This is a question of permutations; the answer is equal to the factorial of 5 (number of digits) divided by the factorial of 3 (number used in each selection), written 5! / 3!. This equals 120 / 6, or 20 ways.
There are 5 digits, and if none can be repeated then the first digit can be chosen in 5 ways, the second can be chosen in 4 ways and the third in 3. 5 x 4 x 3 = 60.
2
Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
9
9*9*9 = 729 ways.
There are 5 digits, and if none can be repeated then the first digit can be chosen in 5 ways, the second can be chosen in 4 ways and the third in 3. 5 x 4 x 3 = 60.
The smallest positive integer is 1357.
Add the last digit plus the sum of all the previous digits. The base case is that if your integer only has a single digit, just return the value of this digit. You can extract the last digit by taking the remainder of a division by 10 (number % 10), and the remaining digits by doing an integer division by 10.
Function sum_odd_digits(ByVal number As Integer) As Integer Dim digit As Integer sum_odd_digits = 0 While number <> 0 digit = number Mod 10 If digit And 1 Then sum_odd_digits = sum_odd_digits + digit number = number / 10 End While End Function
Fifty
9,999,876 is the greatest seven-digit number using four different digits.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
10p is not a digit. A digit is an integer in the range 0-9. Leaving that aside, you can make 31 sums from the 5 coins. If you allow 0 as a valid sum, the answer is 32.
It is -987654. The smallest POSITIVE number is 102345.
Since there are only five different digits, a 6-digit number can only be generated if a digit can be repeated. If digits can be repeated, the smallest 6-digit number is 111111.
#include <stdio.h> int digit_sum(int); int digit_sum_rec(int); main() { int num; printf("Enter 5 digit positive integer. : "); scanf("%d",&num); printf("The sum of the digits is: %d\n",digit_sum(num)); printf("The sum of the digits is (calculated recursively): %d\n", digit_sum_rec(num)); } int digit_sum(int n) { int t=0; while(n) { t+=n%10; n/=10; } return t; } int digit_sum_rec(int n) { if(n==0) return 0; return (n%10)+digit_sum_rec(n/10); }