To determine how many times the digit 7 appears between 1 and 885, we must consider each place value separately. In the units place (1-9), there is one 7. In the tens place (10-99), there are 10 occurrences of 7 (17, 27, 37, 47, 57, 67, 77, 87, 97). In the hundreds place (100-199, 200-299, ..., 800-885), there are 100 occurrences of 7. Therefore, the total number of 7s between 1 and 885 is 1 + 10 + 100 = 111.
6 of them with a remainder of 1
The sixes between 1 and 100 are:6162636465660616263646566676869768696Therefore, there are 20 digit sixes between 1 and 100.
There are 98 numbers are between 1 and 100 that are divisible by five.
There are an infinity of numbers between 1 and 100 by way of fractions or decimals. The normal answer is 100. Between includes the first and last number.
To determine how many times the digit 7 appears between 1 and 885, we must consider each place value separately. In the units place (1-9), there is one 7. In the tens place (10-99), there are 10 occurrences of 7 (17, 27, 37, 47, 57, 67, 77, 87, 97). In the hundreds place (100-199, 200-299, ..., 800-885), there are 100 occurrences of 7. Therefore, the total number of 7s between 1 and 885 is 1 + 10 + 100 = 111.
7,17,27,37,47,57,67,70,71,72,73,74,75,76,77(two sevens),78,79,87,97=20 times
6 of them with a remainder of 1
101
25 numbers are between 1-100.
100
The sixes between 1 and 100 are:6162636465660616263646566676869768696Therefore, there are 20 digit sixes between 1 and 100.
Not including 1 and 100, 99
There are 90 two-digit whole numbers between 1 and 100.
The only consecutive primes between 1 and 100 are 2 and 3.
There are 98 numbers are between 1 and 100 that are divisible by five.
Since 100, 99 & 98 have no sevens we can limit our search to 1 - 97. There will be ten 7's in the ones column and one 7 in the tens column. So it looks like eleven 7's is the answer.