What is the average of all 5-digit numbers that can be formed?

The number will be 57777.2. The average of those digits is 5.2, so we expect said average to be 52000+5200+520+52+5.2=57777.2 since each of them will occur in each position the same number of times in the set of all such permutations.

How many numbers can be formed with the digits 1 3 5 7 9 when taken all at a time find their sum?

= 24 such numbers.

What is the sum of first five digits?

the 1st 5 prime no. s are —– 2,3,5,7,11. now the sum is — 2+3+5+7+11=28………… hope it helps!!!!!!!

How many numbers can be formed from some or all digits 2 3 4 5 if number do not have repeated digits?

Total Number of Numbers which can be formed by numbers 1,2,3,4,5 (without repeating digitsi) = 5*4*3*2*! = 5! = 120.

What is the sum of the first 7 prime numbers?

The first seven primes are 2, 3, 5, 7, 11, 13, and 17. Don’t forget about 2, the smallest prime number, and also the only even prime! Adding these seven numbers gives a sum of 58, and 58/2 = 29.

How many five-digit numbers can be formed from 2 3 5 7?

Question 1021015: Find the sum of all five-digit that can be formed by using the digits 2,3,5,7 and 8,where each of these digits appears in each number. You can put this solution on YOUR website! There are 5! = 120 different five-digit numbers that can be formed from the numbers 2, 3, 5, 7, 8.

What is the sum of four digit numbers in the units?

Four digit numbers $ = 4 \\cdot 3 \\cdot 2 \\cdot 1 = 24$ ways we can form a four digit number. Since it’s a 4 digit number, each digit will appear $6= 24/4$ times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is $6(1+2+5+6)=84$.

What is the sum of all the digits in each position?

So sum of all digits in each position is same = 6* (a+b+c+d). let say the sum is a two digit number say ef. So now the last digit of 1st position is f. The sum of 10th position is ef again but with the carry from 1st position e.

What is the sum of one-fifth of 120?

One 5th of the numbers in each column are 2’s, one-fifth are 3’s, and so on to one-fifth are 8’s, So since one-fifth of 120 is 24, the sum of each column is 24* (2+3+5+7+8) = 24* (25) = 600 So there are 600 ten-thousands + 600 thousands + 600 hundreds + 600 tens + 600 ones or 6000000 + 600000 + 60000 + 6000 + 600 ======== 6666600 Edwin