’m counting correctly

use that noggin of yours

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