To find the total number of digits used in numbers from 51 to 5001, we can break it down into two parts. Numbers from 51 to 99: Each number in this range has 2 digits, so there are 49 numbers in total, resulting in 49 x 2 = 98 digits. Numbers from 100 to 5001: Each number in this range has 3, 4, or 5 digits. a. For numbers from 100 to 999, each number has 3 digits, so there are 900 numbers in total, resulting in 900 x 3 = 2700 digits. b. For numbers from 1000 to 5001, each number has 4 or 5 digits. There are 4002 numbers in total, and if we assume each has 4 digits, it would be 4002 x 4 = 16008 digits. Adding all these together, the total number of digits used in numbers from 51 to 5001 is 98 + 2700 + 16008 = 18706 digits.
Add the digits of the number. If that total is divisible by nine, so is the original number. Example: 12,345,678 The digits total 36. 36 is divisible by 9. So is 12,345,678.
It is 120 if the digits cannot be repeated.
If the last two digits of a number are divisible by 4, the whole number is divisible by 4. If the sum total of the digits in a number are a multiple of 9, the whole number is a multiple of 9.
It you add up all the digit in the number they will total 11. For example the number 425 has digits that sum to 11. (4 + 2 + 5 = 11)
Total number of digits = 5
To find the total number of digits used in numbers from 51 to 5001, we can break it down into two parts. Numbers from 51 to 99: Each number in this range has 2 digits, so there are 49 numbers in total, resulting in 49 x 2 = 98 digits. Numbers from 100 to 5001: Each number in this range has 3, 4, or 5 digits. a. For numbers from 100 to 999, each number has 3 digits, so there are 900 numbers in total, resulting in 900 x 3 = 2700 digits. b. For numbers from 1000 to 5001, each number has 4 or 5 digits. There are 4002 numbers in total, and if we assume each has 4 digits, it would be 4002 x 4 = 16008 digits. Adding all these together, the total number of digits used in numbers from 51 to 5001 is 98 + 2700 + 16008 = 18706 digits.
#include<stdio.h> main () { int number, last_digit, next_digit, total; printf ("Enter the number whose sum of digits is to be calculated: "); scanf ("%d", &number); last_digit = number%10; total = last_digit; next_digit = (number/10) % 10; total = total + next_digit; next_digit = (number/100) % 10; total = total + next_digit; next_digit = (number/1000) %10; total = total + next_digit; next_digit = (number/10000) %10; total = total + next_digit; printf ("The sum of the digits of the entered number is: %d", total); }
I/you/we am 49
The sum of your digits is the total number arrived at after adding two or more numbers.
Add the digits of the number. If that total is divisible by nine, so is the original number. Example: 12,345,678 The digits total 36. 36 is divisible by 9. So is 12,345,678.
Add the digits of the number. If that total is divisible by nine, so is the original number. Example: 12,345,678 The digits total 36. 36 is divisible by 9. So is 12,345,678.
No.
It is 120 if the digits cannot be repeated.
The number seventy billion is written 70,000,000,000 - eleven digits in total.
Human beings have a total of ten digits on their hands. It (therefore) forms the basis of our counting system. It is the atomic number of neon. It is the number of hydrogen atoms in butane, a hydrocarbon.
Is the number even and do its digits total a multiple of 3?