Math

(a) The numbers from 1 to 99 include 9 one-digit numbers (1 to 9), 90 two-digit numbers (10 to 99), and a total of 99 numbers. The one-digit numbers have 1 digit each, the two-digit numbers have 2 digits each. Therefore, the total number of digits is:

9 × 1 + 90 × 2 = 9 + 180 = 189

Sam wrote down 189 digits in total.

(b) There are no 0's among the one-digit numbers. Among the two-digit numbers, there are 10 numbers that have 0 as the second digit (10, 20, 30, ..., 90). Therefore, out of the 99 numbers Sam wrote down, 10 have 0 as the second digit. The probability of choosing one of these numbers is:

P(choosing 0) = 10/99

(c) We can calculate the sum of the digits by adding the sum of the digits in each of the numbers. There are 9 one-digit numbers, which add up to 45. The two-digit numbers can be divided into two groups: those whose first digit is 1 (10 to 19) and those whose first digit is 2 to 9 (20 to 99). The sum of the digits in the first group is:

(1 + 0) + (1 + 1) + (1 + 2) + ... + (1 + 9) = 55

To calculate the sum of the digits in the second group, we first notice that there are 8 numbers whose first digit is 2, and each of these numbers contributes 2 to the sum of the digits. Similarly, there are 8 numbers whose first digit is 3, and each of these numbers contributes 3 to the sum of the digits, and so on up to 9 numbers whose first digit is 9, and each of these numbers contributes 9 to the sum of the digits. Therefore, the sum of the digits in the second group is:

8 × 2 + 8 × 3 + ... + 9 × 9 = 2(8 + 9) + 3(8 + 9) + ... + 9(8 + 9) = (8 + 9)(2 + 3 + ... + 9) = 45 × 44/2 = 990

Therefore, the total sum of the digits that Sam wrote down is:

45 + 55 + 990 = 1090

C)

 

n=1;a=99;s=0;m=n;p=0;n=m;cycle:s=s+(n%10);p=p+1;n=(n div 10);if(n!=0, goto cycle,0);m=m+1;if(m<=a, goto5,0);print"Total Sum =",s;print"Total Num =",p

 

 

Total Sum = 900
Total Num = 189