-4
Answer is 199.
The set of three-digit numbers starts at 000 and goes to 999. So we have to find each number with at least one "7" in this set.
In the first hundred (000 to 099), instances of one or more 7s occur 11 times, for:
007
017
027
037
047
057
067
070
077
087
097
This same sequence happens for the 100s, 200s, 300s, 400s, 500, 600s, 800, and 900s (notice I left out 700s). So far, this totals 11 times 9 = 99 instances.
Then we add the number in 700 through 799. Every one of these has at least one digit of 7, totaling 100 numbers.
So, in total, there are 11 * 9 + 100 = 199 instances.
000-099 has 11
100-199 has 11
200-299 has 11
300-399 has 11
400-499 has 11
500-599 has 11
600-699 has 11
700-799 has 100
800-899 has 11
900-999 has 11
-------------------
TOTAL = 199
2331
use the digit 1234to make two different numbers
The number 8 in that question is the number with the least amount of value (0.008).
If those are two-digit numbers, the answer is 77, 82, 86, 86, 86, 89, 93, 95, 95, 96, 98
.8
It is 9
what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?
100,000
Greatest: 989949 Least: 100000
It is: 1013
0
2000.The smallest four-digit number there can be is 1000. So, add 1000 and 1000 to get 2000, then least sum you can get when adding two four-digit numbers.
252
-98
21
Zero
how does 100 meet the specification? 1 is in the hundreds digit and 0 is in the tens digit. 1 is not at least 0. this is a confusing question?!
The positional place values of digits in negative numbers are in ascending order from least to greatest as for example in the number -987 the least value digit is 9 and the greatest value digit is 7 because -900 < -80 < -7 The positional place values of digits in positive numbers are in descending order from greatest to least as for example in the number 987 the least value digit is 7 and the greatest value digit is 9 because 900 > 80 > 7